Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.
$65^2 = (8^2+1^2)(7^2+4^2) = 8^27^2 + 1^24^2 + 1^27^2 + 8^24^2 = (8\cdot7)^2 + 4^2 + 7^2 + (8\cdot4)^2$ but now I am stuck here, any suggestions!

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.
