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!

If we solve $C=m^2+n^2$ for $n$, we get $n=\sqrt{C-m^2}$. We can then try values of $m$ to see which one(s) yield a positive integer for $n$, where
$$\text{To ensure }n<m:\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor:\text{ to ensure }n\in\mathbb{N}$$
$$\text{In the example of }65\qquad m_{min}=\lceil\sqrt{32.5}\space\rceil = 6\qquad m_{max}=\lfloor\sqrt{65}\rfloor = 8$$
If we test $n=\sqrt{C-m^2}$ using $m=6,7\text{, and }8$, we find positive integers for $n$ where $m=7\Rightarrow n=4$ and $m=8\Rightarrow n=1.$
$$\text{Using Euclid's formula }f(7,4)=(33,56,65)\quad f(8,1)=(63,16,65)$$
