rightways wrote: Let $a$ be a positive integer. Prove that for any pair $(x,y)$ of integer solutions of equation $$x(y^2-2x^2)+x+y+a=0$$we have: $$|x| \leqslant a+\sqrt{2a^2+2}$$ Note that $a+\sqrt{2a^2+2}\ge 3$ since $a$ is a positive integer. So inequality is always true when $|x|\le 3$ Let us consider now $|x|> 3$ We want to prove $|x|-a\le \sqrt{2a^2+2}$ $\iff$ $a\ge |x|$ or $(|x|-a)^2\le 2a^2+2$ $\iff$ $a\ge |x|$ or $x^2-2\le a^2+2a|x|$ $\iff$ $a\ge |x|$ or $\sqrt{2x^2-2}\le a+|x|$ $\iff$ $a\ge |x|$ or $a\ge |x|-\sqrt{2x^2-2})$ And so we want to prove $\frac a{|x|}\ge\sqrt{2-\frac 2{x^2}}-1$ Let $y^2-2x^2=n-1$ where $n$ is an integer $\ne 1$ Equation gives $a=-nx-y$ Since we want to prove in the worst case, we can consider $a=||nx|-|y||=||nx|-\sqrt{2x^2+n-1}|$ And so $\frac a{|x|}=||n|-\sqrt{2+\frac{n-1}{x^2}}|$ And problem is to prove $||n|-\sqrt{2+\frac{n-1}{x^2}}|\ge \sqrt{2-\frac 2{x^2}}-1$ $n=1$ is forbidden If $n=-1$, inequality is true (and in fact is equality) Note that this equality case, with $|x|>3$, is reached for example with $(x,y,a)=(-17,-24,7)$, and infinitely many other. If $|n|\ge 2$, inequality is $|n|-\sqrt{2+\frac{n-1}{x^2}}\ge \sqrt{2-\frac 2{x^2}}-1$ Or $|n|+1\ge \sqrt{2+\frac{n-1}{x^2}}+\sqrt{2-\frac 2{x^2}}$ Since $\sqrt{2-\frac 2{x^2}}\le\sqrt 2$ and $\sqrt{2+\frac{n-1}{x^2}}\le \sqrt 2(1+\frac{n-1}{4x^2})$, enough to prove $|n|+1\ge \sqrt 2+\sqrt 2(1+\frac{n-1}{4x^2})$, which is easily proved. Q.E.D.