Number Thoery Problem

square numbers only have three factors, 1, their squareroot, and themself, apart from 1, which has 1. ( * )

If you instead break it down as

a^2 = c^2 - b^2

and use the fact that

"a^2 has factors 1, a and a^2 only" - which is true as a is prime

then you can uniquely work out b and c from a much as you did, remembering that a,b,c are positive.

Well done, now try this extension:

let [a,b] denote a pair of positive integers such that
a^2 + b^2 = square number

prove that there are as many different values for b as there are proper factors of a (Hence proving that prime numbers a have only one corresponding value for b, having only one proper factor)

Same again, let the square number be c

=> a^2 = (c-b)(c+b)

Now it follows that c-b can be 1,a or any proper factor of a, and so, since c is fixed, the number of possibilities for b is determined by the number of proper factors of a.