Contents of Patterns in Pythagoras

1. Patterns in Right Triangles and Pythagorean Triples

1. Introduction

2. Finding Right Triangles with Integral Sides

3. Primitive Pythagorean Triples and Triangles

4. A Generating Formula for Primitive Pythagorean Triples

5. Properties of Primitive Pythagorean Triples

6. Areas and Perimeters of Complete Right Triangles

2. More Patterns in Right Triangles and Pythagorean Triples

1. Introduction

2. Possible Values for a

3. Possible Values for b

4. Possible Values for c

3. Even More Patterns in Right Triangles and Pythagorean Triple

1. The Relative Sizes of a, b, and c

3A 2. Regular and Inverted Triples

3. Proper Primitive Pythagorean Triples

3B 4. Unit and Constant Difference Triples

3C 5. Complete Primitive Pythagorean Triples

4. Still More Patterns in Right Triangles and Pythagorean Triples

1. Multiplication of Primitive Pythagorean Triples

2. Division of Pythagorean Triples

3. Powers of Primitive Pythagorean Triples

4. The Square Root of Primitive Pythagorean triples

5. Two New Pythagorean Triples (1,0,1) and (0,0,0)

6. Adding Primitive Pythagorean Triples

7. Subtracting Complete Primitive Pythagorean triples

8. Negative Pythagorean triples

9. The Distributive property for Pythagorean Triples

5. Further Patterns in Right Triangles andPythagorean Triples

1. Multiplication Of Pythagorean Triples by a Constant

2. Solving Pythagorean Triple Equations

3. Complete Pythagorean Triples and Vector Spaces

4. Some Notable Patterns (The 3-4-5 and 5-12-13 patterns)

5. Generating Pythagorean Triples from the Fibonacci Sequence

6. Constructing a Triple whose c-value Is c²

Patterns in Pythagoras

4. Still More Patterns in Right Triangles

and Pythagorean Triples

David W. Hansen

In Parts 1 , 2, 3A, 3B, and 3C, we discovered many beautiful patterns existing among the sides of primitive right triangles or the members of primitive Pythagorean triples. But apparently, we are still not through! Yes, there are STILL more patterns to be found in primitive right triangles and Pythagorean triples! But what’s left to discover? Well, c’mon, let’s see!

1. Multiplication of Primitive Pythagorean Triples

Is there a way to multiply two primitive Pythagorean triples together? If so, what should be the rule of multiplication? Can we discover a rule for multiplication that makes sense? Let’s give it a try.

First, we would like the product of two primitive Pythagorean triples to be itself a primitive Pythagorean triple, and it seems natural to want the a-value of the product of the triples to be the product of the two a-values of the triples. For example, if we were to multiply (3,4,5) by (5,12,13), we would want their product to be

(3x5, b, c) = (15, b₁, c₁),

where b₁ and c₁ would be those values which make this product triple a primitive Pythagorean triple. But what should b₁ and c₁ be?

Suppose we multiply two triples together by multiplying their a-values together, their b-values together, and their c-values together. Then, we would have, for example,

(3, 4, 5) x (5, 12, 13) = (3x5, 4x12, 5x13) = (15, 48, 65),

which is not even a Pythagorean triple, since 15² + 48² = 2529 ¹ 65² = 4225. So, this method of multiplication won’t work.

Now, what if we multiply a primitive Pythagorean triple (a, b, c) by itself? We get

(a, b, c) x (a, b, c) = (a², ?, ?).

Since (a, b, c) is a Pythagorean triple, then a² + b² = c², or a² = c² – b², which gives us (c² – b², ?, ?). Now what? Well, doesn’t the c² – b², since it is the a-value of our product, remind us of m² – n², which is the a-value of a primitive triple in terms of m and n? Then, couldn’t the b-value of our product possibly be 2bc, like the 2mn which is the b-value of a primitive triple? And couldn’t the c-value be b² + c², like the m² + n² which is the c-value of a primitive triple? Let’s try it and see!

Accordingly, we have (a, b, c) x (a, b, c) = (a², 2bc, b² + c²),

and if (a, b, c) = (3, 4, 5), we find that

(3, 4, 5) x (3, 4, 5) = (3x3, 2x4x5, 4² + 5²) = (9, 40, 41),

which IS a primitive Pythagorean triple, since 9, 40, and 41 have no common factors, and 9² + 40² = 41².

Let’s look at another example. (5, 12, 13) x (5, 12, 13) = ( 5x5, 2x12x13, 12² + 13²) = (25, 312, 313),

and (25, 312, 313) is a primitive Pythagorean triple. It looks we’ve done it! We have found a way to multiply two primitive triples together and get 1) another primitive triple, and 2) a triple whose a-value is the product of the a-values of our two triples.

However, our method of multiplication is good only for multiplying a triple by itself. How can we multiply two different primitive Pythagorean triples? Let’s take a closer look at the product of (a, b, c) with itself to get some ideas. Now,

(a, b, c) x (a, b, c) = ( a², 2bc, b² + c²),

which can be written as = ( aa, bc + cb, bb + cc ), which says

1) the a-value of the product is the product of the a-value of the first triple(say a₁)and the a-value of the second triple (say a₂). Thus, the a-value of the product is a₁a₂.

2) the b-value of the product is the product of the b value of the first triple (say b₁) and the c-value of the second triple (say c₂) plus the product of the c-value of the first triple (say c₁) and the b-value of the second triple

(say b₂). Thus, the b-value of the product is b₁c₂ + c₁b₂.

3) The c-value of the product is the product of the b value of the first triple (b₁) and the b-value of the second triple (b₂) plus the product of the c-value of the first triple (c₁) and the c-value of the second triple (c₂). Thus, the c-value of the product is b₁b₂ + c₁c₂.

So, using the patterns above for two primitive triples (a₁, b₁, c₁) and (a₂, b₂, c₂), we get

Multiplication of Pythagorean triples

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are two Pythagorean triples, then

(a₁, b₁, c₁) x (a₂, b₂, c₂) = ( a₁a₂, b₁c₂ + c₁b₂, b₁b₂ + c₁c₂ ). (1)

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Will this always work? Let’s see.

Example 1. (3, 4, 5) x (5, 12, 13) = (3x5, 4x13 + 5x12, 4x12 + 5x13) = (15, 112, 113), which is a primitive Pythagorean triple.

Example 2. (21, 20, 29) x (65, 72, 97) = (21x65, 20x97 + 29x72, 20x72 + 29x97) = (1365, 4028, 4253), which is also a primitive Pythagorean triple.

So, yes, it works, at least for these two examples! But, in general, can we expect that the product triple

(a₁a₂, b₁c₂ + c₁b₂, b₁b₂ + c₁c₂ ) from (1) above will always be a primitive Pythagorean triple? Let’s see.

First, we must show that this product triple is a Pythagorean triple; namely, that

(a₁a₂)² + (b₁c₂ + c₁b₂)² = (b₁b₂ + c₁c₂)². (2)

Now, since (a₁, b₁, c₁) and (a₂, b₂, c₂) are Pythagorean triples, we know that

a₁² + b₁² = c₁², or a₁² = c₁² –b₁², (3)

and a₂² + b₂² = c₂², or a₂² = c₂² – b₂². (4)

So, starting with the left-hand side of (2) above, we have

(a₁a₂)² + (b₁c₂ + c₁b₂)² = a₁²a₂² + b₁²c₂² + 2b₁c₂c₁b₂ + c₁²b₂².

Substituting from (3) and (4) for a₁² and a₂², we get

= (c₁² - b₁²)( c₂² – b₂²) + b₁²c₂² + 2b₁c₂c₁b₂ + c₁²b₂².

= c₁²c₂² – c₁²b₂² – b₁²c₂² + b₁²b₂² + b₁²c₂² + 2b₁c₂c₁b₂ + c₁²b₂²

= c₁²c₂² + b₁²b₂² + 2b₁c₂c₁b₂= b₁²b₂² + 2b₁c₂c₁b₂ + c₁²c₂²

= (b₁b₂ + c₁c₂)², which is the right-hand side of (2).

Thus, we have proved (2), and the product of two Pythagorean triples is always a Pythagorean triple.

But is this product of two Pythagorean triples a primitive Pythagorean triple? Sadly, the answer is not always. For example,

a) (5,12, 13) x (15, 8,17) = (5x15, 12x17 + 13x8, 12x8 + 13x17) = (75, 308, 317), which is a

primitive Pythagorean triple, but

b) (3, 4, 5) x (33, 56,65) = (3x33, 4x65 + 5x56, 4x56 + 5x85) = (99, 540, 549) = (9x11, 9x 60, 9x 61), which is NOT a primitive Pythagorean triple even though (3,4,5) and (33, 56, 65) are primitive triples. We see that 9 is a factor of each member of the triple, and we also see the primitive triple (11,60, 61) “embedded” in the product triple.

Let’s write down the products of a few primitive Pythagorean triples and try to find out when the product triples are primitive and when they are not. Using (1), we get

Table 1

Product Common Embedded

Factors Product primitive? factor triple

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1. (3, 4, 5) x (5, 12, 13) = (15, 112, 113) Yes None

2. (3, 4, 5) x (15, 8, 17) = (45, 108, 117) No 9 (5,12,13)

3. (3, 4, 5) x (33, 56, 65) = (99, 540, 549) No 9 (11,60,61)

4. (5,12,13) x (15, 8, 17) = (75, 308, 317) Yes None

5. (5,12,13) x (33, 56, 65) = (165, 1400, 1517) Yes None

6. (5,12,13) x (35,12,37) = (175, 600, 625) No 25 (7,24,25)

7. (7,24,25) x (63,16,65) = (441, 1960, 2009) No 49 (9,40,41)

8. (7,24,25) x (15, 8,17) = (105, 608, 617) Yes None

9. (15, 8,17) x (35,12,37) = (525, 500, 725) No 25 (21,20,29)

10. (15, 8,17) x (21,20,29) = (315, 572, 653) Yes None

11. (15, 8,17) x (21,220,221) = (315, 5508, 5517) No 9 (35,612,613)

12. (15,112,113 x (21,20,29) = (315, 5508, 5517) No 9 (35,612,613)

13. (15,112,113) x (21,220,221) = (315, 49612, 49613) Yes None

Looking at the table above, we can see that for those product triples which are not primitive, their common factors are all perfect squares. (See 2, 3, 6, 7, 9, 11, and 12.) And there is more. Let’s look more closely at three of these products.

In 2, above, we have a complete (c= b+1) triple, (3,4,5), and an incomplete (c ¹ b+1) triple, (15,8,17), with 3 as the common factor of their a-values and 9 = 3² the common factor of their product, (45,108,117). And in the incomplete triple, (15,8,17), c – b = 17 – 8 = 9, the common factor of the product, and its square root is the common factor of the a-values!

In 6, above, the complete triple (5,12,13) and the incomplete triple (35,12,37) have 5 as the common factor of their a-values and 25 = 5² as the common factor of their product, (175, 600, 625). And in the incomplete triple, (35,12,37), c – b = 37 – 12 = 25, the common factor of the product, and its square root is the common factor of the a-values!

In 7, above, the complete triple (7,24,25) and the incomplete triple (63,16,65), have 7 as the common factor of their a-values and 49 = 7² as the common factor of their product, (441,1960, 2009). And in the incomplete triple (63,16,65), c – b = 65 – 16 = 49, the common factor of the product, and its square root is the common factor of the a-values.

It seems that whenever we multiply a complete triple (a₁, b₁, c₁) by an incomplete triple (a₂, b₂, c₂) whose

a-values have a common factor f, equal to the square root of c₂ – b₂, then their product has a common factor of

f² = c₂ –b₂. Let’s see if this is so.

Accordingly, we let (a₁, b₁, c₁) be a complete triple and (a₂, b₂, c₂) be an incomplete triple whose a-values have a common factor of f, where f² = c₂ –b₂. We also let their product be (a, b, c). Then, there exist integers k₁ and k₂ such that a₁ = fk₁ and a₂ = fk₂. Since a₁ and a₂ are both odd, then f, k₁, and k₂ are also odd.

Then, (a₁, b₁, c₁) x (a₂, b₂, c₂) = (fk₁, b₁,c₁) x (fk₂, b₂, c₂)

= ( f²k₁k₂, b₁c₂ + c₁b₂, b₁b₂ + c₁c₂ ).

= (a, b, c).

where a = f²k₁k₂, which shows that f² is a factor of a.

Since (a₁, b₁, c₁) is a complete triple, c₁ = b₁ + 1. Then, a₁² +b₁² = c₁² = (b₁ + 1)²,

or a₁² + b₁² = b₁² + 2b₁ + 1,

and a₁² = 2b₁ + 1,

Solving for b₁, we have b₁ = (a₁² – 1)/2 (5a)

= (f²k₁² – 1)/2 (5b)

Since c₁ = b₁ + 1, we have c₁ = (a₁² – 1)/2 + 1

or c₁ = (a₁² + 1)/2 = (f²k₁² + 1)/2 (6)

Then, b = b₁c₂ + c₁b₂, and substituting from (5b) and (6),

we get = (f²k₁² – 1)/2 x c₂ + (f²k₁² + 1)/2 x b₂,

= (f²k₁²c₂ – c₂)/2 + (f²k₁²b₂ + b₂)/2,

= f²k₁²c₂/2 + f²k₁²b₂/2 – (c₂ – b₂)/2.

And, since c₂ – b₂ = f², we may replace the third term in the equation above by f²/2,

giving us b = f²k₁²c₂/2 + f²k₁²b₂/2 – f²/2 = f²( k₁²c₂/2 + k₁²b₂/2 – ½ ) = f²[ (k₁²c₂ – 1)/2 + k₁²b₂/2 ].

Since k₁ and c₂ are both odd and b₂ is even, then both terms in the brackets above are integers, which shows that f² is a factor of b.

Now, c = b₁b₂ + c₁c₂, and substituting from (5b) and (6),

we get = (f²k₁² – 1)/2 x b₂ + (f²k₁² + 1)/2 x c₂,

= (f²k₁²c₂ – b₂)/2 + (f²k₁²b₂ + c₂)/2,

= f² k₁²c₂/2 + f² k₁²b₂/2 + (c₂ –b₂)/2.

And, since c₂ – b₂ = f², we replace the third term in the equation above by f²/2, giving us

c = f²k₁²c₂/2 + f²k₁²b₂/2 + f²/2 = f²( k₁²c₂/2 + k₁²b₂/2 + ½ ) = f²[ (k₁²c₂ + 1)/2 + k₁²b₂/2 ].

Since k₁ and c₂ are both odd and b₂ is even, then both terms in the brackets above are integers, which shows that f² is a factor of c.

Thus, we have proved that f² is a common factor of (a, b, c); that is,

(a, b, c) = ( f²k₁k₂, f²(k₁²c₂/2 + k₁²b₂/2 – ½ ), f²(k₁²c₂/2 + k₁²b₂/2 + ½ ) ).

If we remove the factor f² from the a, b, and c-values above, we get the “embedded” triple

(a₃, b₃, c₃) = (k₁k₂, k₁²c₂/2 + k₁²b₂/2 – ½ , k₁²c₂/2 + k₁²b₂/2 + ½ ), for which

c₃ = k₁²c₂/2 + k₁²b₂/2 + ½ = k₁²c₂/2 + k₁²b₂/2 + (- ½ + 1)

= (k₁²c₂/2 + k₁²b₂/2 - ½ ) + 1 = b₃ + 1,

showing that the embedded triple (a₃, b₃, c₃) is a complete triple. Since c₃ = b₃ + 1, then b₃ and c₃ are consecutive integers, and thus have no common factors, making (a₃, b₃, c₃) a primitive triple. We have proved

Theorem 40

If (a₁, b₁, c₁) is a complete triple, and (a₂, b₂, c₂) is an incomplete triple, where a₁ and a₂ have

a common factor of f = Ö (c₂ – b₂), then (a₁, b₁, c₁) x (a₂, b₂, c₂) = (f²a₃, f²b₃, f²c₃),

which is not primitive, but the embedded triple (a₃, b₃, c₃) is both primitive and complete.

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Example 3. Without actually carrying out the multiplication, determine if the product of (9, 40, 41) and (21, 20, 29)

is a primitive triple. If not a primitive triple, find its common factor and embedded triple. Since (9, 40, 41) is complete

and (21, 20, 29) is not, we can use Theorem 40. Here, c₂ – b₂ = 29 – 20 = 9, and its square root, 3, is a factor f of

the a-values of both triples. Thus, the product of the two triples will have a common factor of f² = c₂ – b₂ = 9, and is thus not a primitive triple. To find the embedded triple, we must find the product of the two triples. Thus,

(9, 40, 41) x (21, 20, 29) = (189, 1980, 1989) = (9 x 21, 9 x 220, 9 x 221),

giving us the embedded triple (21, 220, 221), which is both primitive and complete.

Example 4. Factor (45, 108, 117) into a product of two triples. Since each term of (45, 108, 117) has a factor of 9,

we know by Theorem 40 that each of the a-values of its factors will have a common factor of 3. Now the factors of 45, the a-value of the given triple, are 45 and 1, 9 and 5, and 3 and 15, and the only factors which have 3 as a common factor are 3 and 15. Thus, we get

(3, b₁, c₁) x (15, b₂, c₂) = (45, 108, 117).

Since the only Pythagorean triple with an a-value of 3 is (3, 4, 5), we must have b₁ = 4, and c₁ = 5. Thus, we get (3, 4, 5) x (15, b₂, c₂) = (45, 108, 117). Multiplying the two factors out gives us

(45, 4c₂ + 5b₂, 4b₂ + 5c₂) = (45, 108, 117).

To find b₂ and c₂, we equate members of the two equal triples above, getting

4c₂ + 5b₂ = 108 5b₂ + 4c₂ = 108

4b₂ + 5c₂ = 117 4b₂ + 5c₂ = 117.

Solving the right-hand set of equations above by determinants, we get

108 4

117 5 540 – 468 72

b₂ = ------------------- = ------------------ = ------ = 8,

5 4 25 – 16 9

4 5

5 108

4 117 585 – 432 153

c₂ = ------------------ = ------------------- = --------- = 17.

5 4 25 – 16 9

4 5

So, b₂ = 8, and c₂ = 17, and we have factored (45,108,117) into (3,4,5) x (15, 8,17).

Now, Theorem 40 tells us what happens when we multiply a complete triple by an incomplete triple. What will happen if we multiply a complete triple by another complete triple? Let’s find out.

Let (a₁, b₁, c₁) and (a₂, b₂, c₂) be two complete triples. Then, c₁ = b₁ +1, c₁ – b₁ = 1, c₂ = b₂ + 1, and

b₂ = c₂ – 1. Will their product be a complete triple? Let’s see.

(a₁, b₁, c₁) x (a₂, b₂, c₂) = ( a₁a₂, b₁c₂ + c₁b₂, b₁b₂ + c₁c₂ ) = (a, b, c),

and to show that the product triple (a, b, c) is complete, we must show that c = b + 1.

Now, c = b₁b₂ + c₁c₂ , and by replacing b₂ by c₂ – 1 and c₂ by b₂ + 1 from above, we have

c = b₁(c₂ –1) + c₁(b₂+1) = b₁c₂ – b₁ + c₁b₂+ c₁ = b₁c₂ + c₁b₂+ (c₁ – b₁), and since c₁ - b₁ = 1,

= b₁c₂ + c₁b₂+ 1 = b + 1.

Thus, the product triple (a, b, c) = ( a₁a₂, b₁c₂ + c₁b₂, b₁b₂ + c₁c₂ ) is complete, and we have proved

Theorem 41

If A and B are two complete triples, then their product, A x B, is complete.

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2. Division of Pythagorean Triples

Let’s now see what we can discover about dividing two primitive Pythagorean triples. Since 6 ¸ 2 = 3 means

2 x 3 = 6, it seems only natural that

(a₁, b₁, c₁) ¸ (a₂, b₂, c₂) = (a₃, b₃, c₃)

should mean that (a₂, b₂, c₂) x (a₃, b₃, c₃) = (a₁, b₁, c₁). (7)

To find the values of the quotient triple, (a₃, b₃, c₃), we multiply out the left-hand side of (7) above, getting

( a₂a₃, b₂c₃ + c₂b₃, b₂b₃ + c₂c₃ ) = (a₁, b₁, c₁).

Equating the a, b, and c-values of the two triples above, we have

a₂a₃ = a₁ or a₃ = a₁ / a₂ (8)

b₂c₃ + c₂b₃ = b₁ (9)

b₂b₃ + c₂c₃ = c₁ (10)

Rearranging (9), and then solving (9) and (10) for b₃ and c₃ by determinants, we get

c₂b₃ + b₂c₃ = b₁

b₂b₃ + c₂c₃ = c₁

b₁ b₂

c₁ c₂ b₁c₂ – c₁b₂ b₁c₂ – c₁b₂ b₁c₂ – c₁b₂

b₃ = ------------------ = -------------------- = -------------------- = ---------------------- (11)

c₂ b₂ c₂c₂ – b₂b₂ c₂² – b₂² a₂²

b₂ c₂

c₂ b₁

b₂ c₁ c₂c₁ – b₂b₁ c₂c₁ – b₂b₁ c₂c₁ – b₂b₁

c₃ = ---------------- = --------------------- = -------------------- = ----------------------- (12)

c₂ b₂ c₂c₂ – b₂b₂ c₂² – b₂² a₂²

b₂ c₂

where c₂² – b₂² = a₂², since a₂² + b₂² = c₂². Substituting (8), (11), and (12) into (7a), we get

(a₁, b₁, c₁) ¸ (a₂, b₂, c₂) = (a₁ /a₂, (b₁c₂ – c₁b₂) / a₂², (c₁c₂ – b₁b₂) / a₂²). (13)

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Now, if we wish the quotient triple (a₃, b₃, c₃) to be a Pythagorean triple, then each of its members must be

a) positive and b) integers.

a) Since a₁ and a₂ are both positive, then a₃ = a₁/a₂ will be positive, and since c₁ > b₁ and c₂ > b₂, then multiplying these two inequalities together, we get c₁c₂> b₁b₂, or c₁c₂– b₁b₂ > 0, so

c₃ = (c₁c₂ – b₁b₂) / a₂² will be positive, but looking at b₃ = (b₁c₂ – c₁b₂) / a₂², we see that

b₁c₂ – c₁b₂ must be positive for b₃ to be positive. Thus, we must have

b₁c₂ – c₁b₂ > 0, b₁c₂ > c₁b₂, or b₁/b₂ > c₁/c₂. (14)

b) Since a₃ = a₁/a₂, then a₁/a₂ must be an integer, say k, where k will be an odd positive integer since both

a₁ and a₂ are odd positive integers. Then,

a₁/a₂ = k, or a₁ = ka₂, for k an odd integer. (15)

However, even though we require that a₁/a₂ be an integer, that is not enough to ensure b₃ and c₃ will be integers. For example,

(39, 80, 89) ¸ (3, 4, 5) = ( 39/3, (80x5 – 89x4)/3², (89x5 – 80x4)/3²)

= (13, 44/9, 125/9),

which shows that b₃ and c₃ can be fractions even though a₁/a₂ = 39/3 = 13 is an integer. However, notice that (39, 80, 89) is not a complete triple, whereas (3, 4, 5) is. Perhaps if both triples were complete, then b₃ and c₃ would always be integers. Let’s see.

Division of Complete Triples

Let (a₁, b₁, c₁) and (a₂, b₂, c₂) be two complete primitive Pythagorean triples with a₁ = ka₂. Then,

c₁ = b₁ + 1, c₂ = b₂ + 1, and from (5a), b₁ = (a₁² – 1)/2, and b₂ = (a₂² – 1)/2.

b₁c₂ – c₁b₂ b₁(b₂ + 1) – (b₁ + 1)b₂ b₁b₂ + b₁ – b₁b₂ – b₂

Then, b₃ = ------------------ = ------------------------------- = --------------------------------

a₂²a₂² a₂²

b₁ – b₂ (a₁² – 1)/2 – (a₂² – 1)/2 a₁² – a₂²

= ------------------- = ------------------------------------- = ---------------------

a₂²a₂² 2a₂²

k²a₂² – a₂² a₂²(k² – 1) k² – 1

= --------------------- = ----------------------- = -------------- (16)

2a₂²2a₂² 2

From (15) above, k is odd, so k² is odd, k² – 1 is even, and thus (k² – 1)/2 is an integer.

Thus, we have proved that b₃ is an integer.

c₁c₂ – b₁b₂ (b₁ + 1)(b₂ + 1) – b₁b₂ b₁b₂ + b₁ + b₂ + 1 – b₁b₂

Next, c₃ = -------------------- = ---------------------------------- = --------------------------------------

a₂²a₂² a₂²

b₁ + b₂ + 1 (a₁² – 1)/2 + (a₂² – 1)/2 + 1 a₁² + a₂²

= -------------------- = ------------------------------------------ = --------------------

a₂²a₂² 2a₂²

k²a₂² + a₂² a₂²(k² + 1) k² + 1

= -------------------- = ------------------- = ------------ (17)

2a₂²2a₂² 2

From (15) above, k is odd, so k² is odd, k² + 1 is even, and thus (k² + 1)/2 is an integer.

Thus, we have proved that c₃ is an integer.

In addition, from (17), we have c₃ = (k² + 1)/2 = (k² – 1 + 2)/2 = (k² – 1)/2 + 2/2

= (k² – 1)/2 + 1 = b₃ + 1, from (16).

Thus, c₃ = b₃ + 1, and the quotient triple (a₃, b₃, c₃) is complete! This gives us

Theorem 42

The quotient of two complete triples is complete.

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All of the above is summarized by the following.

Division of Complete Pythagorean triples (18)

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are two complete Pythagorean triples with a₁/a₂ an integer, and b₁/b₂ > c₁/c₂,

then (a₁, b₁, c₁) ¸ (a₂, b₂, c₂) = ( a₁/a₂, (b₁c₂ – c₁b₂) / a₂², (c₁c₂ – b₁b₂) / a₂²), a complete triple.

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Since for complete triples, c₁ = b₁ + 1 and c₂ = b₂ + 1, we can simplify (18) by writing

b₁/ b₂ > c₁/ c₂ as b₁/ b₂ > (b₁ + 1) / (b₂ + 1).

Then, since b₁ and b₂ are both positive, we may cross-multiply in the last inequality above, giving us

b₁(b₂ + 1) > b₂(b₁ + 1), b₁b₂ + b₁ > b₂b₁ + b₂, or b₁ > b₂.

Also, looking at the expressions for b₃ and c₃ in red above, we have

b₃ = (b₁ – b₂) / a₂², and c₃ = (b₁ + b₂ + 1) / a₂².

Thus, here is a simplified way to divide two complete triples.

Division of Complete Pythagorean triples (Simplified) (19)

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are two complete Pythagorean triples, with a₁/a₂ an integer, and b₁ > b₂,

then (a₁, b₁, c₁) ¸ (a₂, b₂, c₂) = (a₁/a₂, (b₁ – b₂) / a₂², (b₁ + b₂ + 1) / a₂²), a complete triple.

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Here are three examples of dividing complete triples.

1. (15, 112, 113) ¸ (5, 12, 13) = (15 / 5, (112 – 12) / 5², (112 + 12 + 1)/5²)

= (3, 100/25, 125/25) = (3, 4, 5).

(Note that b₁ = 112 is greater than b₂ = 12, and a₁ / a₂ is an integer.)

2. (9, 40, 41) ¸ (3, 4, 5) = (9 /3, (40 – 4) / 3², (40 + 4 + 1) / 3²)

= (3, 36/9, 45/9) = (3, 4, 5).

3. (35, 612, 613) ¸ (7, 24, 25) = (35/7, (612 – 24)/7², (612 + 24 + 1)/7²)

= (5, 588/49, 637/49) = (5, 12, 13)

So, everything goes quite smoothly as long as we use only complete triples in our divisions, and the quotient of their a-values is an integer with b₁ > b₂. But what if all these restrictive conditions are not met? Then what happens? Let’s see. Using (13) above, we have

4. An incomplete triple divided by a complete triple.

(105, 608, 617) ¸ (7, 24, 25) = (105/7, (608x25 – 617x24)/ 7², (617x25 – 608x24)/ 7² )

= (105/7, 392/49, 833 /49) = (15, 8, 17), which is an incomplete Pythagorean triple.

Note that b₁/b₂ = 608/24 = 25.33 > c₁/ c₂ = 617/25 = 24.68.

5. Two complete triples, but a₁/a₂not an integer.

(9, 40, 41) ¸ (5, 12, 13) = (9/5, (40x13 - 41x12)/5², (41x13 - 40x12)/5²) = (9/5, 28/25, 53/25), which is NOT

a Pythagorean triple at all because its members are fractions. In fact, if we write the triple with all of its members

having the same denominator, we get (45/25, 28/25, 53/25), which shows that each member of the triple has a

common factor of 1/25. Amazingly enough, if we remove this common factor, we are left with the triple

(45, 28, 53) which is a primitive Pythagorean triple!

6. An incomplete triple divided by a complete triple.

(33, 56, 65) ¸ (3, 4, 5) = (33/3, (56x5 – 65x4)/ 3², (65x5 – 56x4)/ 3² ) = (33/3, 20/9, 101 /9)

= (11, 20/9, 101/9), which is NOT even a Pythagorean triple because two of its members are fractions. If we

rewrite this triple so that all of its members have a common denominator, then we get (99/9, 20/9, 101/9), showing

that each member of the triple has a common factor of 1/9. Removing this factor, we get the embedded triple

(99, 20, 101), which is an inverted (b < a). incomplete Pythagorean triple with b₁/b₂ = 56/14 = 4 > c₁/c₂

= 65/5 = 13.

If we look closely at the quotient triple, (a₁ /a₂, (b₁c₂ – c₁b₂) / a₂², (c₁c₂ – b₁b₂) / a₂²), we notice that both the b- and c-values have denominators of a₂². If we multiply the numerator and denominator of the a-value by a₂, we get a quotient triple whose members each have the common factor of 1/a₂²; namely,

(a₁a₂ / a₂², (b₁c₂ – c₁b₂) / a₂², (c₁c₂ – b₁b₂) / a₂²). (20)

However, this triple is not a Pythagorean triple because of the common factor 1/a₂². If we remove this common factor, we get the triple (a₁a₂, b₁c₂ – c₁b₂, c₁c₂ – b₁b₂). Let’s see if this “embedded” triple is a Pythagorean triple.

First, we shall require as before that b₁/b₂ > c₁/c₂, thus making b₁c₂ – c₁b₂ positive. Next, we must show that

(a₁a₂)² + (b₁c₂ – c₁b₂)² = (c₁c₂ – b₁b₂)². (21)

Since (a₁, b₁, c₁) and (a₂, b₂, c₂) are Pythagorean triples, we know that

a₁² + b₁² = c₁², or a₁² = c₁² – b₁², (22)

and a₂² + b₂² = c₂², or a₂² = c₂² – b₂². (23)

Starting with the left-hand side of (21) above, we have

(a₁a₂)² + (b₁c₂ – c₁b₂)² = a₁²a₂² + (b₁c₂ – c₁b₂)²

Substituting for a₁² and a₂² from (22) and (23) and expanding, we have

= (c₁² - b₁²)(c₂² - b₂²) + b₁²c₂² – 2b₁c₂c₁b₂ + c₁²b₂²

= (c₁²c₂² – c₁²b₂² – b₁²c₂² + b₁²b₂²) + b₁²c₂² – 2b₁c₂c₁b₂ + c₁²b₂²

= c₁²c₂² – 2c₁c₂b₁b₂+ b₁²b₂²

= (c₁c₂ – b₁b₂)², which is the right-hand side of (21) above.

Thus, the embedded triple is a Pythagorean triple. This gives us

Division of Pythagorean triples

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are two Pythagorean triples, with b₁/ b₂ > c₁/ c₂, then

(a₁, b₁, c₁) ¸ (a₂, b₂, c₂) = (a₁ /a₂, (b₁c₂ – c₁b₂) / a₂², (c₁c₂ – b₁b₂) / a₂²).

= (a₁a₂ / a₂², (b₁c₂ – c₁b₂) / a₂², (c₁c₂ – b₁b₂) / a₂²),

where the embedded triple, (a₁a₂, b₁c₂ – c₁b₂, c₁c₂ – b₁b₂), is a Pythagorean triple.

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Example 7. If (21, 20, 29) is divided by (15, 8, 17), the quotient will not be a Pythagorean triple, since a₁/a₂ = 21/15 = 7/5, which is a fraction. From the division algorithm above, we find the embedded triple to be

(a₁a₂, b₁c₂ – c₁b₂, c₁c₂ – b₁b₂) = (21x15, 20x17 – 29x8, 29x17 – 20x8) = (315, 108, 333),

which is a Pythagorean triple although not primitive, since each of its members has a common factor of 9. Removing this factor, we get the primitive triple (35, 12, 37).

Products and Quotients of Complete and Incomplete Triples

In Theorem 41 above, we proved that the product of two complete triples is a complete triple, and in Theorem 42, we showed that the quotient of two complete triples is also a complete triple. Now, what about the product of two triples, one complete and one incomplete? Will their product be complete or incomplete?

To find out, let’s take a complete triple P and an incomplete triple Q, and form their product R. Then, PQ = R, and R is either complete or incomplete. Let’s assume that R is complete. Since PQ = R, we have Q = P / R, and

Q is the quotient of two complete triples P and R, so by Theorem 42, Q is complete. But this, of course, is not possible, since we know that Q is incomplete. Thus, our assumption that R is complete is false, and R must therefore be incomplete. This gives us

Theorem 43

If (a₁, b₁, c₁) is a complete triple and (a₂, b₂, c₂) is an incomplete triple,

then their product is incomplete.

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Finally, let’s consider the product of two incomplete triples. Will it be complete or incomplete?

Let (a₁, b₁, c₁) and (a₂, b₂, c₂) be two incomplete triples. Then, c₁ ¹ b₁ + 1 and c₂¹ b₂ + 1. Their product

is P = ( a₁a₂, b₁c₂ + c₁b₂, b₁b₂ + c₁c₂ ), which is either complete or incomplete. Let’s assume P is complete. Then its c-value is one more than its b-value, or

b₁b₂ + c₁c₂ = b₁c₂ + c₁b₂ + 1,

b₁b₂ + c₁c₂ – b₁c₂ – c₁b₂ = 1,

c₂(c₁ – b₁) – b₂(c₁ – b₁) = 1,

and (c₂ – b₂)(c₁ – b₁) = 1.

Now in this last equation, b₁, b₂, c₁, and c₂ are all positive integers, with c₂ > b₂, and c₁ > b₁. Thus, c₂ – b₂ and c₁ – b₁ are both positive integers whose product is 1, and the only way this can happen is if

c₂ – b₂ = 1, and c₁ – b₁ = 1.

Then, c₂ = b₂ + 1, and c₁ = b₁ + 1, and thus (a₁, b₁, c₁) and (a₂, b₂, c₂) are both complete triples. But this is, of course, impossible since both triples are incomplete. So our assumption that their product P is complete is false, and P must therefore be incomplete, giving us

Theorem 44

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are both incomplete triples,

then their product is incomplete.

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Theorems 41, 42, 43, and 44 can be summarized as follows:

1) The product (or quotient) of two complete triples is complete.

2) The product of two incomplete triples is incomplete.

3) The product of a complete triple and an incomplete triple is incomplete.

or, The product of two triples is incomplete unless both triples are complete.

Interestingly enough, the quotient of two incomplete triples may be either complete or incomplete! For example, the division of the incomplete triple (75, 308, 317) by the incomplete triple (15, 8, 17) is (5, 12, 13), a complete triple, whereas the quotient of the incomplete triple (315, 572, 653) by (15, 8, 17) is (21, 20, 29), an incomplete triple. How very interesting!

3. Powers of Primitive Pythagorean Triples

Let’s turn now to finding various powers of primitive Pythagorean triples. We have

The square (a, b, c)² = (a, b, c) x (a, b, c)

= (aa, bc + cb, bb + cc)

= (a², 2bc, b² + c²)

Example: (15, 8, 17)² = (15², 2(8)(17), 8² + 17²) = (225, 272, 353)

The cube (a, b, c)³ = (a, b, c)² x (a, b, c)

= (a², 2bc, b² + c²) x (a, b, c)

= (a²a, 2bcc + (b² + c²)b, 2bcb + (b² + c²)c)

= (a³, 2bc² + b³ + bc², 2b²c + b²c + c³)

= (a³, b³ + 3bc², c³ + 3b²c)

Example: (3, 4, 5)³ = ( 3³, 4³ + 3(4)5², 5³ + 3(4²)5 ) = (27, 364, 365)

4^th Power (a, b, c)⁴ = (a, b, c)³ x (a, b, c)

= (a³, b³ + 3bc², c³ + 3b²c) x (a, b, c)

= ( a³a, (b³ + 3bc²)c + (c³ + 3b²c)b, (b³ + 3bc²)b + (c³ + 3b²c)c )

= (a⁴, b³c + 3bc³ + c³b + 3b³c, b⁴ + 3b²c² + c⁴ + 3b²c² )

= (a⁴, 4b³c + 4bc³, b⁴ + c⁴ + 6b²c²)

Example: (3, 4, 5)⁴ = ( 3⁴, 4(4³)5 + 4(4)5³, 4⁴ + 5⁴ + 6(4²)5² ) = (81, 3280, 3281)

5^th Power (a, b, c)⁵ = (a, b, c)⁴ x (a, b, c)

= (a⁴, 4b³c + 4bc³, b⁴ + c⁴ + 6b²c²) x (a, b, c)

= ( a⁴a, (4b³c + 4bc³)c + (b⁴ + c⁴ + 6b²c²)b, (4b³c + 4bc³)b + (b⁴ + c⁴ + 6b²c²)c )

= ( a⁵, 4b³c² + 4bc⁴ + b⁵ + bc⁴ + 6b³c², 4b⁴c + 4b²c³ + b⁴c + c⁵ + 6b²c³ )

= (a⁵, b⁵ + 10b³c² + 5bc⁴, c⁵ + 5b⁴c + 10b²c³ )

Example: (3, 4, 5)⁵ = ( 3⁵, 4⁵ + 10(4³)5² + 5(4)5⁴, 5⁵ + 5(4⁴)5 + 10(4²)5³ )

= ( 243, 1024 + 16000 + 12500, 3125 + 6400 + 20000 ) = (243, 29524, 29525)

Note that to find a power of (a, b, c), we have to calculate all of its preceding powers. As we have just seen, to

find the value of (a, b, c)⁵, we had to calculate its 4 preceding powers, which resulted in many complicated expressions. But are there any patterns in these expressions? Let’s try to find them!

(a, b, c)¹ = (a, b, c)

(a, b, c)² = (a², 2bc, b² + c²)

(a, b, c)³ = (a³, b³ + 3bc², c³ + 3b²c)

(a, b, c)⁴ = (a⁴, 4b³c + 4bc³, b⁴ + c⁴ + 6b²c²)

(a, b, c)⁵ = (a⁵, b⁵ + 10b³c² + 5bc⁴, c⁵ + 5b⁴c + 10b²c³ )

First of all, we see that the a-values are simply powers of a raised to the power of the triple. Thus, the a-value of the 4^th triple is a⁴, and so the a-value of the n^th triple will be aⁿ. Also, if we look carefully, we can see that c raised to the power of the triple appears in the c-value of each triple. Thus, c² is a part of the c-value of the 2^nd triple, c³ is a part of the c-value of the 3^rd triple, c⁴ is a part of the c-value of the 4^th, etc. Let’s rearrange the triples above so that the powers of c are written as the last term in each c-value.

(a, b, c)¹ = (a, b, c)

(a, b, c)² = (a², 2bc, b² + c²)

(a, b, c)³ = (a³, b³ + 3bc², 3b²c + c³)

(a, b, c)⁴ = (a⁴, 4b³c + 4bc³, b⁴ + 6b²c² + c⁴)

(a, b, c)⁵ = (a⁵, b⁵ + 10b³c² + 5bc⁴, 5b⁴c + 10b²c³ + c⁵)

Yes, that looks a bit better, but it still seems very complicated! Are there other patterns?

Well, b² appears as a part of a term in the c-values of all the triples after the first. And

b is in the b-value of the 1^st triple,

b² is in the c-value of the 2^nd triple,

b³ is in the b-value of the 3^rd triple,

b⁴ is in the c-value of the 4^th triple,

b⁵ is in the b-value of the 5^th triple.

It appears that these powers of b alternate between the b-values and the c-values of the triples. Now this pattern is somewhat interesting, but not sufficient for us to discover a pattern that would allow us to write down any power of (a, b, c) directly without first having to calculate all of its preceding powers as done in the lists above.

Now, we have been very fortunate in discovering many, many patterns in the primitive right triangles and Pythagorean triples. Surely, we can find a key pattern here. Let’s continue looking!

(a, b, c)¹ = (a, b, c)

(a, b, c)² = (a², 2bc, b² + c²)

(a, b, c)³ = (a³, b³ + 3bc², 3b²c + c³)

(a, b, c)⁴ = (a⁴, 4b³c + 4bc³, b⁴ + 6b²c² + c⁴)

(a, b, c)⁵ = (a⁵, b⁵ + 10b³c² + 5bc⁴, 5b⁴c + 10b²c³ + c⁵)

Well, there is an interesting pattern in the 2^nd triple above. Its b and c values, 2bc and b² + c², when added together, give us b² + 2bc + c² = (b + c)², a perfect square. Now, this is very nice! Is this pattern in the other power triples? Let’s see!

In the 3^rd triple, its b and c values of b³ + 3bc² and 3b²c + c³, when added together, give us

b³ + 3bc² + 3b²c + c³, or b³ + 3b²c + 3b²c + c³ = (b + c)³.

This is simply amazing! Let’s try the 4^th triple!

In the 4^th triple, its b and c values of 4b³c + 4bc³ and b⁴ + 6b²c² + c⁴, when added together, give us

4b³c + 4bc³ + b⁴ + 6b²c² + c⁴, or b⁴ + 4b³c + 6b²c² + 4bc³ + c⁴,

and using the binomial theorem, we find this is nothing more nor less than (b + c)⁴.

Hooray! We’ve discovered the key to this pattern. It appears that for any triple (a, b, c) raised to a power n, the sum of its b and c values is equal to (b + c)ⁿ. If so, then the sum of the b- and c-values of the 5^th triple above will be (b + c)⁵. Let’s see if it is!

In the 5^th triple, the b- and c-values of b⁵ + 10b³c² + 5bc⁴ and 5b⁴c + 10b²c³ + c⁵, when added together, give us

b⁵ + 10b³c² + 5bc⁴ + 5b⁴c + 10b²c³ + c⁵,

which, when rearranged, is b⁵ + 5b⁴c + 10b³c² + 10b²c³ + 5bc⁴ + c⁵, and this expression,

by the binomial theorem, does equal (b + c)⁵.

Great! Our prediction is correct, and we have discovered a most beautiful pattern for finding the powers of primitive Pythagorean triples and, to our surprise, it is linked to the binomial theorem!

Once again, let’s list the powers of (a, b, c), and this time, we shall also list the appropriate powers of b + c below them. Then, let’s place the letters B and C beneath each term to indicate whether that term belongs either to the b-value (B) or the c-value (C) of the power, as appropriate. The results of this are shown below.

(Note that, for example, in the 3^rd triple, the triple for (a, b, c)³ is listed first, and then beneath it is listed the binomial expansion of (b + c)³. The letter B is placed underneath the b³ because b³ is a part of the b-value

b³ + 3bc² of (a, b, c)³. The letter C is placed underneath the 3b²c to indicate that it is a part of the c-value

3b²c + c³ of (a, b, c)³. In like manner, 3bc² has a B beneath it because it is part of the b-value of (a, b, c)³, and c³ has a C beneath it because it is a part of the c-value of (a, b, c)³.)

1. (a, b, c)¹ = (a, b, c)

(b + c)¹ = b + c

B C

2. (a, b, c)² = (a², 2bc, b² + c²)

(b + c)² = b² + 2bc + c²

C B C

3. (a, b, c)³ = (a³, b³ + 3bc², 3b²c + c³)

(b + c)³ = b³ + 3b²c + 3bc² + c³

B C B C

4. (a, b, c)⁴ = (a⁴, 4b³c + 4bc³, b⁴ + 6b²c² + c⁴)

(b + c)⁴ = b⁴ + 4b³c + 6b²c² + 4bc³ + c⁴

C B C B C

5. (a, b, c)⁵ = (a⁵, b⁵ + 10b³c² + 5bc⁴, 5b⁴c + 10b²c³ + c⁵)

(b + c)⁵ = b⁵ + 5b⁴c + 10b³c² + 10b²c³ + 5bc⁴ + c⁵

B C B C B C

Now notice what we have found in this list. Amazingly, the letters alternate between B and C, or C and B, for each of the powers of (a, b, c)! This means that the terms in the binomial expansion associated with a power of a triple alternate between the b and c values of the power of the triple. And if the power of the triple is odd, as in (a, b, c³, then the alternation starts with a B, and if the power of the triple is even, as in (a, b, c)⁴, the alternation starts with a C. How very interesting! But that is not all.

Take a closer look at the terms in the list with B’s and C’s underneath them. Notice that each term with a B underneath it contains an odd power of b, whereas each term with a C underneath it contains an even power of b. This is really nice. Let’s look at an example.

Example 1. Find (a, b, c)⁶ directly, and then check your work by calculating it from (a, b, c)⁵. We start

with (a, b, c)⁶ = (a⁶, ________ , ________ ), and to find out what goes in the blanks for the b and c values,

we will expand (b + c)⁶ by the binomial theorem and then place alternating B’s and C’s underneath each resulting term. Since the power of the triple is even (6), we will start the alternation with a C. We get

(b + c) ⁶ = b⁶ + 6b⁵c + 15b⁴c² + 20b³c³ + 15b²c⁴ + 6bc⁵ + c⁵.

C B C B C B C

Now the sum of the terms with the B’s under them is 6b⁵c + 20b³c³ + 6bc⁵, and this will be the b-value of

(a, b, c)⁶. Note that all the powers of b are odd.

The sum of the terms with the C’s under them is b⁶ + 15b⁴c² + 15b²c⁴ + c⁵, and this will be the c-value for

(a, b, c)⁶. Note that all the powers of b are even. (We say c⁵ has an even power of b since c⁵ = b⁰c⁵, and 0 is even.)

Thus, (a, b, c)⁶ = (a⁶, 6b⁵c + 20b³c³ + 6bc⁵, b⁶ + 15b⁴c² + 15b²c⁴ + c⁵). (1)

To be sure that (1) is correct, we will find (a, b, c)⁶ by multiplying (a, b, c)⁵ by (a, b, c). Thus,

(a, b, c)⁶ = (a, b, c)⁵ x (a, b, c)

= (a⁵, b⁵ + 10b³c² + 5bc⁴, 5b⁴c + 10b²c³ + c⁵) x (a, b, c)

= ( a⁶, (b⁵+10b³c²+5bc⁴)c + (5b⁴c+10b²c³+c⁵)b, (b⁵+10b³c²+5bc⁴)b + (5b⁴c+10b²c³+ c⁵)c )

= ( a⁶, b⁵c + 10b³c³+ 5bc⁵+ 5b⁵c + 10b³c³+ bc⁵, b⁶+ 10b⁴c²+ 5b²c⁴ + 5b⁴c²+ 10b²c⁴+ c⁶)

= ( a⁶, 6b⁵c + 20b³c³+ 6bc⁵, b⁶+ 15b⁴c²+ 15b²c⁴ + c⁶), which is identical to (1) above.

Thus, our procedure is correct.

Example 2. Find (a, b, c)³ directly. We first use the binomial theorem to expand (b + c)³. Thus,

(b + c)³ = b³ + 3b²c + 3bc² + c³.

Then, the sum of those terms with odd powers of b; namely, b³ and 3bc², will be the b-value of our power, and the sum of those terms with even powers of b; namely, 3b²c and c³, will be the c-value. Thus,

(a, b, c)³ = (a³, b³ + 3bc², 3b²c + c³),

as is confirmed by the 3^rd triple in our list above.

_______________________________________________________________________________

In general, the nth power of the triple (a, b, c) is a Pythagorean triple whose

a-value is aⁿ,

b-value is the sum of those terms with odd powers of b in the expansion of (b + c)ⁿ,

and c-value is the sum of those terms with even powers of b in the expansion of (b + c)ⁿ.

_______________________________________________________________________________

To prove these statements, we would use mathematical induction.

Example 3. An easy way to find, say (a, b, c)⁵, is to think of it as [ a⁵ , (b + c)⁵ ]. Then, expanding

(b + c)⁵ by the Binomial Theorem, we get [ a⁵ , b⁵ + 5b⁴c + 10b³c² + 10b²c³ + 5bc⁴ + c⁵ ].

Now, all we have to do is separate the terms of this expansion, putting those with odd powers of b on the left and those terms with even powers of b on the right, giving us

[ a⁵ , b⁵ + 10b³c² + 5bc⁴ 5b⁴c + 10b²c³ + c⁵ ].

Then, we simply change the brackets to parentheses, put a comma between the separated terms, and we get

( a⁵ , b⁵ + 10b³c² + 5bc⁴ , 5b⁴c + 10b²c³ + c⁵ ). What fun!

Now, let’s take one more look at our list of powers. What can we see?

(a, b, c)¹ = (a, b, c)

(a, b, c)² = (a², 2bc, b² + c²)

(a, b, c)³ = (a³, b³ + 3bc², 3b²c + c³)

(a, b, c)⁴ = (a⁴, 4b³c + 4bc³, b⁴ + 6b²c² + c⁴)

(a, b, c)⁵ = (a⁵, b⁵ + 10b³c² + 5bc⁴, 5b⁴c + 10b²c³ + c⁵)

Well first, for any triple raised to an odd power, we see that the number of terms in its b-value (n_b) equals the number of terms in its c-value (n_c), whereas for one which is raised to an even power, the number of terms in its

c-value is one more than the number of terms in its b-value. For example, in the expansion of (a, b, c)⁴, there are

2 terms in its b-value and 3 terms in its c-value.

Second, can we predict how many terms will be in the b- and c-values of any triple raised to the n^th power?

Well, if we let n equal the power of a triple (a, b, c), n_b equal the number of terms in its b-value, and n_c the number of terms in its c-value, we get the following tables for the odd and even powers of the triple.

Odd powers of n Even powers of n

n + 1 n n_b n_c 2n_c n n_b n_c 2n_b

---------------------------------------- ---------------------------------

2 1 1 1 2 2 1 2 2

4 3 2 2 4 4 2 3 4

6 5 3 3 6 6 3 4 6

n_b = n_c n_b+ 1 = n_c

n + 1 = 2n_c n = 2n_b

Comparing the various columns of these tables, we see that for

Odd powers of n Even powers of n

n_c = n_b n_c = n_b + 1,

2n_c = n + 1 2n_b = n,

or n_c = (n+1)/2 n_b = n/2

Putting all of this together, we get

Theorem 45

The n^th power, (a, b, c)ⁿ, of the Pythagorean triple (a, b, c) is (aⁿ, B, C), where

B = the sum of all the terms with odd powers of b in (b + c)ⁿ, and

C = the sum of all the terms with even powers of b in (b + c)ⁿ.

a) If n is odd, then B and C each have (n+1)/2 terms. Thus, B and C have the same number of terms.

b) If n is even, then B has n/2 terms, and C has n/2 + 1 terms. Thus, C has one more term than B.

____________________________________________________________________________________

Example 4. a) Find the number of terms in the a, b, and c values of (a, b, c)¹⁰ without actually calculating its power, and b) calculate the power to verify your answer.

a) There will be only one term in the a-value of (a, b, c)¹⁰; namely, a¹⁰. Since n = 10, an even number, Theorem 45 tells us that its b-value will have 10/2 = 5 terms, and its c-value one more; namely, 6 terms.

b) (a, b, c)¹⁰ → [ a¹⁰, (b + c)¹⁰ ]. First, we use the binomial theorem to calculate (b + c)¹⁰

= b¹⁰ + 10b⁹c + 45b⁸c² + 120b⁷c³ + 210b⁶c⁴ + 252b⁵c⁵ + 210b⁴c⁶ + 120b³c⁷ + 45b²c⁸ + 10bc⁹ + c¹⁰.

Since the sum of the terms with the odd powers of b is the b-value, the sum of the rest of the terms will be the

c-value, and we have the following for (a, b, c)¹⁰.

1) Its a-value = a¹⁰ (one term)

2) Its b-value = 10b⁹c + 120b⁷c³ + 252b⁵c⁵ + 120b³c⁷ + 10bc⁹ (5 terms)

3) Its c-value = b¹⁰ + 45b⁸c² + 210b⁶c⁴ + 210b⁴c⁶ + 45b²c⁸ + c¹⁰ (6 terms).

These values verify our answers in part a), and we have (a, b, c)¹⁰ =

(a¹⁰, 10b⁹c + 120b⁷c³ + 252b⁵c⁵ + 120b³c⁷ + 10bc⁹, b¹⁰ + 45b⁸c² + 210b⁶c⁴ + 210b⁴c⁶ + 45b²c⁸ + c¹⁰ ).

4. The Square Root of Primitive Pythagorean Triples

Let’s now see if we can find square roots of primitive Pythagorean triples, (a, b, c).

_______

If Ö (a, b, c) = (a₁, b₁, c₁), then (a, b, c) = (a₁, b₁, c₁)² = (a₁, b₁, c₁) x (a₁, b₁, c₁),

or (a, b, c) = (a₁a₁, b₁c₁+c₁b₁, b₁b₁ + c₁c₁) = (a₁², 2b₁c₁, b₁² + c₁²).

Equating the members of each triple in the equation above, we get

a = a₁², or a₁ = Ö a (1)

b = 2b₁c₁, (2)

c = b₁² + c₁² (3)

Adding (3) and (2), we get c + b = b₁² + c₁² + 2b₁c₁ = (c₁ + b₁)². (4a)

Subtracting (2) from (3), we get c – b = b₁² + c₁² – 2b₁c₁ = (c₁ – b₁)². (4b)

_____

From (4a), we have c₁ + b₁ = Ö c + b , (5a)

_____

and from (4b), c₁ – b₁ = Ö c – b . (5b)

_____ _____

Adding (5a) and (5b), we get 2c₁ = Ö c + b + Ö c – b,

_____ _____

or c₁ = ( Ö c + b + Ö c – b ) / 2. (6)

_____ _____

Subtracting (5b) from (5a), we get 2b₁ = Ö c + b – Ö c – b,

____ _____

or b₁ = ( Ö c + b – Ö c – b ) / 2. (7)

Then, substituting the values of a₁, b₁, and c₁ from (1), (7), and (6), into

_______

Ö (a, b, c) = (a₁, b₁, c₁), we have

Theorem 46

The square root of a Pythagorean triple (a, b, c) is given by

_______ _ ____ _____ ____ _____

Ö (a, b, c) = ( Öa, (Ö c + b – Ö c – b ) / 2, (Ö c + b + Ö c – b ) / 2 ).

___________________________________________________________

_____________ ___________________

Example 5. Find a) Ö (225, 272, 353) and b) Ö (225, 25312, 25313). (These two triples have the same

a-values, but different b and c-values. Will their square roots be the same or different?)

Using Theorem 46, we have

_____________ ___ _____ ___ _________ _________ ________

a) Ö (225, 272, 353) = (Ö225, (Ö 353 + 272 – Ö 353 – 272 ) / 2, (Ö 353 + 272 + Ö 353 – 272 ) / 2 )

___ ____ ___ ____ __ ___ ____ __ ____ __

= (Ö225, (Ö 625 – Ö 81 ) / 2, (Ö 625 + Ö81 ) / 2 ) = (Ö225, (Ö 625 – Ö 81 ) / 2, (Ö 625 + Ö81 ) / 2 )

= (15, (25 – 9) / 2, (25+9) / 2 ) = (15, 8, 17).

__________________

b) To find Ö (225, 25312, 25313), let’s first find c + b = 25313 + 25312 = 50625, and c – b =

25313 – 25312 = 1. Then,

_________________ ____ ______ __ ______ _

Ö (225, 25312, 25313) = (Ö 225, (Ö 50625 – Ö 1 ) / 2, (Ö 50625 + Ö1 ) / 2 )

= (15, (225 – 1) / 2, (225+1) / 2 ) = (15, 112, 113), which is a different answer from the one in a) above.

Since the two triples were different (although their a-values were the same), we got different square roots, which is nice!

5. Two New Pythagorean Triples (1,0,1) and (0,0,0)

We know that a number (other than zero) divided by itself is always 1. What do you think will happen if we divide a primitive Pythagorean triple by itself? Well, let’s see. If (a, b, c) is a primitive Pythagorean triple, then

(a, b, c) ¸ (a, b, c) = (a/a, (bc – cb) / (c² – b²), (c² – b²) / (c² – b²) )

= (1, 0, 1).

This is very surprising! We get a new triple, one which we have never seen before! Is it a Pythagorean triple? Is it primitive?

Well, it certainly is primitive, since its members have no common factors (other than 1), and it satisfies the Pythagorean Theorem, a² + b² = c², since 1² + 0² = 1², but it is NOT a Pythagorean triple, since not all of its members are positive integers (b = 0). However, let’s expand our definition of a Pythagorean triple to allow its members to be either positive integers or zero. Then, we can say that

(1, 0, 1) is a primitive Pythagorean triple.

Now in (1, 0, 1), c = 1 and b = 0, so c = b +1, and (1, 0, 1) is a complete triple. Also, the difference between

a and b is 1 – 0 = 1, so (1, 0, 1) is a unit triple. And since a = 1 is greater than b = 0, it is also an inverted triple. However, it is not a proper triple (this is not surprising!), since its a-value of 1 is not divisible by 3, and its c-value of 1 is not divisible by 5, although its b-value of 0 is divisible by 4. Thus,

(1, 0, 1) is a complete, inverted, unit triple.

Now, let’s multiply (1, 0, 1) by itself and see what happens. Accordingly, we have

(1, 0, 1) x (1, 0, 1) = (1 x 1 , 0 x 1 + 1 x 0 , 0 x 0 + 1 x 1) = (1, 0, 1),

and nothing is changed! In multiplying (1, 0, 1) by itself, we find that (1, 0, 1) is left unchanged; that is, its identity is preserved. Let’s look at the properties of (1, 0, 1) more closely.

What if we multiply any triple (a, b, c) by (1, 0, 1)? Will (a, b, c) be changed? Well,

(a, b, c) x (1, 0, 1) = (a x 1 , b x 1 + c x 0 , b x 0 + c x 1) = (a, b, c),

and the triple (a, b, c) is unchanged! Its identity is preserved.

What if we divide any triple (a, b, c) by (1, 0, 1)? Will (a, b, c) be changed? Well,

(a, b, c) ¸ (1, 0, 1) = (a/1 , (b x 1 – c x 0) / (1² – 0²) , (c x 1 – b x 0) / (1² – 0²) ) = (a, b, c),

and the triple (a, b, c) is unchanged!

Apparently, (1, 0, 1) acts on Pythagorean triples just like the number 1 acts on real numbers under multiplication; namely, it leaves them unchanged. Because of this, we shall say that

1 = (1, 0, 1) is the unitary triple for Pythagorean triples.

Now 1 raised to any integral power n is 1, so we would expect that (1, 0, 1)ⁿ would equal (1, 0, 1). Let’s see if it does.

From Theorem 45, we know that (a, b, c)ⁿ is that triple whose a-value is aⁿ, and whose b-value is the sum of the odd powers of b in (b + c)ⁿ, and whose c-value is the sum of the even powers of b in (b + c)ⁿ.

For (1, 0, 1)ⁿ, the a-value will be 1ⁿ = 1.

To find the b- and c-values, we must find (b + c)ⁿ = (0 + 1)ⁿ, which is 1ⁿ = 1 = b⁰. Then the b-value

will be 0 since there are no odd powers of b in (0 + 1)ⁿ = 1, and the c-value will be 1, since the sum of the even powers of b in (0 + 1)ⁿ is b⁰ = 1. Thus, (1, 0, 1)ⁿ = (1, 0, 1).

_______

Example 1. Find Ö (1, 0, 1) . By Theorem 46,

_______ __ _____ ______ _____ __

Ö (1. 0, 1) = ( Ö 1, (Ö 1 + 0 – Ö 1 – 0 ) / 2, (Ö 1 + 0 + Ö 1 – 0 ) / 2 ) = (1, 0, 1).

This example further confirms that the unitary triple (1, 0, 1) behaves just like the integer 1 with respect to multiplication, division, powers and square roots.

Multiplication by the triple (k, 0, k)

Now if multiplication by (1, 0, 1) leaves (a, b, c) unchanged, what would the multiplication of (a, b, c) by

(2, 0, 2), or (3, 0, 3), or (k, 0, k), k an integer, give us? Let’s see.

Now, (2, 0, 2) x (a, b, c) = (2a, 0 x c + 2b, 0 x b + 2c) = (2a, 2b, 2c),

and we have doubled each member of (a, b, c).

(3, 0, 3) x (a, b, c) = (3a, 0 x c + 3b, 0 x b + 3c) = (3a, 3b, 3c),

and we have tripled each member of (a, b, c).

Finally, (k, 0, k) x (a, b, c) = (ka, 0 x c + kb, 0 x b + kc) = (ka, kb, kc),

and we have multiplied each member of (a, b, c) by the positive integer k.

Apparently, (k, 0, k) acts on Pythagorean triples just like the number k acts on real numbers under multiplication; namely, it multiplies them by k. Because of this, we shall say that

K = (k, 0, k) is the k-multiplier triple for Pythagorean triples;

that is, for A = (a, b, c), KA = (k, 0, k) x (a, b, c) = ( ka, kb, kc ).

Thus, multiplying any primitive Pythagorean triple by (k, 0, k) simply introduces k into the triple as a common factor of its members. Unfortunately, the resulting triple will not be primitive (unless k = 1), but then, the triples

(k, 0, k) for k a positive integer ¹ 1, are not primitive either.

Now, what if k = 0 in (k, 0, k)? Then, multiplying any triple by (k, 0, k) with k = 0 will introduce 0 as a common factor in each of its members, resulting in (0 x a, 0 x b, 0 x c) = (0, 0, 0), a new triple. Could this triple act like the integer zero? If so, then we would expect that (a, b, c) x (0, 0, 0) would equal (0, 0, 0), just as 7 x 0 = 0, or 12 x 0 = 0 for integers. Let’s see if it does. Now,

(a, b, c) x (0, 0, 0) = (a x 0, b x 0 + c x 0, b x 0 + c x 0) = (0, 0, 0),

and our prediction is correct, and we shall say that

O = (0, 0, 0) is the zero triple for Pythagorean triples.

In conclusion, we have found that

1 = (1, 0, 1), the unitary triple, acts like the integer 1 for multiplication; that is,

A x 1 = A x (1, 0, 1) = A, for any Pythagorean triple A, and

O = (0, 0, 0), the zero triple, acts like the integer 0 for multiplication; that is,

A x O = A x (0, 0, 0) = O, for any Pythagorean triple A. and

K = (k, 0, k), the k-multiplier triple, acts like the integer k for multiplication; that is,

K x A = (k, 0, k) x A = (ka, kb, kc), for any Pythagorean triple A = (a, b, c).

_______________________________________________________________________

Example 2. 1 x (15, 8, 17) = (1, 0, 1) x (15, 8, 17) = (15, 8, 17)

O x (15, 8, 17) = (0, 0, 0) x (15, 8, 17) = (0, 0, 0)

4 x (15, 8, 17) = (4, 0, 4) x (15, 8, 17) = (4x15, 4x8, 4x17) = (60, 32, 68)

6. Adding primitive Pythagorean triples

To add two primitive Pythagorean triples, it is natural to simply add their a-values together. Thus,

(7, 24, 25) + (15, 8, 17) would equal (22, __ , __ ).

But this gives us a triple with an even a-value of 22, which is not possible, since a must always be odd. In fact, whenever we add the a-values of any two triples together, we will get an even integer, since both a-values of the two triples are odd, and hence their sum will be even. So this method of addition does not work.

Now in section 5 above, we found that (1, 0, 1) acted like the number 1 for multiplication. If (1, 0, 1) also acts like the number 1 for addition, and if we have ordered our triples in terms of their increasing a-values, we would expect that adding (1, 0, 1) to any triple would give us the next higher triple in order. For example, if we were to add

(1, 0, 1) to (5, 12, 13), we should get (7, 24, 25), the next higher triple in order after (5, 12, 13). But, to get

(7, 24, 25), we must add 1 to the sum of the two a-values of (1, 0, 1) and (5, 12, 13) to get this desired sum. Thus, the sum of (5,12,13) and (1,0,1) will be

(5 + 1 + 1, __ , __ ) = (7, 24, 25 ), and in general,

(a, b, c) + (1, 0, 1) will equal (a + 1 + 1, __ , __ ) = (a + 2, __ , __ ),

which is the next triple after (a, b, c) when the triples are ordered in terms of their increasing a-values. So, let’s say that

the a-value of the sum of two triples is one more than the sum of their a-values,

and the b and c values will be those which are associated with the resulting a-value. Then,

a) (7, 24, 25) + (3, 4, 5) will equal (7+3+1, __ , __ ) = (11, 60, 61), and

b) (13, 84, 85) + (5, 12, 13) = (13+5+1, 180, 181) = (19, 180, 181).

Note that we have simply put in the correct b and c values which correspond to the computed a-value. But what if that a-value belongs to more than one triple?

For example, what about the sum of (5, 12, 13) and (9, 40, 41)? It will be (5+9+1, ? , ? ) = (15, ?, ?), but what are we to put in for the b and c values?

There are two triples which have 15 as their a-value; namely, (15, 8, 17) and (15, 112, 113). Which one of these should we pick?

Since (5, 12, 13) and (9, 40, 41) are both complete, let’s pick (15,112,113) as their sum since it is also complete.

Next, let’s consider A = (15, 8, 17), B = (15, 112, 113), and C = (7, 24, 25). Since A ¹ B, we would expect the sum of A and C to be different from the sum of B and C. But, since the a-values of A and B are the same, we get

A + C = (15,8,17) + (7,24,25) = (15+7+1, __ , __ ) = (23,264,265), (1)

and B + C = (15,112,113) + (7,24,25) = (15+7+1, __ , __ ) = (23,264,265), (2)

and we see that A + C = B + C, even though A ¹ B. This is not good!

Now, in (1) above, we added an incomplete and a complete triple together and got a complete triple as their sum, whereas in (2) above, the triples were both complete, and their sum was also complete.

To be consistent, let’s agree to add only those triples which are complete.

Notice that in our rule of addition, we look only at the a-values of the two triples to be added in determining the

a-value of their sum, and then we determine the b- and c- values of this sum based on this resulting a-value. Since we have agreed to restrict our addition to complete triples only, this makes it especially nice, since any complete triple is completely determined by its a-value alone. Let’s see why.

In a complete triple (a, b, c), c = b + 1.

Then, since a² + b² = c²,

we have a² + b² = (b+1)²,

a² + b² = b² + 2b + 1,

and a² = 2b + 1. (3)

Solving (3) for b, we get b = (a² – 1)/2,

giving us

Theorem 47

If (a, b, c) is complete, then b = (a² – 1)/2, and c = b + 1.

__________________________________________________

which is a restatement of Theorem 10 in Part 1.

Example 1. Find the complete triple with a = 7. From Theorem 47, b = (7² – 1)/2 = (49 – 1)/2 = 24, and c = b + 1 = 24 + 1 = 25. Thus, the complete triple is (7, 24, 25).

Example 2. Find the sum of (7, 24, 25) and (13, 84, 85). Since both of these triples are complete, the a-value of their sum will be one more than the sum of their a-values; namely, 7 + 13 + 1 = 21. Then, by Theorem 47,

b = (21² – 1)/2 = (441 - 1)/2 = 220, and c = b + 1 = 221. So, their sum is (21, 220, 221).

Combining our discussion above with Theorem 47, we get

Adding Complete Triples (1^st method)

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are complete Pythagorean triples, then

(a₁, b₁, c₁) + (a₂, b₂, c₂) = (a, b, c), where

a = a₁ + a₂ + 1, b = (a² – 1)/2, and c = b + 1.

_________________________________________________________

From the above, we have

a = a₁ + a₂ + 1,

b = (a² – 1)/2 = [(a₁ + a₂ + 1)² – 1] / 2

= [(a₁² + a₂² + 1 + 2a₁a₂ + 2a₁ + 2a₂) – 1] / 2

= ( a₁² + a₂² + 2a₁a₂ + 2a₁ + 2a₂) / 2, and using (3) above,

we have = [(2b₁ + 1) + (2b₂ + 1) + 2a₁a₂ + 2a₁ + 2a₂] / 2

= b₁ + b₂ + 1 + a₁ + a₁a₂ + a₂,

and c = b + 1

= (b₁ + b₂ + 1 + a₁ + a₁a₂ + a₂) + 1

= (c₁ – 1) + (c₂ – 1) + a₁ + a₁a₂ + a₂ + 2

= c₁ + c₂ + a₁ + a₁a₂ + a₂.

Thus, we get

Adding Complete Triples (2^nd method)

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are complete Pythagorean triples, then

(a₁, b₁, c₁) + (a₂, b₂, c₂) = (a₁ + a₂ + 1, b₁ + b₂ + 1 + d, c₁ + c₂ + d),

where d = a₁ + a₁a₂ + a₂.

_______________________________________________________________

a most complicated expression to say the least! But there ARE patterns in it.

First, to add two triples, we add their a-values, b-values, and c-values, getting

(a₁ + a₂ + . . . , b₁ + b₂ + . . . , c₁ + c₂ + . . . ).

Next, we add 1 to the a- and b-values, giving us

( a₁ + a₂ + 1 + . . . , b₁ + b₂ + 1 + . . . , c₁ + c₂ + . . . ).

Lastly, we add d = a₁ + a₁a₂ + a₂ to the b and c values, getting

(a₁ + a₂ + 1, b₁ + b₂ + 1+ (a₁ + a₁a₂ + a₂) , c₁ + c₂ + (a₁ + a₁a₂ + a₂)).

Example 3. Find the sum of (5, 12, 13) and (7, 24, 25) using the 2^nd method above.

First, let’s calculate d = a₁ + a₁a₂ + a₂ = 5 + 5(7) + 7 = 47. Then, by the 2^nd method,

(5, 12, 13) + (7, 24, 25) = ( 5+7+1, 12+24+1 + d, 13+25 + d )

= (13, 37 + d, 38 + d ) = (13, 37 + 47, 38 + 47)

= (13, 84, 85).

Now, let’s find the sum of any non-zero triple (a₁, b₁, c₁) and the zero triple, O = (0,0,0)

Since O = (0,0,0) is the only Pythagorean triple whose a-value is even, then, when we add O to any non-zero triple, we do not need to increase the sum of the a-values by 1 to get an odd integer for this a-value! We simply add the two a-values together, and then find the b and c values using the 1^st method of addition. So, we have

(a₁, b₁, c₁) + (0, 0, 0) = (a, b, c),

where a = a₁ + a₂ = a₁ + 0 = a₁,

b = (a² – 1)/2 = (a₁² – 1)/2, and, since a₁² = 2b₁ + 1 from (3) above, we have

= (2b₁ + 1 – 1)/2 = b₁,

and c = b + 1 = b₁ + 1 = c₁.

Thus, (a₁, b₁, c₁) + (0, 0, 0) = (a₁, b₁, c₁),

showing us that O = (0,0,0) is the zero triple for addition as well as multiplication since the triple

(a₁, b₁, c₁) was not changed by the addition of (0,0,0). (Note that the 2^nd method for addition could not be used in adding the zero triple to a triple because its expressions for the b and c values were obtained from an a-value of

a₁ + a₂ + 1 rather than the a-value of a₁ + a₂ which we used.)

Now this procedure is fine for adding (0,0,0) to any non-zero triple, but if we add (0,0,0) to the zero triple (0,0,0) by this method, we get

(0,0,0) + (0,0,0) = (a, b, c),

where a = a₁ + a₂ = 0 + 0 = 0,

b = (a² – 1)/2 = (0² – 1)/2 = -1/2 ,

and c = b + 1 = -1/2 + 1 = 1/2,

which is absurd, since b and c are not even integers! This occurred, of course, because the a-value of our sum was 0 and the formulas for b and c only apply to odd integers. Thus, we shall simply state that

O + O = (0,0,0) + (0,0,0) = (0,0,0) = O.

Thus, if A = (a, b, c) is a non-zero Pythagorean triple and O = (0,0,0), we have

A x O = O and A + O = A and O + O = O.

7. Subtracting complete primitive Pythagorean triples

We shall now consider the subtraction of complete Pythagorean triples. If (a, b, c) is the difference between two triples (a₁, b₁, c₁) and (a₂, b₂, c₂), then

(a₁, b₁, c₁) – (a₂, b₂, c₂) = (a, b, c)

means that (a₁, b₁, c₁) = (a₂, b₂, c₂) + (a, b, c) = (a₂ + a + 1, __ , __ ).

Equating the a-values, we have a₁ = a₂ + a + 1,

and solving this for a, we get a = a₁ – a₂ – 1, where a₁ must be greater than a₂ (a₁ > a₂),

so that a will be a positive integer.

Since (a, b, c) is complete, we have (from Theorem 47),

b = (a² – 1)/2

= [ (a₁ - a₂ - 1)² – 1 ] / 2

= [(a₁² + a₂² + 1 - 2a₁a₂ - 2a₁ + 2a₂) – 1] / 2

= (a₁² + a₂² - 2a₁a₂ - 2a₁ + 2a₂) / 2, and from (3) in section 6,

we have = [(2b₁ + 1) + (2b₂ + 1) - 2a₁a₂ - 2a₁ + 2a₂] / 2

= b₁ + b₂ + 1 – (a₁ + a₁a₂ – a₂),

and c = b + 1

= [b₁ + b₂ + 1 - (a₁ + a₁a₂ - a₂)] + 1

= (c₁ – 1) + (c₂ – 1) - (a₁ + a₁a₂ - a₂) + 2

= c₁ + c₂ – (a₁ + a₁a₂ - a₂).

This all gives us

Subtracting Complete Triples

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are complete Pythagorean triples, a₁ > a₂, then

1^st method: (a₁, b₁, c₁) – (a₂, b₂, c₂) = (a, b, c),

where a = a₁ – a₂ – 1, b = (a² – 1)/2, and c = b + 1.

2^nd method: (a₁, b₁, c₁) – (a₂, b₂, c₂)

= (a₁– a₂ –1, b₁+ b₂ + 1 – d , c₁+ c₂ – d ),

where d = a₁ + a₁a₂ – a₂

________________________________________________________________

Example 4. Use both methods of subtraction to find (15, 112, 113) – (11, 60, 61).

1^st method: (15, 112, 113) – (11, 60, 61) = (a, b, c), where a = 15 – 11 – 1 = 3,

b = (3² – 1)/2 = 4, and c = 4 + 1 = 5. Thus, (15, 112, 113) – (11, 60, 61) = (3, 4, 5).

2^nd method: First, let’s calculate d = a₁ + a₁a₂ – a₂ = 15 + 15(11) – 11 = 169. Then,

(15, 112, 113) – (11, 60, 61) = (15 – 11 – 1, 112 + 60 + 1 – d, 113 + 61 – d )

= (3, 173 – d, 174 – d) = (3, 173 – 169, 174 – 169) = (3, 4, 5)

8. Negative Pythagorean triples

Now, what if we subtract the triple (a, b, c) from itself? Let’s see what we get. Using the 2^nd method of subtraction, we first find d = a₁ + a₁a₂ – a₂ = a + a² – a = a². Then,

(a, b, c) – (a, b, c) = (a –a –1, b + b + 1 – d, c + c – d) = ( -1, (2b + 1) – a², 2c – a²),

and since a² = 2b+1 from (3) above, and c = b + 1, we have

= ( -1, a² – a², 2(b+1) – a²) = ( -1, 0, 2b + 2 – a²)

= ( -1, 0, (2b + 1) – a² + 1) = ( -1, 0, a² – a² + 1)

= ( -1, 0, 1), a new triple,

which is both complete (since c = b + 1) and Pythagorean since (-1)²+ 0² = 1².

Now, this is very strange since we would expect to get the (0, 0, 0) triple whenever we subtract a triple from itself ! Instead, we got the new triple (-1, 0, 1), the first triple we have seen which has a negative a-value, and it reminds us of the integer - 1. Now if it acts like - 1, we would expect that multiplying it by itself would give us the triple (1,0,1), which, as we know, acts like the integer 1. Let’s see if this is so.

( -1,0,1) x ( -1,0,1) = ( -1(-1), 0(1) + 1(0), 0(0) + 1(1) ) = (1, 0, 1), which confirms our surmise!

Now if (-1,0,1) acts like the integer -1 in multiplication, let’s see what happens if we multiply it by the triple (3,4,5). Then,

(3,4,5) x (-1,0,1) = ( 3(-1), 4(1)+5(0), 4(0)+5(1) ) = (-3,4,5), and in a like

manner, (5,12,13) x ( -1,0,1) = ( -5,12,13),

and, in general (a, b, c) x ( -1,0,1) = ( -a, b, c). (1)

Now if a > 0, we shall call (a, b, c) a positive triple and (- a, b, c) a negative triple, and we shall say that

(- a, b, c) is the opposite of (a,b,c).

So we have just shown in (1) that multiplying a positive triple (a, b, c) by ( -1, 0, 1) changes it into a negative triple (-a, b, c).

Furthermore, if (-1,0,1) acts like the integer -1, then we would expect that adding it to a triple would give us that triple which is just before it in a list of complete triples ordered by their increasing a-values. Thus, we would expect that (5,12,13) + ( -1,0,1) would equal (3,4,5), the triple just before (5,12,13). However,

(5,12,13) + ( -1,0,1) = (5 + (-1) + 1, __ , __ ) = (5,12,13),

and (5,12,13) is not changed at all, making (-1, 0, 1) seem like the zero triple (0, 0, 0), which, of course, it is not!

Perhaps, in adding the positive triple (5,12,13) to the negative triple (-1,0,1), we should have subtracted rather than added 1 to the sum of their a-values. If so, then we would get

(5,12,13) + ( -1,0,1) = (5 + (-1) – 1, __ , __ ) = (3,4,5),

which IS the triple just before (5,12,13), as expected!

Apparently, when we add a positive and a negative triple, we must add their a-values and then subtract 1, whereas if adding two positive triples, we add their a-values and then add 1. But is this always the case?

To find out, let’s construct a table of Pythagorean triples containing both positive and negative triples along with the zero triple and ordered by their increasing a-values.

Table 2

m n a b c

--------------------------------------------

- 4 - 5 - 9 40 41

- 3 - 4 - 7 24 25

- 2 - 3 - 5 12 13

- 1 - 2 - 3 4 5

0 - 1 - 1 0 1

0 0 0 0 0

1 0 1 0 1

2 1 3 4 5

3 2 5 12 13

4 3 7 24 25

5 4 9 40 41

Now, what should be the rules for addition here? We have found that (1,0,1) acts like the integer 1 and

(-1,0,1) acts like the integer -1. Using these triples as guides to help us, let’s look at

a) (5,12,13) + (1,0,1) b) ( - 5,12,13) + ( -1,0,1) c) ( - 5,12,13) + (1,0,1)

a) Since (1,0,1) acts like the integer 1, it seems natural for (5,12,13) + (1,0,1) to equal the next higher triple in Table 2; namely, (7,24,25). Thus, we must add the a-values of (5,12,13) and (1,0,1), 5 + 1 = 6, and then add 1 more to get (7,24,25). Using this example as a pattern, we can state that, in general, if we are adding two triples

(a₁, b₁, c₁) and (a₂, b₂, c₂), whose a-values are both positive, then the a-value of their sum must be a₁ + a₂ + 1.

b) Since (-1,0,1) acts like the integer -1, it seems natural for ( -5,12,13) + (-1,0,1) to equal the next lower triple in Table 2; namely, ( -7, 24, 25). Thus, we must add the a-values of ( -5,12,13) and (-1,0,1); namely, ( - 5) + ( -1) = - 6,

and then subtract 1 more to get ( - 7, 24, 25). Using this as our pattern, we can say that if we are adding two triples (a₁, b₁, c₁) and (a₂, b₂, c₂), whose a-values are both negative, then the a-value of their sum must be a₁+ a₂ – 1.

However, if one of the triples is positive and the other negative, it gets a little more complicated.

c) Since (1,0,1) acts like the integer 1, it seems natural for (- 5,12,13) + (1,0,1) to be equal to the next higher triple in Table 2; namely, (-3,4,5). Thus, we must add the a-values of (- 5,12,13) and (1,0,1), - 5 + 1 = - 4, and then add 1 more to get (-3,4,5).

Now, consider the table below where, starting with (1,0,1), we have added positive triples with successively higher a-values to (-5,12,13).

Addition a-value of the sum

---------------------------------------------------------------------------------------------------------

(- 5,12,13) + (1,0,1) = (-3,4,5) - 3 = - 5 + 1 + 1 = a₁ + a₂+ 1

(- 5,12,13) + (3,4,5) = (-1,0,1) - 1 = - 5 + 3 + 1 = a₁ + a₂+ 1

(- 5,12,13) + (5,12,13) = (0,0,0) 0 = - 5 + 5 = a₁ + a₂

(- 5,12,13) + (7,24,25) = (1,0,1) 1 = - 5 + 7 – 1 = a₁ + a₂ – 1

(- 5,12,13) + (9,40,41) = (3,4,5) 3 = - 5 + 9 - 1 = a₁ + a₂ – 1

All of these additions are quite reasonable, since by adding successively increasing positive triples to

( -5,12,13), we find that the sums are also successively higher triples.

Looking closer, we see that in the first two additions, where the a-value of the negative triple is larger than the a-value of the positive triple in absolute value, we added 1 to their sum.

In the third addition, where the a-value of the negative triple is equal to the a-value of the positive triple in absolute value, we added nothing to their sum. and

In the last two additions, where the a-value of the negative triple is smaller than the a-value of the positive

triple in absolute value, we subtracted 1 from their sum.

So, using Table 2 and the results found in a), b), and c) above as patterns, we have

Addition of Complete Triples

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are positive or negative complete triples or (0,0,0), then

(a₁, b₁, c₁) + (a₂, b₂, c₂) = (a, b, c), where

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

1. a₁ + a₂ + 1, if a₁ and a₂ are both positive,

2. a₁ + a₂ - 1, if a₁ and a₂ are both negative,

3. a) 0, if a₁ + a₂ = 0,

a = b) a₁ if a₂ = 0,

c) a₂ if a₁ = 0,

4. a₁ + a₂ + 1, if a₁ and a₂ are of opposite sign, and the negative number

is larger than the positive one in absolute value

5. a₁ + a₂ - 1, if a₁ and a₂ are of opposite sign, and the negative number

is smaller than the positive one in absolute value

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

(a² – 1)/2, if a ¹ 0, b + 1, if a ¹ 0,

b = c =

0, if a = 0, 0, if a ¹ 0.

__________________________________________________________________________________

Let’s work a few problems to illustrate each of these rules of addition.

1. Find ( 7, 24, 25) + (11, 60, 61). Since a₁ and a₂ are both positive, we use rule 1. Thus,

a = 7+11+1 = 19, b = (19² – 1)/2 = 180, and c = 180+1 = 181, so

(7,24,25) + (11,60,61) = (19,180,181).

2. Find ( -3,4,5) + ( -9,40,41). Since a₁ and a₂ are both negative, we use rule 2. Thus,

a = -3 + (-9) – 1 = -13, b = ( (-13)² – 1)/2 = 84, and c = 84+1 = 85, so

(- 3,4,5) + (- 9,40,41) = (-13,84,85).

3. Find (15, 112, 113) + ( -15, 112, 113). Since a₁ + a₂ = 0, we use rule 3. Thus,

a = 15 + (-15) = 0, b = 0, and c = 0, so

(15,112,113) + (-15,112,113) = (0,0,0).

4. Find ( -7, 24, 25) + (0, 0, 0). Since a₂ = 0, we use rule 3. Thus,

a = -7, b = ( (-7)² – 1)/2 = 24, and c = 24+1 = 25, so

(-7,24,25) + (0,0,0) = (-7,24,25).

5. Find (-15, 112, 113) + ( 9, 40, 41). Since a₁ and a₂ are of opposite signs, and the negative number (- 15) is larger than the positive one (9) in absolute value, we use rule 4. Thus,

a = - 15 + 9 + 1 = - 5, b = ( (-5)² – 1)/2 = 12, and c = 12 + 1 = 13, so

(-15,112,113) + (9,40,41) = (- 5,12,13).

6. Find (11, 60, 61) + ( - 7, 24, 25). Since a₁ and a₂ are of opposite signs, and the negative number ( - 7) is smaller than the positive one (11) in absolute value, we use rule 5. Thus,

a = 11 + (-7) – 1 = 3, b = (3² – 1)/2 = 4, and c = 4 + 1 = 5, so

(11,60,61) + (-7,24,25) = (3,4,5).

In short, to find the a-value of the sum of two complete triples, find the sum of their two a-values, and

if both a-values are positive, add 1 to this sum,

if both a-values are negative, subtract 1 from this sum,

if one a-value is positive and one is negative,

1) add 1 to this sum if the absolute value of the negative one is larger than the positive one,

2) subtract 1 from this sum if the absolute value of the negative one is smaller than the positive one,

To find the b- and c- values, use b = (a² – 1)/2, and c = b + 1.

If one of the triples is the zero triple (0, 0, 0), then their sum is the other triple, and if the sum of their two

a-values is 0, then the sum of the two triples is (0, 0, 0).

Let’s now consider subtracting both positive and negative triples. For integers, we know that

a – b = a + (-1)b = a + ( - b),

so patterning our subtraction on this, we have

Subtraction of Complete Triples

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are positive or negative complete triples, then

(a₁, b₁, c₁) – (a₂, b₂, c₂) = (a₁, b₁, c₁) + (-1, 0, 1) x (a₂, b₂, c₂)

= (a₁, b₁, c₁) + ( - a₂, b₂, c₂).

________________________________________________________________

To subtract (a₂, b₂, c₂) from (a₁, b₁, c₁), we simply add its opposite to (a₁, b₁, c₁).

a) (7, 24, 25) – (1,0,1) = (7, 24, 25) + ( -1,0,1) = (7 + (-1) – 1, __ , __ )

= (5, 12, 13), by rule 5 above,

b) ( - 9, 40, 41) – ( -3,4,5) = ( -9, 40, 41) + (3,4,5) = ( -9 + 3 + 1, __ , __ )

= ( - 5,12,13), by rule 4 above,

c) ( - 11, 60, 61) – ( - 11, 60, 61) = ( - 11, 60, 61) + (11, 60, 61) = (0, 0, 0), by rule 3 above,

d) (5,12,13) – ( - 3,4,5) = (5,12,13) + (3,4,5) = (5 + 3 + 1, __ , __ )

= (9,40,41), by rule 1 above.

If we wished, we could combine the rule for subtracting triples with the rules for adding complete triples above to come up with a complete list of rules for subtracting complete triples, but it is much easier to perform any subtractions by simply adding the opposite of the triple we wish to subtract.

9. The Distributive Property for Pythagorean triples

The distributive property for integers states that for any integers a, b, and c,

a(b + c) = ab + ac.

Accordingly, we might expect that if A = (a₁, b₁, c₁), B = (a₂, b₂, c₂), and C = (a₃, b₃, c₃), are complete triples,

then A x (B + C) = A x B + A x C. (1)

However, this is not true. If both a₁ and a₂ are positive, we get

A x (B + C) = (a₁, b₁, c₁) x [(a₂, b₂, c₂) + (a₃, b₃, c₃)]

= (a₁, b₁, c₁) x (a₂ + a₃ + 1, __ , __ )

= ([a₁(a₂ + a₃ + 1), __ , __ )

= (a₁a₂ + a₁a₃ + a₁, __ , __ ), (2)

whereas, A x B + A x C = (a₁, b₁, c₁) x (a₂, b₂, c₂) + (a₁, b₁, c₁) x (a₃, b₃, c₃)

= (a₁a₂ , __ , __ ) + (a₁a₃, __ , __ )

= (a₁a₂ + a₁a₃ + 1, __ , __ ). (3)

Since (2) and (3) are NOT identical, we have shown (1) is not true for complete Pythagorean triples.

To make the a-values in (2) and (3) identical, we need to add a₁ to and remove 1 from the a-value of (3).

Adding A = (a₁, b₁, c₁) to (3), we get (a₁a₂ + a₁a₃ + 1, __ , __ ) + (a₁, b₁, c₁)

= (a₁a₂ + a₁a₃ + 1 + a₁ + 1, __ , __ )

= (a₁a₂ + a₁a₃ + a₁ + 2, __ , __ ). (4)

Now in (4), we have the a₁ we needed, but need to get rid of the 2.

Subtracting (1,0,1) from (4), we get (a₁a₂ + a₁a₃ + a₁ + 2, __ , __ ) – (1,0,1)

= (a₁a₂ + a₁a₃ + a₁ + 2, __ , __ ) + ( - 1,0,1), and using Rule 5 for

addition, we get = (a₁a₂ + a₁a₃ + a₁ + 2 - 1 - 1, __ , __ ),

= (a₁a₂ + a₁a₃ + a₁, __ , __ ), which is identical to (2), as desired.

Thus, for positive complete triples, our distributive property is

Distributive Property for Positive Complete Triples

If A, B, and C are positive complete triples, then

A x (B + C) = A x B + A x C + A – (1,0,1).

___________________________________________

a somewhat complicated, but interesting, statement!

Example

Adding inside the brackets first:

(3,4,5) x [ (5,12,13) + (7,24,25) ] = (3,4,5) x (13, 84,85)

= (39, 760,761)

Multiplying first:

(3,4,5) x [ (5,12,13) + (7,24,25) ] = (15,112,113) + (21,220,221) + (3,4,5) – (1,0,1)

= (37,684,685) + (3,4,5) – (1,0,1)

= (41, 840,841) – (1,0,1)

= (39, 760,761)

If A, B, and C are all negative triples, we can show that

A x (B + C) = A x B + A x C – A – (1,0,1).

And we can certainly find other expressions for the distributive property for those A’s, B’s and C’s which have different signs. But I think we’ve done enough for now. There will be other days for these discoveries and more, I’m sure.

So that’s it! We have found many, many patterns in our study of right triangles and Pythagorean triples, and it has been a most fascinating and, above all, an interesting pastime! The joy we have found in discovering these patterns has been wonderful, but you know, the time we have spent and the effort we have made in discovering these patterns have given us the most fun! Here’s to finding other patterns in these triangles and triples in the future, because I know that somewhere, out there, there are more.

October 9, 2005

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