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!
 A: 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.
A: Hint If $m,n$ are integers then $(m^2-n^2, 2mn, m^2+n^2)$ is a Pytagorean triple. 
A: As I said we need to solve a system of equations.  
The system of equations:  
$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$    
the solutions have the form:   $$x=4p^4-s^4$$   $$y=4p^2s^2$$   $$q=4ps(2p^2-s^2)$$   $$t=4p^4-8p^2s^2+s^4$$   $$z=4p^4+s^4$$    
$p,s,k$ - integers.  
Formulas you can write a lot, but will be limited to this. Will make a replacement.
$$a=p^2+s^2-k^2$$
$$b=p^2+s^2+k^2-2pk-2ks$$
$$c=p^2+k^2-s^2+2ps-2kp$$
$$r=s^2+k^2-p^2+2ps-2ks$$
The solution then is.
$$x=2ab$$
$$y=a^2-b^2$$
$$q=2cr$$
$$t=c^2-r^2$$
$$z=a^2+b^2$$
A: 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 $m=8\Rightarrow n=1.$
$$\text{Using Euclid's formula }f(7,4)=(33,56,65)\quad f(8,1)=(63,16,65)$$
