Pythagorean triples

The proof has two halves: the formula always works, and it reaches every primitive triple. Along the way the facts about 3, 4 and 5 fall out. It uses two small tools.

Tool 1: squares and remainders. Every whole number is even or odd. An even number is 2k, and its square is 4k^2, a multiple of 4. An odd number is 2k + 1, and its square is 4k^2 + 4k + 1, which leaves remainder 1 when divided by 4. So a square leaves remainder 0 or 1 when divided by 4, and never anything else. The same idea works for other divisors:

Tool 2: coprime factors of a square. If two positive whole numbers u and v have no common factor and uv is a perfect square, then u and v are each perfect squares. The reason is unique prime factorization: in a perfect square every prime appears an even number of times, and since each prime divides only one of u and v, all its copies sit in that one, so each of u and v has every exponent even.

The formula always gives a triple. Expand and add:

This is the same identity as (x - y)(x + y) = x^2 - y^2 read in reverse, which is the difference of squares pattern. It needs only m > n > 0, so that the first number is positive.

The facts about common factors. Suppose a prime p divides a and b. Then p divides a^2 + b^2 = c^2, and a prime that divides a square divides the number, so p \mid c. Suppose instead p divides a and c. Then p divides c^2 - a^2 = b^2, so p \mid b. The case of b and c is the same. So a prime that divides two of the numbers divides all three.

Every primitive triple has an odd hypotenuse and one even leg. Let (a, b, c) be primitive. If c were even, then c^2 would leave remainder 0 on division by 4, so a^2 + b^2 would too. Each square leaves 0 or 1, and the only way two of those add to a multiple of 4 is 0 + 0. So a and b would both be even, and 2 would divide all three numbers, which a primitive triple does not allow. So c is odd. Then a^2 + b^2 is odd, which means one leg is odd and the other even. (Two odd legs would make a^2 + b^2 leave remainder 2, and c^2 never does.) Say a is odd and b is even.

Every primitive triple comes from the formula. From a^2 + b^2 = c^2,

Both c and a are odd, so c + a and c - a are even. Write c + a = 2u and c - a = 2v, with u > v > 0, and b = 2w. Then 4w^2 = 4uv, so

The numbers u and v have no common factor: a common divisor would divide u + v = c and u - v = a, but a and c share no factor. By Tool 2, both are squares. Write u = m^2 and v = n^2. Then

and m > n > 0 because u > v > 0. The pair is unique, since m^2 = \dfrac{c + a}{2} and n^2 = \dfrac{c - a}{2} are fixed by the triple. The numbers m and n have no common factor, because a common divisor of m and n would divide u and v. And they are not both odd (then a = m^2 - n^2 would be even) and not both even, so one is even and one is odd.

Which pairs give primitive triples. A pair that fails the condition never gives a primitive triple: a common factor d > 1 of m and n makes d^2 divide all three numbers, and two odd numbers m, n make all three numbers even. Conversely, take a pair with no common factor and opposite parity. Then m^2 and n^2 have opposite parity, so a and c are both odd. A prime p dividing both would be odd and would divide c + a = 2m^2 and c - a = 2n^2, so it would divide m and n, which have no common factor. So a and c share no factor, and neither do all three. Scaling by k gives every other triple.

Divisible by 4. In a primitive triple the even leg is 2mn, and exactly one of m, n is even, so 2mn is a multiple of 2 \cdot 2 = 4. Scaling keeps it a multiple of 4.

Divisible by 5. Suppose none of a, b, c is a multiple of 5. By the second table, each square leaves remainder 1 or 4 on division by 5. Two such remainders add to 2, 5 or 8, which leave 2, 0 or 3, and none of those is 1 or 4, the only remainders c^2 can leave. That is a contradiction, so 5 divides at least one of the three numbers.

Divisible by 3. This one is Example 6.

1. Yes, and it is primitive. 12^2 + 35^2 = 144 + 1225 = 1369 = 37^2. The number 37 is prime and does not divide 12, so the three numbers have no common factor.
2. No. 9^2 + 40^2 = 81 + 1600 = 1681, but 42^2 = 1764. Sticks of 9, 40 and 42 could not make a right triangle: the longest, 42, would be the hypotenuse, and then the equation would have to hold.
3. (11, 60, 61) is a primitive triple: 121 + 3600 = 3721 = 61^2, and 61 is prime. (15, 36, 39) is a triple, not primitive: 225 + 1296 = 1521 = 39^2, and it is 3 times (5, 12, 13). (16, 30, 35) is not a triple: 256 + 900 = 1156, which is 34^2, not 35^2 = 1225.
4. The hypotenuse is 26, and the triple is 2 times (5, 12, 13). 10^2 + 24^2 = 100 + 576 = 676 = 26^2, and dividing all three numbers by 2 gives (5, 12, 13).
5. Yes, it is 85 cm. The diagonal squared is 40^2 + 75^2 = 1600 + 5625 = 7225 = 85^2. The sides are 5 times 8 and 5 times 15, so the triple is 5 times (8, 15, 17).
6. (24, 45, 51). The leg 45 is 3 \cdot 15, so multiply by 3. Check: 24^2 + 45^2 = 576 + 2025 = 2601 = 51^2.
7. The legs are 60 and 80. Since 100 = 20 \cdot 5, scale (3, 4, 5) by 20.
8. (14, 48, 50) and (30, 40, 50). Every triple is k times a primitive triple with hypotenuse c', where kc' = 50. A primitive hypotenuse is odd, so c' is 5 or 25. For m^2 + n^2 = 5 the only pair is (2, 1), giving (3, 4, 5) and k = 10. For m^2 + n^2 = 25 the only pair is (4, 3), since 9 + 16 is the only way to write 25 as two positive squares, giving (7, 24, 25) and k = 2.
9. Four triples: (16, 63, 65), (25, 60, 65), (33, 56, 65) and (39, 52, 65). The odd divisors of 65 above 1 are 5, 13 and 65. For c' = 5 there is (3, 4, 5) with k = 13, giving (39, 52, 65). For c' = 13 there is (5, 12, 13) with k = 5, giving (25, 60, 65). For c' = 65, m^2 + n^2 = 65 has the two pairs (8, 1) and (7, 4), since 65 = 64 + 1 = 49 + 16, and both pairs qualify, giving (16, 63, 65) and (33, 56, 65).
10. (48, 55, 73). m^2 - n^2 = 64 - 9 = 55, 2mn = 48, m^2 + n^2 = 73. Check: 48^2 + 55^2 = 2304 + 3025 = 5329 = 73^2.
11. (33, 56, 65). m^2 - n^2 = 49 - 16 = 33, 2mn = 56, m^2 + n^2 = 65. Check: 33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2.
12. The triple is (12, 16, 20). It is not primitive because m = 4 and n = 2 are both even, so they share the factor 2, which breaks the condition. It is 4 times (3, 4, 5), which comes from (m, n) = (2, 1). Doubling both m and n multiplies the triple by 4.
13. (m, n) = (7, 2). m^2 = (53 + 45) / 2 = 49 and n^2 = (53 - 45) / 2 = 4. Check: 2mn = 2 \cdot 7 \cdot 2 = 28.
14. (m, n) = (7, 6). m^2 = (85 + 13) / 2 = 49 and n^2 = (85 - 13) / 2 = 36. Check: 2mn = 2 \cdot 7 \cdot 6 = 84.
15. Four triples: (11, 60, 61), (91, 60, 109), (221, 60, 229) and (899, 60, 901). The even leg is 2mn = 60, so mn = 30. The pairs with m > n and product 30 are (30, 1), (15, 2), (10, 3) and (6, 5). Each has no common factor and opposite parity, so all four count, giving (899, 60, 901), (221, 60, 229), (91, 60, 109) and (11, 60, 61).
16. (15, 112, 113). With d = 1, 15^2 = 2b + 1, so 225 = 2b + 1, b = 112 and c = 113. Check: 225 + 12544 = 12769 = 113^2.
17. (16, 63, 65). With d = 2, 16^2 = 4b + 4, so 256 = 4b + 4, b = 63 and c = 65. Check: 256 + 3969 = 4225 = 65^2.
18. (8, 15, 17), with perimeter 8 + 15 + 17 = 40. A primitive triple from (m, n) has perimeter 2m^2 + 2mn = 2m(m + n), so a triple that is k times it has perimeter 2km(m + n) = 40, and km(m + n) = 20. The values of m(m + n) for the smallest qualifying pairs are 6 for (2, 1), 15 for (3, 2), 20 for (4, 1), and then 28 or more. Only 20 divides 20 and k = 1 works, giving (m, n) = (4, 1), legs 15 and 8, and hypotenuse 17.
19. Every triple is k times a primitive triple (a', b', c'), and a primitive triple has exactly one even leg. So ab = k^2 a'b' is even, and the area \dfrac{ab}{2} is a whole number. For example, (3, 4, 5) has area 6 and (5, 12, 13) has area 30.
20. For odd n, the border method gives \big(n,\ (n^2 - 1)/2,\ (n^2 + 1)/2\big): both new numbers are whole because n^2 is odd, and (n^2 - 1)/2 \ge 4. For even n = 2t with t \ge 2, take (2t,\ t^2 - 1,\ t^2 + 1). It works because (2t)^2 + (t^2 - 1)^2 = t^4 + 2t^2 + 1 = (t^2 + 1)^2, and t^2 - 1 \ge 3 is positive. For example n = 6 gives (6, 8, 10) and n = 10 gives (10, 24, 26).
21. Suppose 3 \mid c. Then c^2 leaves remainder 0 on division by 3, so a^2 + b^2 does too. Each square leaves 0 or 1, and the only way two of those add to a multiple of 3 is 0 + 0. So 3 divides a and b as well, and 3 divides all three numbers, which a primitive triple does not allow.
22. The even leg 12 would be 2mn, so mn = 6, and the whole-number pairs with m > n and mn = 6 are (6, 1) and (3, 2). They give (35, 12, 37) and (5, 12, 13), and neither is (9, 12, 15). The triple has the common factor 3, so it is not primitive. It is reached by the multiplier: (9, 12, 15) = 3 \cdot (3, 4, 5), which uses k = 3 and (m, n) = (2, 1).
23. Suppose 1^2 + b^2 = c^2. Then 1 = c^2 - b^2 = (c - b)(c + b). Since b \ge 1 and c > b, c + b \ge 3 and c - b \ge 1, so the product is at least 3, not 1. No such whole numbers exist.