Proof that $y^2=x^3+x$ has a unique integer solution Prove that the equation $y^2=x^3+x$ has only one integer solution, namely $x=y=0$.

Write $y^2=x(x^2+1)$. If $x=0$ then $y=0$ and if $y=0$ then $x=0$. We can then assume that $xy\not=0$, and replace $x$ with $-x$ or $y$ with $-y$ so that $x>0$, $y>0$. Observe that $x$ and $x^2+1$ are coprime, so since $x(x^2+1)$ is a square, so is $x$ and so is $x^2+1$. This is impossible because $x^2+1=a^2$ is impossible as $x>0$ and $x,a\in \mathbb{Z}$.
