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.

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 $F(7,6)=(13,84,85)$
