A [[Proof]] that is [[Truth Value|proved]] by checking all cases. >[!example] > >Prove that $x^2+y^2=3$ has no integer solutions. > >*Proof*: > >We only need to check positive values $x, y$, since if $x,y$ is a solution, then $-x, y$, $x,-y$, $-x,-y$ are all solutions. > >Because $x^2>0\wedge y^2>0$, $x^2\leq 3$ and $y^2\leq{3}$. > >If $x \ge 2 \or y\ge 2$, then $x^2+y^2\ge 4 \ge3$. > >If $x<2,y<2$, $x^2+y^2 = \set{0, 1}$, therefore we have checked all possible cases and the proposition is true.