Proof that $y^2=x^3+x$ has a unique integer solution

Question

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

Answer 1

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 $x,a\in \mathbb{Z}$.

Answer 2

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 $a=0$.