## There are only finitely many Pythagorean triples (x,y,z) with any one parameter fixed

Suppose $(x,y,z)$ is a solution to $x^2 + y^2 = z^2$ in $\mathbb{N}^+$. Show that $x^2 \geq 2y+1$. Deduce that if we fix any of $x$, $y$, or $z$, then there are only finitely Pythagorean triples $(x,y,z)$.

Suppose $x^2 + y^2 = z^2$. Now $x^2 = z^2 - y^2$, and $z > y$. (We may assume that $xyz \neq 0$.) Now $x^2 = (z+y)(z-y) \geq z+y > 2y$. Since $x^2$ and $2y$ are integers, $x^2 \geq 2y+1$.

With $x$ fixed, there are only finitely many possible $y$, as they must satisfy $2y+1\leq x^2$. With $x$ and $y$ fixed, $z$ is determined, so there are only finitely many solutions with fixed $x$. Symmetrically, there are finitely many solutions with fixed $y$.

With $z$ fixed, we have $z > x,y$, so there are certainly only finitely many solutions.