CatalinBordea wrote: Prove that the number of solutions in $ \left( \mathbb{N}\cup\{ 0 \} \right)\times \left( \mathbb{N}\cup\{ 0 \} \right)\times \left( \mathbb{N}\cup\{ 0 \} \right) $ of the parametric equation $$ \sqrt{x^2+y+n}+\sqrt{y^2+x+n} = z, $$is greater than zero and finite, for nay natural number $ n. $ 1) Set of solution is non empty Just consider $(x,y,z)=(n-1,n-1,2n)$ Q.E.D. 2) Set of solutions is finite Equation implies that both $x^2+y+n$ and $y^2+x+n$ are perfect squares And so that $y+n\ge 2x+1$ and $x+n\ge 2y+1$ Adding, we get $x+y\le 2n-2$ Q.E.D.