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 $x=y=0$.

We see that $x$ divides $y^2$. But $x^2+1$ is prime with $x$. Therefore any prime dividing $x$ must appear with an even power. Therefore $x=a^2$ and $a$ divides $y$.
Putting this into the equation we get $y=ay_1$ and
$$a^2y_1^2=a^2(a^4+1).$$
If $a\neq 0$, then $$y_1^2=a^4+1$$
or $$(y+a^2)(y-a^2)=1.$$
Therefore $a=0$.
