Pythagorean triples

Use the Brahmagupta-Fibonacci Identity $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$ This identity can be verified by multiplying out each side, and in nicer ways.

From the Identity, we get $$65^2=(8^2+1^2)(7^2+4^2)=(8\cdot 7-1\cdot 4)^2 +(8\cdot 4+1\cdot 7)^2.$$

We can get another representation of $65^2$ as the sum of two squares by letting $c=4$ and $d=7$.

Remark: The Identity gives the useful result that the product of two numbers, each the sum of two squares, is itself the sum of two squares.

Hint If $m,n$ are integers then $(m^2-n^2, 2mn, m^2+n^2)$ is a Pytagorean triple.