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.

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