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 $z$?

Question

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.

Answer 1

$$x^2+y^2=z^2$$ $$169=z^2-y^2$$ $$169=(z+y)(z-y)$$ $$169=169(169-2y)$$ $$2y=168$$ $$y=84$$

So $y=84,z=85$ is the only solution.

Answer 2

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 $\mathbb{R}$.