How to find nonprimitive pythagorean triples given hypotenuse?

Question

The integer $2015$ is the largest integer in $4$ different pythagorean triples, none of which is a primitive triple. Can someone explain how to find the $4$ triples?

Answer 1

Since they’re not primitive, each of them will have a common factor greater than $1$. Since the prime factorization of $2015$ is $5\cdot13\cdot31$, the possible values of the long side in the primitive triples are $5,13,31,65,155$, and $403$. You can probably already pick out two of the four by eye. If you know that the largest number in a primitive Pythagorean triple is the sum of two squares, you shouldn’t have too much trouble finding the other two. I’ve added an extra hint in the spoiler-protected block below.

HINT: What is the smallest positive integer that is the sum of two different squares in two different ways?