In the course of exploring proofs of that result, I suggested a proof using the representation , which generates all primitive Pythagorean triples (triples which have no common factors). Proving that 60 divides their product proves the general result, as every Pythagorean triple has the form , where k is a positive integer and is primitive.
In the ensuing discussion, a question arose as to how we know that all possible Pythagorean triples can be found in this way. Since it is easier for me to post LaTeX code to this blog rather than in the comments on the jd2718 blog, I’ll present the proof here.
Suppose that , with positive integers having no common factors. This implies that exactly one of a and b is even (since if both are even, then so is c, leading to a contradiction, while if both a and b are odd, then both and are congruent to 1 mod 4, while would be congruent to 0 mod 4, not 2 mod 4).
So without loss of generality, assume that a is even. Thus is also even, and since and differ by an even number, both of those factors must be even.
Furthermore, exactly one of and is 2x(an odd number), since if both factors are divisible by 4, it would follow that both b and c would be even (since b and c are .) Likewise, any odd common factor of and would also then be a common factor of both b and c, from which it follows that
- and have no odd common factors
- one of these expressions is of the form (for some odd u)
- and hence the other expression is of the form , where , where u and v are positive integers
It now follows that , and . We can find a using , and thus .
Conclusion: every primitive Pythagorean triple has the form , for positive integers u and v.