Patterns in Pythagoras
3A. Even More Patterns in Right Triangles
and Pythagorean Triples
David W. Hansen
© 2008
In Parts 1 and 2, we discovered many beautiful patterns existing among the sides of primitive right triangles or the members of primitive Pythagorean triples. But apparently we are not through! There are more patterns to discover than we had previously thought, so c’mon, let’s get to it!
1. The Relative Sizes of a, b, and c
Table 1
Primitive Right Triangles and Pythagorean Triples
a b c Perimeter
m n m2 - n2 2mn m2 + n2 a+b+c
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2 1 3 4 5 12
3 2 5 12 13 30
4 1 15 8 17 40
4 3 7 24 25 56
5 2 21 20 29 70
5 4 9 40 41 90
6 1 35 12 37 84
6 5 11 60 61 132
7 2 45 28 53 126
7 4 33 56 65 154
7 6 13 84 85 182
8 1 63 16 65 144
8 3 55 48 73 176
8 5 39 80 89 208
8 7 15 112 113 240
9 2 77 36 85 198
9 4 65 72 97 234
9 8 17 144 145 306
10 1 99 20 101 220
Looking at Table 1 above, we can clearly see that both a and b are always less than c. We would expect this to be true since c represents the length of the hypotenuse of a right triangle, which is always greater than the lengths of each of its two legs. We can also show this by using the formulas from Theorem 5, which was proved in Part 1.
Here’s the theorem.
Theorem 5
Let m and n be positive integers of opposite parity with no common factors and m > n. Then,
for all primitive Pythagorean triples (a, b, c), a = m2 - n2, b = 2mn, and c = m2 + n2.
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a < c Informally, if a < c, then m2 - n2 < m2 + n2. Subtracting m2 from both sides of this inequality, we have - n2 < n2. Adding n2 to both sides of this inequality gives us 0 < 2n2, which is always true for n ¹ 0. However, in this proof we assumed that a < c, which is what we are trying to prove! Here is a correct proof.
Since n > 0, 0 < 2n2 Subtracting n2 from both sides of this
inequality gives us - n2 < n2. Adding m2 to both sides of this
inequality, we get m2 - n2 < m2 + n2,
or a < c.
b < c Informally, if b < c, then 2mn < m2 + n2 , 0 < m2 - 2mn + n2, and 0 < (m – n)2, which is always true for m ¹ n. However, as before, we assumed what we were trying to prove! Writing our statements above in reverse order, we have 0 < (m – n)2, which is always true for m ¹ n.
Expanding, we get 0 < m2 - 2mn + n2.
Adding 2mn to both sides, we get 2mn < m2 + n2,
or b < c. Thus, we have proved
Theorem 21
In any primitive right triangle or Pythagorean triple (a, b, c), both a < c, and b < c.
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2. Regular and Inverted Triples
Now, what about a and b? In some primitive Pythagorean triples, such as (3,4,5) or (5,12,13), a < b. When this occurs, we call these triples regular. In other triples, such as (15,8,17) and (77,36,85), a > b, and we call these triples inverted. Let’s see if we can find out just when a primitive Pythagorean triple is regular and when it is inverted.
Regular triples (a < b). If a < b, then we have m2 - n2 < 2mn,
or m2 - 2mn - n2 < 0.
Adding 2n2 to both sides of this inequality gives us m2 - 2mn + n2 < 2n2,
or (m - n)2 < 2n2.
Taking the positive square root of this last inequality, we get m - n < nÖ2.
Then, m < n + nÖ2 = n(1 + Ö2) ≈ 2.414n,
or m/n < 1 + Ö2 ≈ 2.414.
Thus, whenever m/n < 1 + Ö2 ≈ 2.414, our triple is regular.
Inverted triples (a > b). By reversing the direction of the inequality symbols in the proof above, we can show that whenever m/n > 1 + Ö2 ≈ 2.414, our triple is inverted.
Theorem 13a in Part 2 tells us that if a = pq, where p and q are odd positive integers (p > q) with no common factors, then m = (p + q)/2 and n = (p - q)/2. Thus,
(p+q/2 p + q
m/n = ------------ = ------------- > 1 + Ö2 for an inverted triple.
(p - q/2 p – q
Multiplying both sides of this last inequality by p – q and reading from left to right, we get
p + q > (1 + Ö2)(p - q), p + q > p - q + pÖ2 - qÖ2, 2q + qÖ2 > pÖ2,
pÖ2 < (2 + Ö2)q, p < (2 + Ö2)q /Ö2, p < ( 2 /Ö2 + Ö2 /Ö2 )q,
p < (Ö2 + 1)q, or p / q < 1 + Ö2.
Thus, whenever p / q < 1 + Ö2, the associated triple will be inverted. In a similar fashion, by reversing the direction of the inequality symbols above, we can show that whenever p / q > 1 + Ö2, then the associated triple will be regular. All of this gives us the following theorem.
Theorem 22 (Regular and inverted triples)
In a primitive Pythagorean triple P = (a, b, c), if a < b, then P is called regular.
If a > b, then P is called inverted.
a) P will be regular (a < b ) if m/n < (1+Ö2) ≈ 2.414.
b) P will be inverted (a > b) if m/n > (1+Ö2) ≈ 2.414.
If a = pq, where p and q are odd positive integers (p > q) with no common factors, then
c) P will be regular (a < b ) whenever p/q > (1+Ö2) ≈ 2.414.
d) P will be inverted (a > b) whenever p/q < (1+Ö2) ≈ 2.414.
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Example 1. Find all inverted primitive Pythagorean triples for m = 13. The possible values for m and n, together with their respective values for m/n, are shown below.
m n m/n m n m/n
----------------------------------------------------------------
13 2 6.50 13 8 1.63
13 4 3.25 13 10 1.30
13 6 2.17 13 12 1.08
Theorem 22 b) tells us that for an inverted triple, we must have m/n > 2.414, and only the first two sets of m
and n satisfy this requirement. For m = 13, n = 2, we get (165, 52, 173). For m = 13, n = 4, we get
(153,104, 185), and both triples are inverted ( a > b ).
Example 2. Find all regular primitive Pythagorean triples for a =105. The factors for a = 105 together with their respective values for p/q are shown below.
p q p/q p q p/q
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105 1 105 21 5 4.20
35 3 11.67 15 7 2.14
For a regular triple, we must have p/q > 2.414, and the first three sets of p and q satisfy this requirement. Using Theorem 13b from Part 2, we get for
p = 105, q = 1, b = (1052 – 12 )/2, c = (1052 + 12 )/2, giving us (105, 5512, 5513),
p = 35, q = 3, b = (352 – 32 )/2, c = (352 + 32 )/2, giving us (105, 608, 617),
p = 21, q = 5, b = (212 – 52 )/2, c = (212 + 52 )/2, giving us (105, 208, 233).
1. Does every value of a belong to at least one regular triple? Yes, and here’s why. Any a can be written as a = pq = a(1). Then p = a, q = 1, p/q = a/1 = a ³ 3 > 2.414. Thus, by Theorem 22 c), a is a member of a regular triple.
2. Does every value of a belong to at least one inverted triple? No. 3, 5, and 49 are examples of values of a for which no inverted triples exist, since the only way to write these values as a product of two factors p and q with no common factors is 3 = 3(1), 5 = 5(1), and 49 = 49(1), and in each of these cases, the ratio p/q is greater than 2.414. Thus, by Theorem 22 d), they cannot be members of an inverted triple. In fact, if a is a prime or a power of a prime, then a cannot be a member of an inverted triple. Let’s see why.
Let a be a power of an odd prime P. Then a = Pk, where k is a positive integer. The only way to factor a into a product of two factors p and q with no common factors is a = Pk(1), since any other factorizations, such as
Pk – 3(P3), will all have a common factor of some power of P. Thus, a = pq = Pk(1) has only one distinct prime factor; namely, P, so by Theorem 14 of Part 2, a is a member of only one triple. By Theorem 22 c), this triple will be regular if p/q > 2.414. Now, p/q = Pk / 1 = Pk. Since k ³ 1 and P ³ 3, then Pk must be greater than or equal to 3. Thus, p/q ³ 3 > 2.414, and the triple is regular. Furthermore, this triple is complete (c = b + 1), since for
a = pq = Pk(1), we have
m = (p + q)/2 = (Pk + 1) / 2 and n = (p - q)/2 = (Pk - 1) / 2.
Then, b = 2mn = 2 (Pk + 1)/2 (Pk - 1)/2 = (P2k – 1)/2, (1)
and c = m2 + n2 = [ (Pk + 1)/2 ]2 + [ (Pk - 1)/2 ]2
= ( P2k + 2Pk + 1 + P2k - 2Pk + 1 ) / 4
= (2P2k + 2) / 4 = (P2k + 1) / 2. (2)
Now, from (1), b + 1 = (P2k – 1)/2 + 1 = (P2k – 1)/2 + 2 /2 = (P2k + 1) / 2 = c. [ from (2) ]
So, c = b + 1, and the triple is complete. All of this gives us
Theorem 23
All a’s are members of at least one regular primitive Pythagorean triple. If a is a power of an odd
prime, then it is a member of one and only one triple, which is both regular and complete, and it
cannot be a member of any inverted primitive Pythagorean triple.
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Theorem 23 tells us that all a’s are members of at least one regular primitive Pythagorean triple, whereas not all a’s are members of inverted ones. In this sense, we may say that there are more regular triples than inverted triples.
3. Are there any a’s which belong to more than one inverted triple? For a = pq to be a member of an inverted triple, p/q must be small; that is, p/q must be less than 2.414. To accomplish this, we must have factors p and q which are close to each other in size, so that p/q will be small. Now a = pq = 33 can be written only as 33(1) and 11(3), but in both of these cases, p and q are not close to each other in size, and thus p/q is larger than 2.414.
(33/1 = 33, and 11/3 = 3.67.) But if we try, a = 11(13)(17)(19) = 46,189, whose prime factors are all close together in size, we can get values for p and q which are also all close in size, giving us the following:
a = 11(13)(17)(19) = 46,189
Factors chosen for m n
p q p q p/q (p+q)/2 (p-q)/2 a b c
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13(17) 11(19) 221 209 1.06 215 6 46,189 2,580 46,261
13(19) 11(17) 247 187 1.32 217 30 46,189 13,020 47,989
17(19) 11(13) 323 143 2.26 233 90 46,189 41,940 62,389
Thus, a = 46,189 is a member of three inverted primitive Pythagorean triples! So, yes, there are some a’s which belong to more than one inverted triple.
Now suppose that P = (a, b, c) is a complete (c = b + 1) primitive Pythagorean triple. Will it be a regular or inverted triple, or perhaps sometimes a regular and sometimes an inverted triple? Let’s see. Theorem 6 in Part 1 tells us that if P is complete, then m = n + 1. To see if P is regular or inverted, let’s calculate the value of
m/n = (n+1)/n = 1 + 1/n.
We know that n ³ 1. Dividing both sides of this inequality by n, we get 1 ³ 1/n. Adding 1 to both sides of this inequality gives us 2 ³ 1 + 1/n, or 1 + 1/n £ 2. Thus, m/n = 1 + 1/n £ 2 < 2.414, and by Theorem 22 a),
P is a regular triple. This gives us
Theorem 24
Every complete primitive Pythagorean triple is regular.
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The converse of Theorem 24 (if P is a regular triple, then P is complete) is false, since (39, 80, 89) is regular
(a < b), but not complete; that is, c ¹ b + 1.
3. Proper Primitive Pythagorean Triples
In the very first primitive Pythagorean triple; that is, the smallest one, (3,4,5), a is divisible by 3, b is divisible by 4, and c is divisible by 5. This is an example of a proper primitive Pythagorean triple; namely, a triple in which a is divisible by 3, b is divisible by 4, and c is divisible by 5. (5,12,13) is not a proper triple, since a is not divisible by 3 and c is not divisible by 5. (63,16, 65) and (153,104,185) are both proper Pythagorean triples. Are there others?
To find out, let’s construct a table of primitive Pythagorean triples showing the values of b = 2mn in ascending order, together with b/2 = mn, and m and n.
Table 2
b b/2 m n a b c Proper? Why not?
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4 2 2 1 3 4 5 yes
8 4 4 1 15 8 17 no (c not divisible by 5)
12 6 6 1 35 12 37 no (c not divisible by 5)
3 2 5 12 13 no (c not divisible by 5)
16 8 8 1 63 16 65 yes
20 10 10 1 99 20 101 no (c not divisible by 5)
5 2 21 20 29 no (c not divisible by 5)
24 12 12 1 143 24 145 no (a not divisible by 3)
4 3 7 24 25 no (a not divisible by 3)
28 14 14 1 195 28 197 no (c not divisible by 5)
7 2 45 28 53 no (c not divisible by 5)
32 16 16 1 255 32 257 no (c not divisible by 5)
36 18 18 1 323 36 325 no (a not divisible by 3)
9 2 77 36 85 no (a not divisible by 3)
40 20 20 1 399 40 401 no (c not divisible by 5)
5 4 9 40 41 no (c not divisible by 5)
44 22 22 1 483 44 485 yes
11 2 117 44 125 yes
48 24 24 1 575 48 577 no (a not divisible by 3)
8 3 55 48 73 no (a not divisible by 3)
52 26 26 1 675 52 677 no (c not divisible by 5)
13 2 165 52 173 no (c not divisible by 5)
56 28 28 1 783 56 785 yes
7 4 33 56 65 yes
60 30 30 1 899 60 901 no (c not divisible by 5)
15 2 221 60 229 no (c not divisible by 5)
64 32 32 1 1023 64 1025 yes
68 34 34 1 1155 68 1157 no (c not divisible by 5)
17 2 285 68 293 no (c not divisible by 5)
72 36 36 1 1295 72 1297 no (a not divisible by 3)
9 4 65 72 97 no (a not divisible by 3)
76 38 38 1 1443 76 1445 yes
19 2 357 76 365 yes
80 40 40 1 1599 80 1601 no (c not divisible by 5)
8 5 39 80 89 no (c not divisible by 5)
84 42 42 1 1763 84 1765 no (a not divisible by 3)
21 2 437 84 445 no (a not divisible by 3)
14 3 187 84 205 no (a not divisible by 3)
7 6 13 84 85 no (a not divisible by 3)
88 44 44 1 1935 88 1937 no (c not divisible by 5)
11 4 105 88 137 no (c not divisible by 5)
92 46 46 1 2115 92 2117 no (c not divisible by 5)
23 2 525 92 533 no (c not divisible by 5)
96 48 48 1 2303 96 2305 no (a not divisible by 3)
16 3 247 96 265 no (a not divisible by 3)
100 50 50 1 2499 100 2501 no (c not divisible by 5)
25 2 621 100 629 no (c not divisible by 5)
104 52 52 1 2703 104 2705 yes
13 4 153 104 185 yes
108 54 54 1 2915 108 2917 no (a not divisible by 3)
27 2 725 108 733 no (a not divisible by 3)
112 56 56 1 3135 112 3137 no (c not divisible by 5)
8 7 15 112 113 no (c not divisible by 5)
116 58 58 1 3363 116 3365 yes
29 2 837 116 845 yes
120 60 60 1 3599 120 3601 no (a not divisible by 3)
20 3 391 120 409 no (a not divisible by 3)
15 4 209 120 241 no (a not divisible by 3)
12 5 119 120 169 no (a not divisible by 3)
124 62 62 1 3843 124 3845 yes
31 2 957 124 965 yes
Examining the table closely, it seems that only those b’s which end in 4 or 6 are members of a proper Pythagorean triple. However, 24, 36, 84, and 96 all end in 4 or 6 but are not proper triples because their a-values are not divisible by 3. Thus, their b-values must be divisible by 3, since Theorem 7a of Part 1 tells us that 3 must be a factor of either a or b, but not both.
Now, writing down just those b’s which end in 4 or 6 and are not divisible by 3, we get
b 4 16 44 56 64 76 104 116 124
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differences 12 28 12 8 12 28 12 8
These differences do form a pattern, but not one that is wholly satisfying! Let’s write down all the values of the b’s which end in 4 or 6, paying no attention to whether or not they are divisible by 3. We get
b 4 16 24 36 44 56 64 76 84 96 104 116 124
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differences 12 8 12 8 12 8 12 8 12 8 12 8
Wow! Now we really get a pretty pattern! Apparently, each successive b differs from its predecessor by 12, then by 8, then by 12, then by 8, repeatedly. How very nice! But that’s not all ! If we write down the first value of b, the third value of b, the fifth value, the seventh, and so on, we get
4 24 44 64 84 104 124
and each value differs from its predecessor by 20! This suggests that
b = 4 + 20 k, for k = 0, 1, 2, 3, . . .
Similarly, if we write down the second value of b, the fourth value of b, the sixth, and so on, we get
16 36 56 76 96 116
and again, each value differs from its predecessor by 20! Thus, it also seems that
b = 16 + 20 k, for k = 0, 1, 2, 3, . . .
Now to get these nice patterns, we included those b’s which were divisible by 3. However, in a proper Pythagorean triple, only the a’s are divisible by 3. So, we must exclude any of the b’s we found that are divisible by 3. All of this suggests:
Theorem 25 (proposed)
b will be a member of a proper primitive Pythagorean triple if and only if b is not divisible
by 3, and b = 4 + 20k or b = 16 + 20k, for k = 0, 1, 2, 3, . . .
Alternatively, we may say that b is NOT a member of a proper primitive Pythagorean triple
unless b ends in a 4 or 6 and is not divisible by 3.
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Now, to prove Theorem 25, we must find the values of all b’s which are divisible by 4, but not 3 or 5. Let’s start with b = 4 and attempt to add to it a multiple of k = 0, 1, 2, 3, . . . , to obtain another b. If we use b = 4 + k, for k = 0, 1, 2, 3, . . ., then b will not be divisible by 4, since there will not be a factor of 4 in both terms of b (unless k itself is a multiple of 4). So, let’s try b = 4 + 4k, for k = 0, 1, 2, 3, . . . Then, b will always be divisible by 4, since there is now a factor of 4 in each term of b = 4 + 4k.
But there is trouble, for if k =1, then b = 8, which, as shown in Table 2, is not a member of a proper triple! Furthermore, if k =2, then b = 12, and is thus divisible by 3, which is not possible in a proper triple. However, if we introduce 3 as a factor in the multiple of k, then we get b = 4 + 4(3)k = 4 + 12k, for k = 0, 1, 2, 3, . . . Now, is b = 4 + 12k divisible by 3? No, because there is not a factor of 3 in both terms of b. While 3 is a factor of 12k, it certainly is not a factor of 4. Thus, 3 is NOT a divisor of b = 4 + 12k, for k = 0, 1, 2, . . .
Of course, b cannot be divisible by 5 either. But, if k = 8, then b = 4 + 12(8) = 4 + 96 = 100, which is divisible by 5! So, let’s introduce 5 as a factor into our multiple of k. Then, b = 4 + 12(5)k = 4 + 60k, for k a positive integer or 0. Now, 5 is a factor of 60k, but it is certainly not a factor of 4. Thus, 5 is NOT a divisor of
b = 4 + 60k, for k a positive integer or 0.
So, the formula, b = 4 + 3(4)(5)k = 4(1 + 15k) = 4 + 60k meets our requirements that b be divisible by 4, but not by 3 or 5, but does it generate all the values of b found in Table 2 above; namely,
4 16 44 56 64 76 104 116 124 ?
Let’s see. Using b = 4 + 60k with k = 0, 1, 2, we get only the values 4, 64, and 124:
4 = 4 + 60(0), 64 = 4 + 60(1), 124 = 4 + 60(2).
We are missing the values 16, 44, and 56, which lie between 4 and 64, and also the values 76, 104, and 116. To remedy this, we write b in the following four ways.
For k = 0, 1, 2, 3, . . . 1) b = 4 + 60k, 2) b = 16 + 60k,
3) b = 44 + 60k, 4) b = 56 + 60k.
If k = 0, we get b = 4, 16, 44, and 56.
If k = 1, we get: b = 64, 76, 104, and 116.
If k = 2, we get b = 124 (from formula 1).
Thus, the four formulas above generate all the values of b found in Table 2.
But, do these values of b obtained by using formulas 1) through 4) generate proper triples? That is, will the triples (a, b, c) generated by these values of b produce values of a and c which are divisible by 3 and 5, respectively? Let’s see.
First, we can write formulas 1) to 4) as 1) b - 60k = 4, 2) b - 60k = 16, 3) b - 60k = 44, and 4) b - 60k = 56. Then, if 3 were a divisor of b, then 3 would be a divisor b - 60k, and thus a divisor of 4, 16, 44, and 56. But this is impossible. So, 3 is not a divisor of b in formulas 1) to 4). Thus, 3 is a divisor of a, since we know that 3 must be a divisor of a or b, but not both.
Second, by writing formulas 1) to 4) as
1) b = 4 + 60k = 4( 1 + 15k), 2) b = 16 + 60k = 4( 4 + 15k)
3) b = 44 + 60k = 4(11 + 15k), 4) b = 56 + 60 k = 4(14 + 15k),
we see that 4 is a factor of each of these formulas. Thus, 4 is a divisor of b.
Finally,
1) If b = 4 + 60k, then b/2 = 2 + 30k = mn. Choosing m = 2 + 30 k, n = 1, we get
c = m2 + n2 = (2+30k)2 + 12 = 4 + 120k + 900k2 + 1 = 900k2 + 120k + 5
= 5(180k2 + 24k + 1). Thus, c has a factor of 5, and 5 is a divisor of c.
2) If b = 16 + 60k, then b/2 = 8 + 30k = mn. Choosing m = 8 + 30 k, n = 1, we get
c = m2 + n2 = (8+30k)2 + 12 = 64 + 480k + 900k2 + 1 = 900k2 + 120k + 65
= 5(180k2 + 24k + 13). Thus, c has a factor of 5, and 5 is a divisor of c.
3) If b = 44 + 60k, then b/2 = 22 + 30k = mn. Choosing m = 22 + 30 k, n = 1, we get
c = m2 + n2 = (22+30k)2 + 12 = 484 + 1320k + 900k2 + 1 = 900k2 + 120k + 485
= 5(180k2 + 24k + 97). Thus, c has a factor of 5, and 5 is a divisor of c.
4) If b = 56 + 60k, then b/2 = 28 + 30k = mn. Choosing m = 28 + 30 k, n = 1, we get
c = m2 + n2 = (28+30k)2 + 12 = 784 + 1680k + 900k2 + 1 = 900k2 + 120k + 785
= 5(180k2 + 24k + 157). Thus, c has a factor of 5, and 5 is a divisor of c.
Thus, in the triple (a,b,c), where b is determined by formulas 1) through 4) above, 3 divides a, 4 divides b, and 5 divides c. So, formulas 1) through 4) do generate proper triples! And since b must be in one of the forms 1) through 4) for it to be divisible by 4 but not 3 or 5, formulas 1) through 4) generate all proper triples.
It is interesting to note that formulas 3) and 4) can be rewritten as follows:
3) b = 44 + 60k, for k = 0, 1, 2, . . .
= 44 - 60 + 60 + 60k = -16 + 60(k+1) = -16 + 60k, for k = 1, 2, 3, . . .
4) b = 56 + 60k, for k = 0, 1, 2, . . .
= 56 - 60 + 60 + 60k = - 4 + 60(k+1) = - 4 + 60k, for k = 1, 2, 3, . . . . . .
We have thus proved Theorem 25.
Theorem 25
b belongs to a proper primitive Pythagorean triple if and only if for k = 0, 1, 2, . . .
1) b = 4 + 60k, 2) b = 16 + 60k,
3) b = 44 + 60k, 4) b = 56 + 60k.
( b = ± 4 + 60 k, or b = ± 16 + 60k, b > 0 )
Alternatively, b belongs to a proper primitive Pythagorean triple if and only if
b is divisible by 4, ends in a 4 or 6, and is not divisible by 3.
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Example 1. Determine if b = 232 is a member of a proper primitive Pythagorean triple. If b is a member of a proper triple, it must be capable of being written as one of the formulas 1) through 4) of Theorem 25. However, since 232 = 52 + 60(3), there are NO values for k which allow us to write b = 232 in one of the forms 1) through 4). So, b = 232 is NOT a member of any proper primitive Pythagorean triple. Alternatively, we may note that while b = 232 is divisible by 4 (and is thus a member of a primitive Pythagorean triple) and is not divisible by 3, it does not end in a 4 or 6, and thus by Theorem 25 is NOT a member of any proper triple.
Example 2. Determine if b = 616 is a member of a proper primitive Pythagorean triple. If so, then find all proper triples to which it belongs. Since b = 616 is divisible by 4 and thus a member of a primitive Pythagorean triple and is not divisible by 3 and ends in a 6, we know by Theorem 25, that it is a member of a proper triple. To find these proper triples, we construct the following table:
b b/2 m n a b c
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616 308 308 1 94,863 616 94,865
77 4 5913 616 5945
44 7 1887 616 1985
28 11 663 616 905
Thus, we have four proper (and inverted) triples:
(94,863, 616, 94,865), (5913, 616, 5945), (1887, 616, 1985), (663, 616, 905)
Now, what about a? Can we find conditions on a that will ensure it is a member of a proper primitive Pythagorean triple? Here is a list of all the values in Table 2 in which a belongs to a proper triple.
3 63 483 117 783 33 1023 1443 357 2703 153 3363 837 3843 957 (1)
Notice that the last digits are always 3 or 7, and that they are all divisible by 3. Putting those values of a whose last digit is a 3 in ascending order, we get
3 33 63 153 483 783 1023 1443 3363 3843 (2)
The first three numbers differ successively by 30. Is it possible that all the values of a have successive differences of 30? If so, then the next numbers after 63 should be 93 and 123 as shown below.
3 33 63 93? 123? 153 483 783 1023 1443 3363 3843
Is 93 a member of a proper triple? Yes, since 93 = pq = 93(1). Then, m = (93+1)/2 = 47, n = (92-1)/2 = 46, and we have the triple (93, 4324, 4325), which is a proper triple! Also, 93 = pq = 31(3), so m = (31+3)/2 = 17, n = (31-3)/2 = 14, and we get another proper triple (93, 476, 485). 123 is also a member of a proper triple, since 123 = pq = 123(1), so m = (123+1)/2 = 62, n = (1234-1)/2 = 61, giving us (123,7564,7565), a proper Pythagorean triple. Also, 123 = pq = 41(3), so m = (41+3)/2 = 22, and n = (41-3)/2 = 19, giving us (123, 836, 845). Great! With these triples as confirmation of our assumption, let’s state that
a = 3 + 30k, for k = 0, 1, 2, . . .
and see if this formula gives us all the values in list (2) above.
If k = 5, then a = 3 + 30(5) = 153 which is the fourth number in list (2). For 483, we get 3 + 30k = 483, 30k = 480, and k = 16, so the fifth number in the list can be obtained from our formula. For the rest, we have 783 = 3 + 30(26), 1023 = 3 + 30(34), 1443 = 3 + 30(48), 3363 = 3 + 30(112), and 3843 = 3 + 30(128).
And what about the numbers ending in 7 in list (1)? That is, 117, 357, 837, and 957? If all these numbers also differ successively by 30, then the numbers before 117 would be 117 - 30 = 87, 87 - 30 = 57, and
57 - 30 = 27. The list for these values ending in 7 would then be
27? 57? 87? 117 . . . 357 . . . 837 . . . 957
and so on, and we would have the formula a = 27 + 30k, for k = 0, 1, 2, . . .
Now, does this formula produce proper triples? Let’s see.
For a = 27, we get the triple (27, 364, 365), a proper triple. In a similar fashion, we can show that 57 and 87 are also members of proper triples. We do get 357, 837, and 957 by using our formula. With k = 11, we have 27 + 30(11) = 357; with k = 27, we have 27 + 30(27) = 837; and with k = 31, we have 957 = 27 + 30(31). This suggests the following:
Theorem 26
a is a member of a proper primitive Pythagorean triple if and only if
1) a = 3 + 30k, or a = 27 + 30k, for k = 0, 1, 2, 3, . . .
(or, a = ± 3 + 30k, for k = 0, 1, 2, . . . , a > 0 ) or
2) a ends in 3 or 7 and is divisible by 3.
_________________________________________________________
To prove Theorem 26, we must find all values of a which are divisible by 3, but not by 4 or 5. Following the proof of Theorem 25, we let a = 3 + some multiple of k, for k = 0, 1, 2, … Let’s start with a = 3 + 3k. Then k is clearly divisible by 3, since 3 is a divisor of both 3 and 3k. But, if k =1, then a = 3 + 3(1) = 6, which is an even integer, but a must be odd. Accordingly, we introduce a factor of 2 into the multiple of k, getting a = 3 + 3(2)k = 3 + 6k. Then, a is not divisible by 2 because there is not a factor of 2 in both terms of a.
However, if k = 2, then a = 3 + 6(2) = 15, which is divisible by 5, which is not possible for a proper triple.
Accordingly, we introduce a factor of 5 into the multiple of k, giving us a = 3 + 6(5)k = 3 + 30k. Then, a is not divisible by 5 because there is not a factor of 5 in both terms of a.
Thus, the formula, a = 3 + 3(2)(5)k = 3 + 30k meets our requirements that a be divisible by 3, but not by 4 or 5, but does it generate all of the values of a shown below; namely,
3 27 33 57 63 ?
Let’s see. Using a = 3 + 30k with k = 0, 1, 2, we get only the values 3, 33, and 63:
3 = 3 + 30(0), 33 = 3 + 30(1), 63 = 3 + 30(2).
We are missing the value 27, which lies between 3 and 33, and the value 57, which lies between 33 and 63. To remedy this, we write a in the following two ways.
For k = 0, 1, 2, 3, . . . 1) a = 3 + 30k or 2) a = 27 + 30k.
For k = 0, we get: a = 3 and a = 27.
For k = 1, we get: a = 33 and a = 57.
For k = 2, we get a = 63.
But, do the values obtained by using formulas 1) and 2) generate proper triples? Let’s see.
A. a = 3 + 30k, for k = 0, 1, 2, . . .
Since a = 3 + 30k = 3(1 + 10k), 3 is a divisor of a.
Furthermore, a = pq = (3 + 30k) x 1. Choosing p = 3 + 30k and q = 1, we have
m = (p + q)/2 = [(3+30k) + 1] / 2 = 2 + 15k,
and n = (p - q)/2 = [(3+30k) – 1] / 2 = 1 + 15k.
Now, b = 2mn = 2(2+15k)(1+15k), and since 2 + 15k and 1 + 15k are consecutive integers, one of them must be even and so divisible by 2. Thus, b is a product of three factors, two of which are even, and thus b is divisible by 4. Thus, 4 is a divisor of b.
Finally, c = m2 + n2 = (2 + 15k)2 + (1 + 15k)2 = 4 + 60k + 225k2 + 1 + 30k + 225k2
= 450k2 + 90k + 5 = 5(90k2 + 18k + 1), which has a factor of 5. Thus, 5 is a divisor of c.
B. a = 27 + 30k, for k = 0, 1, 2, . . .
Since a = 27 + 30k = 3(9 + 10k), 3 is a divisor of a.
Furthermore, a = pq = (27 + 30k) x 1. Choosing p = 27 + 30k and q = 1, we have
m = (p + q)/2 = [(27 + 30k) + 1] / 2 = 14 + 15k,
and n = (p - q)/2 = [(27 + 30k) – 1] / 2 = 13 + 15k.
Now, b = 2mn = 2(14+15k)(13+15k), and since 14 + 15k and 13 + 15k are consecutive integers, one of them must be even and so divisible by 2. Thus, b is a product of three factors, two of which are even, and thus b is divisible by 4. Thus, 4 is a divisor of b.
Finally, c = m2 + n2 = (14+15k)2 + (13+15k)2 = 196 + 420k + 225k2 + 169 + 390k + 225k2
= 450k2 + 810k + 365 = 5(90k2 + 78k + 73), which has a factor of 5. Thus, 5 is a divisor of c.
So, formulas 1) and 2) above do generate proper triples since each formula generates a triple (a, b, c), where 3 divides a, 4 divides b, and 5 divides c. And since a must be in one of the forms 1) or 2) for it to be divisible by 3 but not 4 or 5, formulas 1) and 2) generate all proper triples.
It is interesting to note that formula 2) can be rewritten as follows:
2) a = 27 + 30k, for k = 0, 1, 2, . . .
= 27 - 30 + 30 + 30k = - 3 + 30(k+1),
= - 3 + 30k, for k = 1, 2, 3, . . .
Thus, we have that a = ± 3 + 30k, for k = 0, 1, 2, . . . , a > 0, and we have proved Theorem 26.
Example 3. Determine if a = 117 is a member of a proper primitive Pythagorean triple. If it is, find all the proper triples to which it belongs. Since a = 117 ends in a 7 and is divisible by 3, Theorem 26 tells us that it is a member of a proper triple. To find those proper triples to which a belongs, we write a = 117 = pq = 117(1) = 39(3) = 13(9). This gives us
p q m n a b c
--------------------------------------------------------------------------
117 1 59 58 117 6844 6845
13 9 11 2 117 44 125
Thus, a =117 belongs to two proper primitive Pythagorean triples, (117, 6844, 6845) and (117, 44, 125). Note that the first is regular, and the second inverted.
Finally, what conditions must we impose upon c in order for it to be a member of a proper triple? Writing in ascending order all those values of c which belong to a proper triple as found in Table 2 above, we
get
5 65 125 185 365 485 785 845 965 1025 1445 2705 3365 3845
Note that the last digit in each of these numbers is 5, and that the first 4 numbers differ successively by 60. Following the same procedure as used for a and b, we predict that c = 5 + 60k, for k = 0, 1, 2, . . . and verify that each number in our list satisfies this formula. Thus,
5 = 5 + 60(0), 65 = 5 + 60(1), 125 = 5 + 60(2), 185 = 5 + 60(3),
365 = 5 + 60(6), 485 = 5 + 60(8), 785 = 5 + 60(13), 845 = 5 + 60(14),
965 = 5 + 60(16), 1025 = 5 + 60(17), 1445 = 5 + 60(24), 2705 = 5 + 60(45),
3365 = 5 + 60(56), 3845 = 5 + 60(64).
However, for k = 4, we get c = 5 + 60(4) = 245, which is NOT in our list and is NOT a member of a proper triple because it is not a member of any primitive Pythagorean triple, proper or not! This is because 245
= 5(7)(7), and its factors of 7 are not of the form 4L+1 for some integer L. This suggests the following theorem.
Theorem 27
c will be a member of a proper primitive Pythagorean triple if and only if it is a member
of a primitive Pythagorean triple and is of the form c = 5 + 60k, for k = 0, 1, 2, 3, . . .
______________________________________________________________________
To prove Theorem 27, we must find all values of c which are divisible by 5, but not 3 or 4. Following the pattern of the previous two proofs, we start with c = 5 + 5k, for k = 0, 1, 2, . . . Clearly, c = 5 + 5k = 5(1+k), so 5 is a divisor of c, but for k = 2, we get c = 15, which is divisible by 3, and for k = 3, we get c = 20, which is divisible by 4, which is not possible for a proper triple! Accordingly, we will incorporate both 3 and 4 as factors in our multiple of k, getting c = 5 + 5(3)(4)k = 5 + 60k. Then, c is not divisible by 3 or 4, because there is not a factor of 3 nor a factor of 4 in both terms of c.
Next, we must show that if c = 5 + 60k, then it will be a member of a proper triple in which b is divisible by 4, but not 3, and a is divisible by 3, but not 4.
We know that b + c = (m + n)2, and since m and n have opposite parity, m + n must be an odd integer. Thus,
m + n = 2L + 1 for some integer L, so
b + c = (2L+1)2,
and b = (2L+1)2 – c
= 4L2 + 4L + 1 – (5 + 60k)
= 4L2 + 4L – 4 – 60k = 4(L2 + L – 1 – 15k),
which shows that b has a factor of 4, and thus 4 is a divisor of b.
Now, is b divisible by 3? Let’s hope not! We have b = 4(L2 + L – 1 – 15k)
= 4H,
where H = L2 + L – 1 – 15k. Then, if b is divisible by 3, 3 must be a factor of H, since it surely is not a factor of 4. Now, either a) L has a factor of 3, or b) L doesn’t have a factor of 3
a) If L has a factor of 3, it can be written as L = 3j, for some integer j. Then,
H = L2 + L – 1 – 15k = 9j2 + 3j – 1 – 15k = 3(3j2 + j – 5k) – 1,
which shows that H does not have a factor of 3.
b) If L does not have a factor of 3, it can be written as either 1) L = 3j+1 or 2) L = 3j+2.
1) If L = 3j+1, then H = L2 + L – 1 – 15k = (3j+1)2 + (3j+1) – 1 – 15k
= 9j2 + 6j + 1 + 3j + 1 – 1 – 15k = 9j2 + 9j – 15k – 1
= 3(3j2 + 3j – 5k) – 1, which shows that H does not have a factor of 3.
2) If L = 3j+2, then H = L2 + L – 1 – 15k = (3j+2)2 + (3j+2) – 1 – 15k
= 9j2 + 12j + 4 + 3j + 2 – 1 – 15k = 9j2 + 15j – 15k + 3 + 2
= 3(3j2 + 5j – 5k + 1) + 2, which shows that H does not have a factor of 3.
Now in both cases a) and b), we have shown that H does not have a factor of 3, so b = 4H does not have a factor of 3, and thus for c = 5 + 60k, b is divisible by 4 but not 3.
Finally, since a is always odd, it cannot be divisible by 4. We also know that in any primitive Pythagorean triple, either a or b, but not both, is divisible by 3. Since we have just proved that b is not divisible by 3, then a must be divisible by 3. Thus, a is divisible by 3 but not 4,
Thus, we have shown that if c = 5 + 60k, it will be a member of a proper triple in which b is divisible by 4, but not 3, and a is divisible by 3, but not 4, and c is divisible by 5. This proves Theorem 27. Below are Theorems 25 through 27 combined into one theorem.
Theorem 28
a is a member of a proper primitive Pythagorean triple if and only if
1) a = 3 + 30k, or a = 27 + 30k, for k = 0, 1, 2, . . . or
2) a = ± 3 + 30k, for k = 0, 1, 2, . . . , b > 0. or
3) a ends in 3 or 7 and is divisible by 3.
b belongs to a proper primitive Pythagorean triple if and only if
1) b = 4 + 60k or b = 16 + 60k or b = 44 + 60k or b = 56 + 60k, for k = 0, 1, 2, . . . or
2) b = ± 4 + 60 k or b = ± 16 + 60k, for k = 0, 1, 2, . . ., b > 0, or
3) b ends in 4 or 6 and is divisible by 4, but not 3.
c belongs to a proper primitive Pythagorean triple if and only if c is a member of
a primitive Pythagorean triple and is of the form c = 5 + 60k, for k = 0, 1, 2, . . .
___________________________________________________________________________
To see what we’ve got, let’s now construct tables of proper primitive Pythagorean triples for each of the formulas for a, b, and c from Theorem 28 above, showing the values of k, m, n, a, b, and c (and other appropriate values as needed).
Table a
Proper Pythagorean Triples
a = 3 + 30k and a = 27 + 30k, for k = 0, 1, 2, . . . ,7
k a p q m n a b c
-----------------------------------------------------------------------------------------------
0 3 3 1 2 1 3 4 5
27 27 1 14 13 27 364 365
1 33 33 1 17 16 33 544 545
11 3 7 4 33 56 65
57 57 1 29 28 57 1624 1625
19 3 11 8 57 176 185
2 63 63 1 32 31 63 1984 1985
9 7 8 1 63 16 65
87 87 1 44 43 87 378 3785
29 3 16 13 87 416 425
3 93 93 1 47 46 93 4324 4325
31 3 17 14 93 476 485
117 117 1 59 58 117 6844 6845
13 9 11 2 117 44 125
4 123 123 1 62 61 123 7564 7565
41 3 22 19 123 836 845
147 147 1 74 73 147 10804 10805
49 3 26 23 147 1196 1205
5 153 153 1 77 76 153 11704 11705
17 9 13 4 153 104 185
177 177 1 89 88 177 15664 15665
59 3 31 28 177 1736 1745
6 183 183 1 92 91 183 16744 16745
61 3 32 29 183 1856 1865
207 207 1 104 103 207 21424 21425
23 9 16 7 207 224 305
7 213 213 1 107 106 213 22684 22685
71 3 37 34 213 2516 2525
237 237 1 119 118 237 28084 28085
79 3 41 38 237 3116 3125
------------------------------------------------------------------------------------------------
Table b
Proper Pythagorean Triples
b = 4 + 60k or b = 16 + 60k or b = 44 + 60k or b = 56 + 60k for k = 0, 1, 2, 3, 4.
k b b/2 m n a b c
--------------------------------------------------------------------------------
0 4 2 2 1 3 4 5
16 8 8 1 63 16 65
44 22 22 1 483 44 485
11 2 117 44 125
56 28 28 1 783 56 785
7 4 33 56 65
1 64 32 32 1 1023 64 1025
76 38 38 1 1443 76 1445
19 2 357 76 365
104 52 52 1 2703 104 2705
13 4 153 104 185
116 58 58 1 3363 116 3365
29 2 837 116 845
2 124 62 62 1 3843 124 3845
31 2 957 124 965
136 68 68 1 4623 136 4625
17 4 273 136 305
164 82 82 1 6723 164 6725
41 2 1677 164 1685
176 88 88 1 7743 176 7745
11 8 57 176 185
3 184 92 92 1 8463 184 8465 23 4 513 184 545
196 98 98 1 9603 196 9605
49 2 2397 196 2405
224 112 112 1 12543 224 12545
16 7 207 224 305
236 118 118 1 13923 236 13925
59 2 3477 236 3485
4 244 122 122 1 14883 244 14885
61 2 3717 244 3725
256 128 128 1 16383 256 16385
284 142 142 1 20163 284 20165
71 2 5037 284 5065
296 148 148 1 21903 296 21905
37 4 1353 296 1385
--------------------------------------------------------------------------------------
Table c
Proper Pythagorean triples
c = 5 + 60k, for k = 0, 1, 2, 3, . . . , 16.
k c m n a b c
--------------------------------------------------------------------------------------------
0 5 2 1 3 4 5
1 65 8 1 63 16 65
7 4 33 56 65
2 125 11 2 117 44 125
3 185 13 4 153 104 185
11 8 87 176 185
4 245 = 5(7)(7) (not a member of a primitive Pythagorean
triple since 7 is not of the form 4L+1)
5 305 17 4 273 136 305
16 7 207 224 305
6 365 19 2 357 76 365
14 13 27 364 365
7 425 19 8 297 304 425
16 13 87 416 425
8 485 22 1 483 44 485
17 14 93 476 485
9 545 23 4 513 184 545
17 16 33 544 545
10 605 = 5(11)(11) (not a member of a primitive Pythagorean
triple since 11 is not of the form 4L+1)
11 665 = 5x7x19 (not a member of a primitive Pythagorean
triple since 7 and 19 are not of the form 4L+1)
12 725 26 7 627 364 725
23 14 333 644 725
13 785 28 1 783 56 785
23 16 273 736 785
14 845 29 2 837 116 845
22 19 123 836 845
15 905 29 8 777 464 905
28 11 663 616 905
16 965 31 2 957 124 965
26 17 387 884 965
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