For Pythagorean triple $x^2+y^2=z^2$, if $x=13$ and $y+z=169$, then how can I determine all possible $y$ and $z$? 
If I know that $x=13$, that $x^2+y^2=z^2$ and $y+z=169$, how can I determine all possible values for $y$ and $z$?

I know that one possibility (if not only one) is $84$ and $85$, but was curious as to how this would be found.
 A: $$x^2+y^2=z^2$$
$$169=z^2-y^2$$
$$169=(z+y)(z-y)$$
$$169=169(169-2y)$$
$$2y=168$$
$$y=84$$
So $y=84,z=85$ is the only solution.
A: Here's how I  would do it.
With the first equation:
\begin{gather}
x = 13\\
x^2 = 169 \\
\end{gather}
Hence the second equation becomes:
\begin{gather}
169 + y^2 = z^2 \\
y^2 - z^2 = -169\\
(y + z)\cdot(y - z) = -169\\
169 (y - z ) = -169\\
y -z = -1\\
z = y+1\\
\end{gather}
And for the third equation, we obtain:
\begin{gather}
y + (y+1) = 169\\
2y = 168\\
y = 84\\
\end{gather}
Hence we selected only one solution, $(x=13,y=84,z=85)$, even over $\mathbb{R}$.
A: Another method is to solve the $A$-function of Euclid's equation for $n$ and test a defined range of $m$-values to see which, if any, yield integers.
$$A=m^2-n^2\implies n=\sqrt{m^2-A}\quad\text{where}\quad \lceil\sqrt{A+1}\space\rceil\le m \le \biggl\lceil\frac{A}{2}\bigg\rceil$$
$$\text{For }A=13\qquad n=\sqrt{m^2-13}\quad\text{where}\quad \lceil\sqrt{13+1}\space\rceil=4\le m \le \biggl\lceil\frac{13}{2}\bigg\rceil=7$$
Only $m=7$ yields and integer: $n=\sqrt{49-13}=6$ and $F(7,6)=(13,84,85)$
