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 and Pythagorean 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

5. Further Patterns in Right Triangles

and Pythagorean Triples

David W. Hansen

In Part 4, we learned how to add complete Pythagorean triples. The results are given below.

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,

b) a₁ if a₂ = 0,

a = 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 = and c =

0, if a = 0, 0, if a = 0.

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To subtract complete Pythagorean triples, we found

Subtraction of Complete Triples

If (a₁, b₁, c₁) and (a₂, b₂, c₂) are 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).

_________________________________________________________

and we also found out how to multiply, divide, take square roots of, and find integral powers of Pythagorean triples. However, we never considered what it would mean to multiply a Pythagorean triple by an integer. Let's consider it now.

Multiplying a Positive Complete Triple by a Positive Integer

In multiplying 6 by 2, we mean adding up two sixes; 2(6) = 6 + 6, and in multiplying 5 by 3, we mean adding up three 5’s; 3(5) = 5 + 5 + 5. Thus, in multiplying the triple (3, 4, 5) by 2, we shall mean adding up two (3, 4, 5) triples; thus, 2(3, 4, 5) = (3, 4, 5) + (3, 4, 5). Adding these two triples together by using Rule 1 in the table of addition above, we get

2(3, 4, 5) = (3, 4, 5) + (3, 4, 5) = (3 + 3 + 1, b, c)

= (2(3) + 1, b, c) = (7, 24, 25).

Note that the a-value is twice the a-value of the original triple, 2(3), plus 1 more, since we carried out one addition.

To multiply (5, 12, 13) by 4, we have

4(5, 12, 13) = (5, 12, 13) + (5, 12, 13) + (5, 12, 13) + (5, 12, 13),

and using Rule 1 in the table above for each of the three additions, we get

4(5, 12, 13) = (5, 12, 13) + (5, 12, 13) + (5, 12, 13) + (5, 12, 13),

= (5 + 5 + 1, __ , __ ) + (5 + 5 + 1, __ , __ ),

= ((5 + 5 + 1) + (5 + 5 + 1) + 1, __ , __ )

= ( 4(5) + 3, __ , __ ) = (23, 264, 265).

Note that the a-value is four times the a-value of the original triple, 4(5), plus 3 more, since we carried out three additions.

In general, if (a, b, c) is a positive complete triple, then k(a, b, c), for k a positive integer, means the sum of k of the (a, b, c)’s; that is k(a, b, c) = (a, b, c) + (a, b, c) + . . . + (a, b, c) for k triples, and in finding the a-value of this sum, we simply add up the k a’s and then add 1 more for each addition. Since adding up k triples involves carrying out k –1 additions, we must add k –1 to the sum of our k a’s, giving us ka + (k – 1) for the a-value of the sum. To find the b- and c-values, we use those given in the addition table above.

Since we are not multiplying two triples together, but rather are multiplying one triple by an integer, we shall call this scalar multiplication, and the multiplication of one triple by another, we shall call triple multiplication. Thus, we have

Scalar Multiplication of Positive Complete Triples

If k is a positive integer, and (a, b, c) is a positive complete triple, then

k(a, b, c) = (a₁, b₁, c₁),

where a₁ = ka + (k -1), b₁ = (a₁² – 1)/2, and c₁ = b₁ + 1.

___________________________________________________________ (1)

and from section 4, we have

Triple Multiplication of 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 ),

which is a Pythagorean triple.

_____________________________________________________

Examples.

1. 3(5, 12, 13) = (a₁, b₁, c₁), where a₁ = 3(5) + 2 = 17, b₁ = (17² – 1)/2 = 144, and c₁ = 144 + 1 = 145.

Thus, 3(5, 12, 13) = (17, 144, 145), which is scalar multiplication.

2. (3, 4, 5) x (5, 12, 13) = (3x5, 4x13 + 5x12, 4x12 + 5x13) = (15, 112, 113), which is triple multiplication.

3. If X = (9, 40, 41) and Y = (5, 12, 13), then

4X – 5Y + (7, 24, 25) = 4(9, 40, 41) – 5(5, 12, 13) + (7, 24, 25)

= (4(9)+3, __ , __ ) – (5(5)+4, __, __ ) + (7, 24, 25)

= (39, (39²- 1)/2, __ ) + ( - 29, (29²- 1)/2, __ ) + (7, 24, 25)

= (39, 760, 761) + ( - 29, 420, 421) + (7, 24, 25)

= (39 – 29 – 1, __, __ ) + (7, 24, 25) = (9, (9²- 1)/2, __ ) + (7, 24, 25)

= (9, 40, 41) + (7, 24, 25) = (9 + 7 + 1, __, __ )

= (17, (17² - 1)/2, __ ) = (17, 144, 145).

Here we have shown the full solution, using the addition and subtraction rules given on page 1 above, and filling in the blank values for b and c as we moved along. But it is not necessary to find the b- and c-values for these intermediate triples in our solution. Since we are working with complete triples, we need find only the a-values of the intermediate triples as we go along, and then calculate the b- and c-values, b = (a² – 1)/2, c = b + 1, at the end. Here’s how.

4X – 5Y + (7, 24, 25) = 4(9, 40, 41) – 5(5, 12, 13) + (7, 24, 25)

a-value = 4(9)+3, – (5(5)+4), + 7

= 39, – 29, + 7

= 39 – 29 – 1, + 7 = 9, + 7

= 9 + 7 + 1 = 17.

Thus, a = 17, b = (17² - 1)/2 = 144, and c = 144+1 = 145, giving us (17, 144, 145). (For clarity, we used commas to separate the a-values for each of the three triples.)

4. Solve X – 4(3, 4, 5) = 2(11, 60, 61) for the complete triple X. Adding 4(3, 4, 5) to both sides of the given

equation, we get X = 2(11, 60, 61) + 4(3, 4, 5).

Calculating the a-value of X for the sum of the two triples above, we have

2(11)+1, + 4(3)+3 = 23, + 15 = 23 + 15 + 1 = 39.

Thus, a = 39, b = (39² – 1)/2 = 760, c = b + 1 = 761, and X = (39, 760, 761).

Check: Substituting X into the left-hand side of the equation above, we get

(39, 760, 761) – 4(3, 4, 5) = (39, 760, 761) – (15, 112, 113) = (39 – 15 – 1, __ , __ )

= (23, 264, 265) = (2(11)+1, __ , __ ) = 2(11, 60, 61). Ö

Multiplying a Negative Complete Triple by a Positive Integer

Now what if our complete triple is a negative triple, say (a, b, c), where a is negative and b and c are positive integers? This is no problem. 3( - 5, 12, 13) simply means three of the triples ( - 5, 12, 13) added together,

( - 5, 12, 13) + ( - 5, 12, 13) + ( - 5, 12, 13),

and using Rule 2 in the table above for each of the two additions, we have

3( - 5, 12, 13) = ( - 5, 12, 13) + ( - 5, 12, 13) + ( - 5, 12, 13)

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

= (( - 5 + (- 5) - 1) + (- 5 ) - 1 , __ , __ )

= ( 3(-5) - 2 , __ , __ ) = ( - 17, 144, 145).

Note that the a-value is three times the a-value of the original triple, 3(- 5), minus 2, since we carried out two additions.

In general, k(a, b, c), where a is negative, means the sum of k of the (a, b, c)’s,

k(a, b, c) = (a, b, c) + (a, b, c) + . . . + (a, b, c) (for k triples),

and in finding the a-value of this sum, we add up the k a’s and then subtract 1 for each addition since a is negative. Since adding up k triples involves carrying out k –1 additions, we must subtract k –1 from the sum of our k a’s, giving us ka - (k -1) for the a-value of the sum. This gives us

Scalar Multiplication of Negative Complete Triples

If k is a positive integer, and (a, b, c) is a negative complete triple, then

k(a, b, c) = (a₁, b₁, c₁)

where a₁ = ka – (k -1), b₁ = (a₁² – 1)/2, and c₁ = b₁ + 1. (2)

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Examples.

5. If X = ( - 7, 24, 25) and Y = (- 3, 4 5), then 5X – 2Y = 5( - 7 , 24, 25) – 2( - 3, 4, 5)

= ( 5(-7) – 4, __ , __ ) – ( 2(-3) – 1, __ , __ ) = ( - 39, __ , __ ) – ( - 7, __ , __ ).

Using our method of subtraction from above, we change the subtraction to the addition of the negative of the triple ( - 7, __ , __ ) and get

= ( - 39, __ , __ ) + (- ( - 7), __ , __ ) = ( - 39, __ , __ ) + ( 7, __ , __ )

= ( - 39 + 7 + 1, __ , __ ) = (- 31, ((-31)² – 1)/2), __ ) = ( - 31, 480, 481).

Multiplying a Complete Triple by a Negative Integer

Let’s see what happens if we multiply a complete triple by a negative integer. Just as (- 2)(5) = 2(- 5), it seems reasonable for - 2(5, 12, 13) to equal 2( - 5, 12, 13). Let’s see if it does.

Let (a, b, c) be any complete triple. Then,

(a, b, c) – 2(5, 12, 13) = (a, b, c) – (2(5)+1, __ , __ )

= (a, b, c) – (11, 60, 61)

= (a, b, c) + ( - 11, 60, 61)

= (a, b, c) + ( 2(-5) – 1, 60, 61)

= (a, b, c) + 2( - 5, 12, 13).

Thus, (a, b, c) – 2(5, 12, 13) = (a, b, c) + 2( - 5, 12, 13).

Subtracting (a, b, c) from both sides of the above equation, we get

- 2(5, 12, 13) = 2( - 5, 12, 13),

and our surmise is correct. In general, we shall say that

- k(a, b, c) = k(-a, b, c). (3)

Now, (3) tells us, for example, that - 3(3, 4, 5) = 3( - 3, 4, 5). However, if we use the rule for scalar multiplication of negative triples from (2) above to check this, we find that the

a-value of - 3(3, 4, 5) = - 3(3) – ( - 3 - 1) = - 9 – ( - 4) = - 5,

but the a-value of 3( - 3, 4, 5) = 3(- 3) – ( 3 -1) = - 9 -– 2 = - 17.

What went wrong? Well, for 3( - 3, 4, 5) , the k-value, 3, is positive and tells us that there are 3 triples to be added together. However, for - 3(3, 4, 5), the k-value, - 3, is negative, and so does not tell us the number of triples to be added. To correct this, we must use the absolute value of k when determining the number of triples to be added. Thus, we must rewrite the a-value for k(a, b, c) in (2) as ka – ( |k| - 1).

Similarly, just as (-2)(-5) = 2(5), it seems reasonable that - 2( - 5, 12, 13) should equal 2(5, 12, 13). Let’s see if it does.

Let (a, b, c) be any complete triple. Then,

(a, b, c) – 2( - 5, 12, 13) = (a, b, c) – (2(-5) - 1, __ , __ )

= (a, b, c) – ( - 11, 60, 61)

= (a, b, c) + ( 11, 60, 61)

= (a, b, c) + ( 2(5) + 1, 60, 61)

= (a, b, c) + 2(5, 12, 13).

Thus, (a, b, c) – 2( - 5, 12, 13) = (a, b, c) + 2( 5, 12, 13).

Subtracting (a, b, c) from both sides of the above equation, we get

- 2( - 5, 12, 13) = 2( 5, 12, 13),

and our surmise is correct. Thus, in general, we shall say that

- k(-a, b, c) = k(a, b, c). (4)

Now, (4) tells us, for example, that - 3( -3, 4, 5) = 3( 3, 4, 5). However, if we use the rule for scalar multiplication of negative triples from (2) above to check this, we find that the

a-value of - 3( -3, 4, 5) = - 3( -3) – ( - 3 - 1) = 9 – ( - 4) = 13,

but the a-value of 3( 3, 4, 5) = 3(3) – ( 3 -1) = 9 – 2 = 7.

What went wrong? Well, for 3( 3, 4, 5) , the k-value, 3, is positive and tells us that there are 3 triples to be added together. However, for - 3( -3, 4, 5), the k-value, - 3, is negative, and so does not tell us the number of triples to be added. To correct this, we must use the absolute value of k when determining the number of triples to be added. Thus, we must rewrite the a-value for k(a, b, c) in (1) as ka – ( |k| - 1).

If k = 0, then we shall interpret 0(a, b, c) as meaning none or zero triples, and thus

0(a, b, c) = (0, 0, 0). (5)

Multiplying a Complete Triple by an Integer

Combining (3), (4), and (5) with our observations about using |k| in determining the number of triples, we get

Scalar Multiplication of Complete Triples by Integers

If k is a non-zero integer, and (a, b, c) is a complete triple, then

k(a, b, c) = (a₁, b₁, c₁), where

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

ka + ( |k| - 1) if ka > 0,

a₁ =

ka – ( |k| - 1) if ka < 0,

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

b₁ = (a₁² – 1)/2, and c₁ = b₁ + 1.

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

If k = 0, then k(a, b, c) = (0, 0, 0)

_______________________________________________________________ (6)

Examples.

6. If a > 0, then -1(a, b, c) = ( -1(a) – (|-1| - 1), b₁, c₁) = ( - a – 0, b₁, c₁) = ( - a, b₁, c₁),

and if a < 0, then -1(a, b, c) = ( -1(a) + (|-1| - 1), b₁, c₁) = ( - a + 0, b₁, c₁) = ( - a, b₁, c₁),

where b₁ = (a² – 1)/2 = b, and c₁ = b + 1 = c. So,

- 1(a, b, c) = ( - a, b, c). (7)

7. The a-value of 2( - 3, 4, 5) – 3(7, 24, 25) – 6( - 5, 12, 13)

= 2(-3) – (|2| -1), - 3(7) – (|-3| -1), (-6)(-5) + (|- 6| -1)

= 2(-3) – 1, - 3(7) – 2, 6(5) + 5

= - 7, - 23, 35

= - 7 – 23 – 1, 35

= - 31, 35 = - 31 + 35 - 1 = 3.

Thus 2( - 3, 4, 5) – 3(7, 24, 25) – 6( - 5, 12, 13) = (3, 4, 5).

Multiplying a Complete Triple by a Rational Number

The rules for scalar multiplication of complete triples as given in (6) above apply only when we multiply a complete triple by an integer. There is utterly no reason to think that these rules should apply when multiplying a complete triple by a rational number. But, let’s see.

We know that if we multiply (5, 12, 13) by 3, we get 3(5, 12, 13) = (17, 144, 145). So, if we multiply

(17, 144, 145) by 1/3, we should get back to (5, 12, 13), our original triple, shouldn’t we? Let’s see.

Using the rules for scalar multiplication of complete triples by integers in (6) above, we have

(1/3)(17, 144, 145) = ((1/3)17 + (|1/3)| - 1), __ , __ ) = (17/3 + (1/3 – 1), __ , __ )

= (17/3 – 2/3, __ , __ ) = (15/3, __ , __ ) = (5, 12, 13) ,

which is our original triple! Thus, it seems that our rules for scalar multiplication of complete triples may apply for some rational numbers as well as all integers! Let’s try more examples.

8. Using (6) with k = 4/3, we get (4/3) (5, 12, 13) = ((4/3) 5 + 4/3 – 1, __ , __ )

= (20/3 + 4/3 – 1, __ , __ ) = (7, 24, 25), which is a complete triple.

9. Using (6) with k = 3/5, we get (3/5) (7, 24, 25) = ((3/5)7 + 3/5 – 1, __ , __ )

= (21/5 + 3/5 – 1, __, __ ) = (24/5 – 1, __, __ ) = (19/5, __ , __ ), which is NOT a complete triple.

(In fact, it is not even a Pythagorean triple because its a-value is not an integer.)

Clearly, we must come up with some restrictions on the values of k so that our scalar multiplication using rational numbers will always yield a complete triple.

Let a > 0, and k = p/q, where p and q are integers > 0 with no common factors. Then, from (6), we have

a₁ = ka + ( |k| - 1) = (p/q)a + |p/q| - 1 = (p/q)a + p/q - 1 = (p/q)(a + 1) - 1,

or a₁ = p(a + 1) - 1. (8)

a) Since a₁ must be an integer, then p(a + 1) must be an integer, and since p and q have no common

factors, then q must be a divisor of a + 1. However, if q = a + 1, then q is even, since it is one greater than a, which is always odd. And, if q = a + 1, then (8) tells us that a₁ = p – 1, so p = a₁ + 1, and p is also even, since a₁ is always odd. But p and q cannot both be even because they have no common factors. Thus, q cannot equal a + 1, and q must be a proper divisor of a + 1.

b) Since a₁ must be odd, then p(a + 1) – 1 must be odd. Then, p(a + 1) must be even, and so

q q

either (a + 1)/q or p must be even (or both).

Similar analyses of (6) using various combinations of the signs of a, p, and q yield the following result.

Scalar Multiplication of Complete Triples by Rational Numbers

If k = p/q is a rational number, p and q have no common factors, (a, b, c) is a complete triple or O,

q is a proper divisor of |a|+1, and either p or (|a|+1)/q is even, then k(a, b, c) = (a₁, b₁, c₁), where

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

ka + ( |k| - 1) if ka > 0,

a₁ = ka - ( |k| - 1) if ka < 0,

0 if ka = 0

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

(a₁² – 1)/2 if a₁ ¹ 0 b₁ + 1 if b₁ ¹ 0

b₁ = and c₁ =

0 if a₁ = 0 0 if b₁ = 0

_______________________________________________________________________________ (9)

Examples.

10. If we multiply ( - 9, 40, 41) by 3/5, will we obtain a complete triple? Yes, because q = 5 is a proper divisor of

|a| + 1 = 9 + 1 = 10, and (|a| + 1)/q = (9+1)/5 = 2 is even.

Check: (3/5) ( - 9, 40, 41) = ((3/5)(-9) – (|3/5| – 1), _ , _ ) = ( - 27/5 - 3/5 + 1, _ , _ )

= ( - 30/5 + 1, _ , _ ) = ( - 6 + 1, _ , _ ) = ( - 5, 12, 13).

11. Will (3/2) (13, 84,85) be a complete triple? No. q = 2 is a proper divisor of |a| + 1 = 13 + 1 = 14, but since

p = 3 is odd, (|a|+1)/q must be even by (9). However, (|a|+1)/q = (13+1)/2 = 7, which is odd.

Check: (3/2) (13, 84, 85) = ((3/2)13 + |3/2| – 1, _ , _ ) = (39/2 + 3/2 – 1, _ , _ ) = (20, _ , _ ), and we get

a triple whose a-value is even, not odd as required!

12. Will ( - 7/4) (5, 12, 13) be a complete triple? No, because q = 4 is NOT a proper divisor of |a| + 1

= 5 + 1 = 6.

Check: ( - 7/4) (5, 12, 13) = (( - 7/4)5 – (| - 7/4| - 1), _ , _ ) = ( - 35/4 - 7/4 + 1, _ , _ ) = ( - 42/4 + 1, _ , _ )

= ( - 38/4, _ , _ ) = ( - 19/2, _ , _ ), which is NOT a triple at all!

2. Solving Pythagorean Triple Equations

Using the rules for addition, subtraction, and multiplication of complete triples as given above at the start of section 1, and the rules for scalar multiplication of complete triples by rational numbers as given just above Example 10 , let’s solve some equations.

a) Linear equations and pure quadratic equations

1. (2/3)X = (7, 24, 25). Multiplying both sides of the equation by 3, we have

3(2/3)X = 3(7, 24, 25), 2X = (3(7) + 2, __ , __ ), 2X = (23, 264, 265).

Multiplying both sides of this last equation by 1/2 , we get

(1/2)2X = (1/2)(23, 264, 265) = (23/2 + 1/2 - 1, __ , __ ),

or X = (11, 60, 61).

Check: (2/3)(11, 60, 61) = ((2/3)11 + |2/3| - 1, _ , _ ) = (24/3 – 1, _ , _ ) = (7, 24, 25).

Alternative solution: Multiplying both sides of the given equation by 3/2, we have

(3/2)(2/3)X = (3/2)(7, 24, 25), X = (21/2 + 3/2 - 1, __ , __ ),

or X = (11, 60, 61).

2. 2X – (1, 0, 1) = ( - 5, 12, 13). Adding (1, 0, 1) to both sides of the equation, we get

2X = ( - 5, 12, 13) + (1, 0, 1) = ( - 5 + 1 + 1, __ , __ ) = ( - 3, 4, 5),

or 2X = ( - 3, 4, 5). Multiplying both sides by 1/2, we get

X = (1/2)( - 3, 4, 5) = ( - 3/2 – 1/2 + 1, __ , __ ), or X = ( - 1, 0, 1).

Check: 2(-1, 0, 1) – (1, 0, 1) = (2(-1) – (|2| - 1), _ , _ ) – (1, 0, 1) = (- 3, _ , _ ) – (1, 0, 1)

= ( - 3 – 1 – 1, _ , _ ) = ( - 5, 12, 13).

3. 5X² + (7, 24, 25) = (57, 1624, 1625). Subtracting (7, 24, 25) from both sides of the equation, we get

5x² = (57, 1624, 1625) – (7, 24, 25) = (57 – 7 – 1, __ , __ ),

5X² = (49, 1200, 1201). Multiplying both sides by 1/5, we get

X² = (1/5)(49, 1200, 1201) = (49/5 + 1/5 – 1, __, __ ), or X² = (9, 40, 41).

Taking the square root of both sides of this equation by Theorem 36 of Part 4, we find

_________ __ _______ _______ _______ _______

X = Ö (9, 40, 41) = ( Ö9, (Ö 41 + 40 - Ö 41 – 40 )/2, (Ö 41 + 40 + Ö 41 – 40 )/2 ) .

= (3, (9 – 1)/2, (9 + 1)/2) = (3, 4, 5), so X = (3, 4, 5).

Check: 5(3,4,5)² + (7, 24, 25) = 5(9, 40, 41) + (7, 24, 25) = (5(9) + 4, _, _ ) + (7, 24, 25)

= (49, _, _ ) + (7, 24, 25) = (49 + 7 + 1, _ , _ ) = (57, 1624, 1625).

b) Scalar Quadratic Equations X² + kX + A = O

Let’s now solve general quadratic Pythagorean triple equations. We shall begin with a scalar quadratic equation of the form X² + kX + A = O, where X = (x, y, z), k is an integer, A = (a, b, c) is a complete triple, and O = (0, 0, 0) is the zero triple.

Substituting these values of X, k, and A into the quadratic equation, we get

(x, y, z)² + k(x, y, z) + (a, b, c) = (0, 0, 0).

Now, if x, k, and a are all positive, then the left-hand side of this equation will yield a positive triple and thus can never equal the zero triple on the right. Accordingly, at least one of the x, k, or a must be negative for the quadratic equation to have any solutions.

A is a positive triple. ( a > 0, kx < 0 )

Let’s assume that a is positive and that x and k are of opposite signs, so kx < 0. Then, we have

X² + kX + (a, b, c) = (0, 0, 0),

or (x, y, z)² + k(x, y, z) + (a, b, c) = (0, 0, 0),

and (x², __ , __ ) + (kx - ( |k| - 1), __ , __ ) + (a, b, c) = (0, 0, 0).

Adding the first and third triples and simplifying the second triple in this last equation, we have

(x² + a + 1, __ , __ ) + (kx - |k| + 1, __ , __ ) = (0, 0, 0),

and using rule 3a from the Addition of Complete Triples in section 1, we get

(x² + a + 1) + ( kx - |k| + 1) = 0,

or x² + kx + (a - |k| + 2) = 0, (1)

which is the auxiliary equation of X² + kX + A = O. The solutions to (1) give us the a-value(s) of X.

To solve (1), we must know the values of k and a, but these are easy to find, since k is the scalar multiple of X in

X² + kX + A = O, and a is the a-value of the triple A .

Example 1. Solve X² + 7X + (15, 112, 113) = (0, 0, 0). (2)

For this equation, k = 7 and a = 15, so the auxiliary equation (1) associated with this quadratic equation is

x² + kx + ( a - |k| + 2) = 0,

or x² + 7x + (15 - |7| + 2) = 0,

which simplifies to x² + 7x + 10 = (x + 2)(x + 5) = 0. So x = - 2 or x = - 5.

Note that since kx < 0 and k = 7 is positive, then the x-values will be negative. Furthermore, since these values of x are the a-values of the solution triple X, they must be odd, and we thus have only one solution; x = - 5. Our solution triple is then

X = ( - 5, 12, 13)

To check this, we simply substitute X into the left-hand side of our original equation (2). Thus,

( - 5, 12, 13)² + 7( - 5, 12, 13) + (15, 112, 113)

= (25, 312, 313) + ( - 41, 840, 841) + (15, 112, 113)

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

Example 2. Solve X² – 8X + (21,220,221) = (0, 0, 0). Here k = - 8 and a = 21. Since k is negative, the x’s will be positive. The auxiliary equation x² + kx + (a - |k| + 2) = 0 becomes x² – 8x + (21 – | - 8 | + 2) = 0,

which simplifies to x² – 8x + 15 = (x – 3)(x – 5) = 0. So, x = 3 and x = 5 are the a-values of X. Since both solutions are odd, we get two solutions to our equation;

X = (3, 4, 5), and X = (5, 12, 13).

Now, as shown in Examples 1 and 2, we sometimes get one and sometimes two solutions for a scalar quadratic equation depending on whether x² + kx + (a – |k| + 2) = 0 has one or two odd solutions. However, if

x² + kx + (a – |k| + 2) = 0 had two even solutions, then the scalar quadratic equation would have no solutions.

But this is not possible. Here’s why.

Suppose that e₁ and e₂ are two even integral solutions of x² + kx + (a – |k| + 2) = 0. Then,

x² + kx + (a – |k| + 2) = (x – e₁)(x – e₂) = x² – (e₁ + e₂)x + e₁e₂.

Equating coefficients, we see that k = - (e₁ + e₂), and a – |k| + 2 = e₁e₂. So,

a = e₁e₂ + |k| – 2, and substituting for k,

we get a = e₁e₂ + | - (e₁ + e₂)| – 2,

which is a sum of even integers. Thus, a is even. But this is not possible since a is the a-value of a complete triple and thus odd. So, x² + kx + (a – |k| + 2) = 0 CANNOT have two even integral solutions; at least one of them must be odd. Therefore, X² + kX + A = O will always have at least one solution. This gives us

Solving X² + kX + A = O for X if A is positive.

If A = (a, b, c) is a positive complete triple, k is an integer, and kx < 0, then any odd integral

value of x which is a solution of x² + kx + (a – |k| + 2) = 0 will be the a-value of a solution triple

X = (x_,y, z) of X² + kX + A = O, where y = (x² – 1)/2, and z = y + 1.

Furthermore, if x² + kx + (a – |k| + 2) = 0 has integral solutions, at least one of them must be

odd, and thus X² – kX + A = O will have at least one solution.

_______________________________________________________________________________ (3)

Example 3. Solve 3X² – 48X + (233, 27144, 27145) = (0, 0, 0). Multiplying both sides of the equation by 1/3, we get

X² – 16X + (233/3 + 1/3 – 1, __, __ ) = (0, 0, 0) ,

or X² – 16X + (77, _ , _ ) = (0, 0, 0).

The auxiliary equation is then x² – 16x + (77 – |-16| + 2) = x² – 16x + 63 = (x – 7)(x – 9) = 0.

So, x = 7 and x = 9, giving us X = (7, 24, 25) and X = (9, 40, 41).

Check: For (7, 24, 25), we have 3(7, 24, 25)² – 48(7, 24, 25) + (233, 27144, 27145)

= 3(49, _ , _ ) – (48(7) + 47, _ , _ ) + (233, _ , _ ) = (3(49) + 2, _ , _ ) – (383, _ , _ ) + (233, _ , _ )

= (149, _ , _ ) – (383, _ , _ ) + (233, _ , _ ) = (383, _ , _ ) – (383, _ , _ ) = (0, 0, 0). Ö

Check: For (9, 40, 41), we have 3(9, 40, 41)² – 48(9, 40, 41) + (233, 27144, 27145)

= 3(81, _ , _ ) – (48(9) + 47, _ , _ ) + (233, _ , _ ) = (3(81) + 2, _ , _ ) – (479, _ , _ ) + (233, _ , _ )

= (245, _ , _ ) – (479, _ , _ ) + (233, _ , _ ) = (479, _ , _ ) – (479, _ , _ ) = (0, 0, 0). Ö

The General Method

So far, in solving X² + kX + A = O, we have considered the cases where k < 0 and x > 0 , or k > 0 and x < 0,

(kx < 0), and A is a positive triple. We could now consider the cases where A is a negative triple, or where

k and x are both negative and A is either a positive or negative triple, but this would be tedious. It is much easier to use a general method to solve scalar quadratic equations. This method consists of simply replacing X by

(x, y, z) in the original equation to be solved and then solving the resulting auxiliary equation by using the rules for addition, subtraction, and multiplication of complete Pythagorean triples as given above. Here are four examples of this method, one for each possible sign of k or a.

Example 6a): Solve X² + 4X + (5, 12, 13) = (0, 0, 0). k = 4, a = 5

(k > 0 , a > 0)

(x, y, z)² + 4(x, y, z) + ( 5, 12, 13) = (0, 0, 0).

If x > 0, we get (x², _ , _ ) + ( 4x + 3, _ , _ ) + ( 5, _, _ ) = O

(x²+ 4x + 3 + 1, _ , _ ) + ( 5, _ , _ ) = O.

By rule 3a) for addition, the sum of the two a-values of these triples must equal zero, so (x² + 4x + 4) + 5 = x² + 4x + 9 = 0. Since this auxiliary equation cannot be factored in integers, we get no solution for x > 0.

If x < 0, we get (x², _ , _ ) + ( 4x – (|4| - 1), _ , _ ) + ( 5, _ , _ ) = O. Adding the first and

third triples gives us (x² + 5 + 1, _ , _ ) + (4x - 3, _ , _ ) = O.

Again, by rule 3a), the sum of the two a-values of these triples must equal zero, so

(x² + 5 + 1) + (4x – 3) = x² + 4x + 3 = (x + 1)(x + 3) = 0. Thus, x = - 1 or x = - 3, and we have

X = ( - 1, 0, 1) and X = ( - 3, 4, 5).

Example 6b): Solve X² – 4X + (5, 12, 13) = (0, 0, 0). k = - 4, a = 5

(k < 0, a > 0)

(x, y, z)² – 4(x, y, z) + (5, 12, 13) = (0, 0, 0).

If x > 0, we get (x², _ , _ ) + (- 4x - 3, _ , _ ) + (5, _, _ ) = O. Adding the first and third

triples gives us (x²+ 5 + 1, _ , _ ) + ( - 4x - 3, _ , _ ) = O,

or X² + 5 + 1 + ( - 4x - 3) = x² - 4x + 3 = (x – 1)(x – 3) = 0. Thus, x = 1 or x = 3

and we have X = (1, 0, 1) and X = (3, 4, 5).

If x < 0, then - 4X will be a positive triple, and X² – 4X + (5, 12, 13) will be a sum of three positive triples and thus cannot equal the zero triple O. So, there is no solution for x < 0.

Example 6c): Solve X² + 4X – (25, 312, 313) = (0, 0, 0). k = 4, a = - 25

(k > 0, a < 0)

(x, y, z)² + 4(x, y, z) – (25, 312, 313) = (0, 0, 0).

If x > 0, we get (x², _ , _ ) + ( 4x + 3, _ , _ ) + ( - 25, _ , _ ) = O. Adding the first and

second triples gives us (x²+ 4x + 3 + 1, _ , _ ) + ( - 25, _ , _ ) = O,

or X² + 4x + 4 + ( - 25) = x² + 4x - 21 = (x – 3)(x + 7) = 0. Thus, x = 3 or x = - 7.

Since x must be positive, we discard x = - 7 to get X = (3, 4, 5).

If x < 0, we get (x², _ , _ ) + ( 4x - 3, _ , _ ) + ( - 25, _, _ ) = O.

Since 4x - 3 is negative (x < 0), adding the second and third triples gives us

(x², _ , _ ) + (4x – 3 – 25 – 1 , _ , _ ) = O, or x² + 4x – 29 = 0.

Since this auxiliary equation cannot be factored in integers, we get no solution for x < 0.

Example 6d): Solve X² – 4X – (25, 312, 313) = (0, 0, 0). k = - 4, a = - 25

(k < 0, a < 0)

(x, y, z)² – 4(x, y, z) + ( - 25, 312, 313) = (0, 0, 0).

If x > 0, we get (x², _ , _ ) + ( - 4x - 3, _ , _ ) + ( - 25, _, _ ) = O

(x², _ , _ ) + ( - 4x – 3 – 25 - 1, _ , _ ) = O, or x² – 4x – 29 = 0.

This auxiliary equation cannot be factored in integers, so we get no solution for x > 0.

If x < 0, we get (x², _ , _ ) – ( 4x - 3, _ , _ ) + ( - 25, _, _ ) = O

(x², _ , _ ) + ( - 4x + 3, _ , _ ) + ( - 25, _, _ ) = O

Since - 4x + 3 is positive (x < 0), adding the first two triples gives us

(x² - 4x + 3 + 1, _ , _ ) + ( - 25, _ , _ ) = O,

or x² – 4x + 4 – 25 = x² – 4x – 21 = (x – 7)(x + 3) = 0. Thus, x = 7 or x = - 3.

Since x must be negative, we discard x = 7 and use x = - 3 to get X = ( - 3, 4, 5).

Of course we can check any of these solutions. Let’s check the solution for 6d). Since X = ( - 3, 4, 5),

then X² – 4X – (25, 312, 313) = ( - 3, 4, 5)² – 4( - 3, 4, 5) – (25, 312, 313)

= (9, 40, 41) + (15, _ , _ ) – (25, 312, 313) = (0, 0, 0). Ö

c) Triple Quadratic Equations X² + AX + B = O

We shall now apply our general method to solving those Pythagorean triple equations in which the coefficient of X is a Pythagorean triple rather than a scalar; namely, an equation of the form X² + AX + B = O, where A and B are complete triples.

Example 7. Solve X² + (7, 24, 25)X – (17, 144, 145) = O. Letting X = (x, y, z), we have

(x, y, z)² + (7, 24, 25)(x, y, z) – (17, 144, 145) = O,

If x > 0, we get (x², _ , _ ) + (7x, _ , _ ) + ( - 17, _ , _ ) = O,

(x²+ 7x + 1, _ , _ ) + ( - 17, _ , _ ) = O,

and by rule 3a), we have x² + 7x + 1 + ( - 17) = x² + 7x – 16 = 0, which is not factorable in integers, and thus there is no solution for x > 0.

If x < 0, we get (x², _ , _ ) + (7x, _ , _ ) + ( - 17, _ , _ ) = O, and adding the second and third triples, since both 7x and – 17 are negative, we have

(x², _ , _ ) + (7x - 17 - 1 , _ , _ ) = O.

By rule 3a), we get x² + (7x – 17 – 1) = x² + 7x – 18 = (x + 9)(x – 2) = 0, so x = - 9 or x = 2. Since x must be negative, we discard x = 2 to get X = ( - 9, 40, 41).

Check: ( - 9, 40, 41)² + (7, 24, 25) ( - 9, 40, 41) - (17, 144, 145)

= (81, _ , _ ) + ( - 63, 1984, 1985) + ( - 17, 144, 145) = (81, 3280, 3281) + ( - 63 – 17 – 1, _ , _ )

= (81, 3280, 3281) + ( - 81, 3280, 3281) = (0, 0, 0). Ö

Now, if X = (x, y, z), A = (a₁, b₁, c₁), and B = (a₂, b₂, c₂), then X² + AX + B = O becomes

(x, y, z)² + (a₁, b₁, c₁) (x, y, z) + (a₂, b₂, c₂) = (0, 0, 0),

or (x², _ , _ ) + (a₁x, _ , _ ) + (a₂, b₂, c₂) = (0, 0, 0). (1)

Since we do not know whether a₂ is positive or negative, when we add the first and third triples together, we shall get either (x² + a₂ + 1, _ , _ ) or (x² + a₂ – 1, _ , _ ). Thus, (1) becomes

(x²+ a₂ ± 1, _ , _ ) + (a₁x, _ , _ ) = (0, 0, 0), and using rule 3a),

we get x² + a₁x + a₂ ± 1 = 0. (2)

Now, if (2) is factorable, we have x² + a₁x + a₂ ± 1 = (x – r₁)(x – r₂) for some integers r₁ and r₂. Multiplying out the right-hand side of the last equation, we get

x² + a₁x + a₂ ± 1 = x² – r₁x – r₂x + r₁r₂ .

Equating the coefficients of x gives us a₁ = – r₁ – r₂ = - (r₁ + r₂). Now a₁ is odd, so r₁ and r₂ must be of opposite parity (one odd and one even). So, only one root x of equation (2) will be odd, and thus X² + AX + B = O will have only one solution X. Thus, we have

Solving X² + AX + B = O

To solve X² + AX + B = O for X, where X = (x, y, z), A = (a₁, b₁, c₁), and B = (a₂, b₂, c₂) are all complete triples, simply replace each triple by its components to get

(x, y, z)² + (a₁, b₁, c₁)(x, y, z) + (a₂, b₂, c₂) = (0, 0, 0)

(x², _ , _ ) + (a₁x, _ , _ ) + (a₂, _ , _ ) = (0, 0, 0).

Then, considering the cases where x > 0 and x < 0, and using the appropriate rules for the addition and subtraction of complete Pythagorean triples, solve the resulting quadratic equation for integral x, if possible. Then, since X is complete, y = (x² – 1)/2, and z = y + 1, and there will be at most one solution.

_____________________________________________________________________________________________

3. Complete Pythagorean Triples and Vector Spaces

It is of some interest to note that the set V of all complete Pythagorean triples together with the zero triple O and the operations of addition and of scalar multiplication for integers as defined above forms a vector space. Listed below are the ten axioms of a vector space and the proofs that V satisfies them. In 7, 8, and 9 below, only one or two of the possible cases are proved. The rest are easily proved in a similar manner.

Let A , B, and C be either complete Pythagorean triples or the zero triple O. Then,

1. If A and B are in V, then A + B is in V. Parts b) and c) of the definition of addition above ensure us that if

A and B are complete triples or the zero triple O, then A + B is either a complete triple (c = b+1) or O. Thus, A + B is in V.

2. A + (B + C) = (A + B) + C. In finding the a-values for the sums A + (B + C) and (A + B) + C, we are simply adding integers, and the set of integers satisfies the associative axiom. Thus, A + (B + C) = (A + B) + C.

3. A + B = B + A. Since rules 1, 2, 4, and 5 for addition all involve a₁ + a₂, which equals a₂ + a₁, then

A + B = B + A.

4. There is a triple O in V such that A + O = O + A = A for any triple in V. The triple O is (0, 0, 0), which is in V.

Rule 3 for addition shows that A + O = A for any complete triple A and that O + O = O. Axiom 3 above shows that A + O = O + A.

5. If A is in V, then there is a triple - A in V, such that A + ( - A) = O. If A = (a, b, c) is a complete triple, then

let - A be ( - a, b, c), and Rule 3a) for addition shows that A + ( - A) = (a, b, c) + ( - a, b, c) = (0, 0, 0) = O.

6. If k is an integer, then kA is in V. By the rules of scalar multiplication, if k ¹ 0, kA is a complete triple (c = b + 1) and thus in V, and if k = 0, kA = O, which is in V. Thus, kA is in V.

7. If k and m are integers, then k(mA) = (km)A).

a) If k, m, a > 0, then k(mA) = k [m(a, b, c)] = k (ma + (m – 1), __ , __ )

= ( k(ma + m – 1) + (k – 1), __ , __ ) = (kma + km – 1, __ , __ ), (1)

and (km)A = (km)(a, b, c) = ( (km)a + (km – 1), __ , __ )

= ( kma + km – 1, __ , __ ). (2)

Since (1) and (2) are identical, we have (km)A = k(mA) for k, m, a > 0.

b) If k, a < 0, m > 0, then k(mA) = k [m(a, b, c)] = k (ma – (m – 1) , __ , __ )

We know that k < 0, but to carry out the scalar multiplication above, we must also know the sign of

ma – (m – 1). Now, ma is negative, since a is negative and m is positive. Also, m ³ 1, since m is a

positive integer, so m – 1 ³ 0. Thus, ma, a negative integer, is less than m – 1, a non-negative

integer, or ma < m –1. Subtracting m – 1 from both sides of this last inequality gives us

ma – (m – 1) < 0, so ma – (m – 1) is negative. Thus,

k (ma – (m – 1) , __ , __ ) = ( k(ma – m + 1) + (|k| – 1), __ , __ ),

and since |k| = – k if k < 0, = (kma – km + k + ( – k – 1), __ , __ )

= (kma – km – 1, __ , __ ). (3)

Since km < 0, (km)A = (km)(a, b, c) = ( (km)a + (|km| – 1), __ , __ )

= ( kma + (– km – 1, __ , __ ) = (kma – km – 1, _ , _ ). (4)

Now (3) and (4) are identical, so we have (km)A = k(mA) for k, a < 0, m > 0.

8. If k and m are integers, then (k + m)A = kA + mA.

a) If k, m, a > 0, then (k + m)A = (k + m)(a, b, c) = ( (k+m)a + (k+m) – 1, _ , _ )

= (ka + ma + k + m – 1, _ , _ ), (5)

and kA + mA = k(a, b, c) + m(a, b, c) = (ka + (k – 1), _ , _ ) + (ma + (m – 1), _ , _ )

= (ka + (k – 1) + ma + (m – 1) + 1, _ , _ ) = (ka + ma + k + m – 1, _ , _ ) (6)

Since (5) and (6) are identical, we have (k + m)A = kA + mA for k, m, a > 0.

. b) If k, m, a < 0, then (k + m)A = (k + m)(a, b, c) = ( (k+m)a + (|k+m|– 1), _ , _ )

= (ka + ma + ( - (k + m) – 1, _ , _ ) = (ka + ma – k – m – 1, _ , _ ) (7)

and kA + mA = k(a, b, c) + m(a, b, c)

= (ka + (|k| – 1), _ , _ ) + (ma + (|m| – 1), _ , _ )

= (ka + ( - k – 1) + ma + ( - m – 1) + 1, _ , _ ) = (ka + ma – k – m – 1, _ , _ ) (8)

Since (7) and (8) are identical, we have (k + m)A = kA + mA for k, m, a < 0.

9. If k is an integer, then k(A + B) = kA + kB.

a) If k, a, d > 0, then k(A + B) = k[(a, b, c) + (d, e, f)] = k(a + d + 1, _ , _ )

= ( k(a + d + 1) + (k – 1), _ , _ ) = (ka + kd +2k – 1, _ , _ ) (9)

and kA + kB = k(a, b, c) + k(d, e, f) = (ka + (k – 1), _ , _ ) + (kd + (k – 1), _ , _ )

= (ka + (k – 1) + kd + (k – 1) + 1, _ , _ ) = (ka + kd + 2k + 1, _ , _ ) (10)

Since (9) and (10) are identical, we have k(A + B) = kA + kB for k, a, d > 0.

b ) If k, d > 0, a < 0, |a| < d, then k(A + B) = k[(a, b, c) + (d, e, f) ] = k( a + d – 1, _ , _ ).

To carry out this multiplication, we must know whether a + d – 1 is positive or negative. Since |a| < d,

then - d < a < d, and adding d – 1 to this inequality gives us

- d + d – 1 < a + d – 1 < d + d – 1,

or -1 < a + d – 1 < 2d – 1,

showing us that a + d – 1 is greater than - 1 and so not negative. Thus,

k(A + B) = k(a + d – 1, _ , _ ) = ( k(a + d – 1) + (k – 1), _ , _ ) = (ka + kd – 1, _ , _ ). (11)

Now, kA + kB = k(a, b, c) + k(d, e, f) = (ka – (k – 1), _ , _ ) + (kd + (k – 1), _ , _ ). (12)

Since the a-values of the first and second triples are negative and positive, respectively, in order to

find their sum, we must know whether |ka – (k-1)| is greater or less than kd + (k-1). We have

- d < a < d

- kd < ka < kd

- kd – (k-1) < ka – (k-1) < kd – (k-1)

- kd – (k-1) < ka – (k-1) < kd + (k-1)

- [kd + (k-1)] < ka – (k-1) < kd + (k-1)

|ka – (k-1)| < kd + (k-1).

Thus, the absolute value of the negative a-value of the first triple is less than the positive

a-value of the second triple, and we get from (12),

kA + kB = (ka – (k – 1), _ , _ ) + (kd + (k – 1), _ , _ )

= (ka + kd – 1, _ , _ ). (13)

Since (11) and (13) are identical, k(A + B) = kA + kB for k, d > 0, a < 0, |a| < d

10. 1A = A. From the table for scalar multiplication on page 9, we have that

If a > 0, then 1A = 1(a, b, c) = (1a + (|1| - 1), _ , _ ) = (a, b, c) = A.

If a < 0, then 1A = 1(a, b, c) = (1a – (|1| - 1), _ , _ ) = (a, b, c) = A.

If a = 0, then 1A = 1(0,0,0) = (0, 0, 0) = A.

Thus, 1A = A.

Thus, the ten axioms are satisfied, and V is a vector space.

4. Some Notable Patterns

The 3 - 4 - 5 Pattern

In looking at a table of primitive Pythagorean triples, one might notice a rather interesting pattern in those complete triples whose a-values all end in the digit 3. Here is a list.

a b c m n

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

3 4 5 2 1

13 84 85 7 6

23 264 265 12 11

33 544 545 17 16

43 924 925 22 21

Note that the last digit of the a-values is always 3, the last digit of the b-values is always 4, and the last digit of the c-values is always 5. Thus, the 3, 4, 5 pattern is preserved in each of these triples whose a-values end in 3. Now, why is this?

Well, first we notice that the m-values start at 2 and then increase by 5 as we proceed from one triple to the next in the list; thus, m = 2 + 5k for k = 0, 1, 2, . . . , and n is always one less than m, so that n = 1 + 5k.

Now, b = 2mn, so we get

b = 2mn = 2(2 + 5k)(1 + 5k) = 2(25k² + 15k + 2) = 50k² + 30k + 4 = 10(5k² + 3k) + 4.

Now 10 times any integer always yields an integer whose last digit is zero, so 10(5k² + 3k) will be an integer whose last digit is zero. Adding 4 to this integer then gives us an integer whose last digit is 4. Thus, the last digit of b will always be 4. Since all of these triples are complete, then c is always one more unit than b, so the last digit of c will always be 5. This gives us

The 3 - 4 - 5 Pattern

If the a-value of a complete Pythagorean triple ends in a 3,

then its b- and c-values will end in a 4 and a 5, respectively.

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It is interesting to note that those digits preceding the 4’s or 5’s in the table above form two identical series,

namely, 8, 26, 54, 92, . . .

The 5 -12 -13 Pattern

Look at the following list of Pythagorean triples.

a b c m n

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

5 12 13 3 2

15 112 113 8 7

25 312 313 13 12

35 612 613 18 17

45 1012 1013 23 22

55 1512 1513 28 27

65 2112 2113 33 32

75 2812 2813 38 37

Note that the last digit of the a-values is always 5, the last two digits of the b-values are always 12, and the last two digits of the c-values are always 13. Thus, the 5, 12, 13 pattern is preserved in each of these triples whose a-values end in 5. Now, why is this?

Well, first we notice that the m-values start at 3 and then increase by 5 as we proceed from one triple to the next in the list; thus, m = 3 + 5k for k = 0, 1, 2, . . . , and n is always one less than m, so that n = 2 + 5k.

Now b = 2mn, so we get

b = 2mn = 2(3 + 5k)(2 + 5k) = 2(25k² + 25k + 6) = 50k² + 50k + 12 = 50k(k+1) + 12.

Now either k is even or odd.

If k is even, then k = 2r for some integer r, and we have

b = 50k(k+1) + 12 = 100r(2r+1) + 12.

Since 100 times any integer always yields an integer whose last two digits are zero, then 100r(2r+1) will be an integer whose last two digits are zero. Adding 12 to this integer then gives us an integer whose last two digits are 12. Thus, the last two digits of b will always be 12 if k is even.

If k is odd, then k = 2r + 1 for some integer r, and we have

b = 50k(k+1) + 12 = 50(2r + 1)(2r + 2) + 12 = 100(2r + 1)(r + 1) + 12.

As before, 100 times any integer always yields an integer whose last two digits are zero, so 100(2r + 1)(r + 1) will be an integer whose last two digits are zero. Adding 12 to this integer then gives us an integer whose last two digits are 12. Thus, the last two digits of b will again always be 12 if k is odd.

Thus, in either case, the last two digits of b will be 12. Since all of these triples are complete, then c is always one more unit than b, so the last two digits of c will always be 13. This gives us

The 5 -12 -13 Pattern

If the a-value of a complete Pythagorean triple ends in a 5,

then its b- and c-values will end in a 12 and 13, respectively.

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Example 1. If the a-value of a complete triple is 135, determine if its b - and c - values end in 12 and 13, respectively. Since the triple is complete, then b = (a² – 1)/2 = (135² – 1)/2 = 9112, which ends in 12, and

c = b + 1 = 9113, which ends in 13.

It is interesting to note that those digits preceding the 12’s or 13’s in the table above form the well-known sequence of triangular numbers 0, 1, 3, 6, 10, 15, 21, 28. How appropriate!

5. Generating Pythagorean Triples from the Fibonacci Sequence

The Fibonacci sequence is made up of positive integers whose first two terms are a₁ = 1 and a₂ = 1 and whose succeeding terms are defined by a_n+2 = a_n + a_n+1 for n ³ 1. Thus, we get

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . ,

as the Fibonacci sequence. Let’s look at its first four terms: 1, 1, 2, 3, and see if we can’t generate the Pythagorean triple (3, 4, 5) from them.

First, If we multiply the first and last terms together, we get 1(3) = 3, which is the a-value of (3, 4, 5). So far, so good. Next, If we multiply the two middle terms together, we get 1(2) = 2, which when doubled gives us 4, the

b-value of (3, 4, 5). Not bad! But how can we get the c-value of 5? Well, since c = m² + n², a sum of two squares, let’s find the sum of the squares of the two middle terms, 1² + 2², which gives us 5, the c -value of (3, 4, 5), and we’re done!

We’ve generated a primitive Pythagorean triple from a 4-term Fibonacci sequence by using

a) the product of the first and last terms (product of the extremes) as the a-value,

b) twice the product of the middle terms (twice the product of the means) as the b-value, and

c) the sum of the squares of the middle terms (sum of the squares of the means) as the c-value.

Will this procedure yield other primitive triples? Let’s see.

Consider the second through the fifth terms of the Fibonacci sequence 1, 2, 3, 5.

a) The product of the extremes is 1(5) = 5, our a-value.

b) Twice the product of the means is 2(2)(3) = 12, our b-value, and

c) the sum of the squares of the means is 2² + 3² = 13, our c-value,

and we have generated the primitive triple (5, 12, 13). Nice!

Alas, when we apply this procedure to the third through the sixth terms of the Fibonacci sequence

2, 3, 5, 8,

we do NOT get a primitive Pythagorean triple. Here’s what happens.

The product of the extremes is 2(8) = 16 for the a-value; twice the product of the means is 2(3)(5) = 30 for the

b-value, and the sum of the squares of the means is 3² + 5² = 34 for the c-value, giving us (16, 30, 34), which is not even primitive because each of its members has a factor of 2. Removing this factor of 2, we get (8, 15, 17), a primitive Pythagorean triple, although its a - and b - values are interchanged. Reversing these values, we get

(15, 8, 17), an inverted primitive Pythagorean triple. Now, what’s happening here?

Well, since the first term, 2, is even, the product of the extremes is even, giving us an even rather than an odd integer for a. Furthermore, twice the product of the means gives us an even integer for b, and thus a and b have a common factor of 2, making the resulting triple non-primitive, and requiring us to divide by 2.

So whenever the first term in a 4-term Fibonacci sequence is even, we shall modify the procedure above by

a) taking half the product of the extremes,

b) taking half of twice the product of the means,

c) taking half the sum of the squares, and then

d) interchanging the values of a and b.

Then, the a-value will be half of twice the product of the means (or simply the product of the means), the b-value will be half the product of the extremes, and the c-value will be half the sum of the squares of the means.

Applying this procedure to 2, 3, 5, 8 we get the

a) product of the means is 3(5) = 15 for the a-value,

b) half the product of the extremes is (1/2)(2)(8) = 8 for the b-value, and

c) half the sum of the squares of the means is (1/2)(3² + 5²) = 17 for the c-value,

generating the inverted primitive triple (15, 8, 17). Let’s look at some examples.

4 - term Fibonacci

sequence Triple Procedure

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

1, 1, 2, 3 a = 1(3) = 3 (product of the extremes)

b = 2(1)(2) = 4 (twice the product of the means)

(1^st term odd) c = 1² + 2² = 5 (sum of the squares of the means)

1, 2, 3, 5 a = 1(5) = 5

b = 2(2)(3) = 12 (same as above)

(1^st term odd) c = 2² + 3² = 13

2, 3, 5, 8 a = 3(5) = 15 (product of the means)

b = (1/2)2(8) = 8 (half the product of the extremes)

(1^st term even) c = (1/2)(3² + 5²) = 17 (half the sum of the squares of

the means)

3, 5, 8, 13 a = 3(13) = 39 (product of the extremes)

b = 2(5)(8) = 80 (twice the product of the means)

(1^st term odd) c = 5² + 8² = 89 (sum of the squares of the means)

5, 8, 13, 21 a = 5(21) = 105

b = 2(8)(13) = 208 (same as above)

(1^st term odd) c = 8² + 13² = 233

8, 13, 21, 34 a = 13(21) = 273 (product of the means)

b = (1/2)8(34) = 136 (half the product of the extremes)

(1^st term even) c = (1/2)(13² + 21²) = 305 (half the sum of the squares of

the means)

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Looking at these examples, it seems that the following is true.

Generating Triples from Fibonacci Sequences

For the four consecutive terms a_n, a_n+1, a_n+2, and a_n+3 of a Fibonacci sequence,

(a, b, c) will be a primitive Pythagorean triple if:

1) a_n is odd, and a = a_na_n+3, b = 2a_n+1a_n+2, and c = a_n+1² + a_n+2².

In this case, (a, b, c) will be regular (a < b).

2) a_n is even, and a = a_n+1a_n+2, b = (1/2)a_na_n+3, and c = (1/2) (a_n+1² + a_n+2²).

In this case, (a, b, c) will be inverted (a > b).

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Let’s prove that (a, b, c) is a Pythagorean triple for case 1) above.

From the definition of a Fibonacci sequence, we know that

a_n+2 = a_n + a_n+1 (1a)

and that a_n+3 = a_n+1 + a_n+2 = a_n+1 + (a_n + a_n+1) = a_n + 2a_n+1 (1b)

Using case 1) and (1b), we get

a = a_na_n+3 = a_n(a_n + 2a_n+1) = a_n² + 2a_na_n+1 (2a)

Using case 1) and (1a), we get

b = 2a_n+1a_n+2 = 2a_n+1(a_n + a_n+1) = 2a_na_n+1 + 2a_n+1², (2b)

and c = a_n+1² + a_n+2² = a_n+1² + (a_n + a_n+1)² = a_n² + 2a_na_n+1 + 2a_n+1². (2c)

To show that (a, b, c) is a Pythagorean triple, we must show that a² + b² = c².

Using (2a) and (2b), we have

a² + b² = ( a_n² + 2a_na_n+1 )² + (2a_na_n+1 + 2a_n+1² )²

= a_n⁴ + 4a_n³a_n+1 + 4a_n²a_n+1² + 4a_n+1²a_n² + 8a_na_n+1³ + 4a_n+1⁴

= a_n⁴ + 4a_n³a_n+1 + 8a_n²a_n+1² + 8a_na_n+1³ + 4a_n+1⁴, (3)

and from (2c), we have

c² = [a_n² + 2a_na_n+1 + 2a_n+1²]²

= a_n⁴ + 4a_n²a_n+1² + 4a_n+1⁴ + 2(2a_n³a_n+1) + 2(2a_n²a_n+1²) + 2(4a_na_n+1³)

= a_n⁴ + 4a_n²a_n+1² + 4a_n+1⁴ + 4a_n³a_n+1 + 4a_n²a_n+1² + 8a_na_n+1³

= a_n⁴ + 4a_n³a_n+1 + 8a_n²a_n+1² + 8a_na_n+1³ + 4a_n+1⁴. (4)

Since (3) and (4) are identical, we have proved that a² + b² = c², and thus (a, b, c), as given by case 1), is a Pythagorean triple.

Now, is (a, b, c), as given by case 1), a primitive triple; that is, do a, b, and c have no common factors? Let’s see.

Subtracting (2b) from (2c), and then (2a) from (2c), we get

c – b = a_n² and c – a = 2a_n+1².

Now, suppose a, b, and c have a common even factor of k. Then k is a factor of c – b, and since 2 is a factor of k, then 2 is a common factor of c – b and thus a common factor of a_n², and so a common factor of a_n. Thus, a_n is even, but this is false since in case 1) above, a_n is odd. Thus, a, b, and c do not have any even common factors.

Suppose a, b, and c have a common odd factor k, and k₁ is a prime factor of k. Then, k₁ will be odd since there can be no even factors of the odd factor k, and k₁ will be a common factor of c – b and c – a, and so a common factor of a_n² and a_n+1². Since k₁ is prime, it is thus a common factor of a_n and a_n+1. But this is false, since a common property of a Fibonacci sequence is that a_n and a_n+1 have NO common factors. ( Since a_n+1 = a_{n -1} + a_n, we can solve for a_{n – 1} to get a_{n – 1} = a_{n +1} – a_n . Then, If a_n and a_{n +1} had a common factor of f, then a_n -1 would also have a common factor of f for any positive integer n > 2. But this says that if n = 2, then a₃ = 2, a₂ = 1, and a₁ = 1 would all have a common factor (other than 1), which they don’t. So, no two consecutive terms of a Fibonacci sequence have a common factor.) Thus, a, b, and c do not have any odd common factors.

Thus, a, b, and c have no common factors, and (a, b, c), as given by case 1), is a primitive Pythagorean triple.

Finally, is (a, b, c) a regular triple (a < b)? Since we have a Fibonacci series,

a_n < a_n+1. Squaring both sides of this inequality, we get

a_n² < a_n+1², and adding 2a_na_n+1 to both sides gives us

a_n² + 2a_na_n+1 < a_n+1² + 2a_na_n+1. (5)

From (2a), we see that the left-hand side of (5) is a, so we have

a < a_n+1² + 2a_na_n+1. (6)

Adding a_n+1² to just the right-hand side of (6), we get

a < a_n+1² + 2a_na_n+1 + a_n+1²,

or a < 2a_na_n+1 + 2a_n+1² = b from (2b).

Thus, we have shown that a < b, and (a, b, c), as given by case 1), is a regular triple.

A similar analysis using case 2) above will show that (a, b, c) is an inverted (a > b) primitive Pythagorean triple.

Example 1. For the four-term Fibonacci sequence 13, 21, 34, 55, determine whether the Pythagorean triple generated by this sequence will be regular or inverted, and then find the generated triple. Since the first term is odd, we shall get a regular triple. The product of the extremes gives us a = 13(55) = 715, twice the product of the means gives us b = 2(21)(34) = 1428, and the sum of the squares of the means gives us c = 21² + 34² = 1597. So the regular primitive Pythagorean triple generated by the four-term Fibonacci sequence above is (715, 1428, 1597).

6. Constructing A Triple whose c-value is c².

If (a, b, c) is a regular primitive Pythagorean triple, then a and b have no common factors, are of opposite parity, and b > a. Thus, we may generate a new primitive Pythagorean triple by using b and a for our values of m and n, and since c² = b² + a², we get

(m² – n², 2mn, m² + n²) = (b² – a², 2ba, b² + a²) = (b² – a², 2ba, c²).

If (a, b, c) is an inverted primitive Pythagorean triple, then a and b have no common factors, are of opposite parity, and a > b. Thus, we may generate a new primitive Pythagorean triple by using a and b for our values of m and n, and since c² = a² + b², we get

(m² – n², 2mn, m² + n²) = (a² – b², 2ab, a² + b² ) = (a² – b², 2ab, c²). Thus,

For any primitive Pythagorean triple (a, b, c), there exists a primitive

Pythagorean triple, (|a² – b²|, 2ab, a² + b²), whose c-value is c².

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Examples.

(3, 4, 5) is a regular triple, so using m = b = 4 and n = a = 3, we get

( 4² – 3², 2(4)(3), 4² + 3²) = (7, 24, 25),

whose c-value, 25 = 5², is the square of the c-value of the original triple (3, 4, 5).

(15, 8, 17) is an inverted triple, so using m = a = 15 and n = b = 8, we get

( 15² – 8², 2(15)(8), 15² + 8²) = (161, 240, 289),

whose c-value, 289 = 17², is the square of the c-value of the original triple (15, 8, 17).

May 17, 2009

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