Rational solutions of $y^2 = x^3 - x$

If the equation $y^2=x^3-x$ has rational soloutions in which $x,y$ are each nonzero, we may put $y=tx$ and after division by $x$ we have the quadratic equation for $x$ $$x^2-t^2x-1=0 \tag{1}$$ whose discriminant $t^4+4$ must be square. Put $t=m/n$ with $\gcd(m,n)=1.$ Here $m,n$ are nonzero and $m \neq n$ since $m=n$ means $t=1$ in which case (1) has no rational solution. Then $m^4+4n^4=(m^2)^2+(2n^2)^2=w^2$ for some $w.$ Now here we may assume that $m$ is odd, otherwise put $m=2k$ and after a few steps get to $4(k^2)^2+(n^2)^2=s^2$ where $s=w/2.$ So we have $$(m^2)^2+(2n^2)^2=w^2$$ where $m$ is odd, and since $m,n$ are nonzero with gcd $1$ this is a primitive Pythagorean triple. So we may write $2n^2=2pq,\ m^2=p^2-q^2$ with $\gcd(p,q)=1$ and $p,q$ of opposite parity. Then from this, $n^2=pq$ forces each of $p,q$ to be squares, say $p=u^2,q=v^2.$ This when put into the equation for $m^2$ then gives the equation $u^4-v^4=m^2.$ That the latter equation has no solutions in nonzero and unequal $u,v$ has proofs using Fermat's descent method. As I recall one proof lies in his proof that the aea of a right triangle with integer length sides cannot be a square, at any rate see for example Dicksons History of the Theory of Numbers, Volume II Ch XXII starting at page 615 in the Dover edition. There are probably good treatments of this equation via descent at websites.

Here is a link to the wiki page, skip down to something like "proofs for specific exponents" for the statement re. difference of fourth powers a square.

For the sake of any future readers, I thought I would extend coffeemath’s excellent answer a little to give a self-contained proof, since it’s quite easy.

Primitive Pythagorean triples

We briefly recall the classification of primitive Pythagorean triples. Let $x^2+y^2=z^2$ with $x$, $y$ coprime natural numbers. An odd square is congruent to $1$ modulo $4$, and an even square is congruent to $0$, so the sum of two odd squares can never be a square. Therefore one of $x$ and $y$ must be even. Since they are coprime, it follows that the other is odd. So suppose $x=2k$ for some $k$, and $y, z$ are odd. Then we have $4k^2+y^2=z^2$, hence $$k^2=(\frac{z+y}{2})(\frac{z-y}{2}).$$It follows (since they are coprime) that $\frac{z+y}{2}$ and $\frac{z-y}{2}$ are both square, so let $a^2=\frac{z-y}{2}$ and $b^2=\frac{z+y}{2}$. Then $$x=2ab,\qquad y=b^2-a^2,\qquad z=a^2+b^2$$ with $a<b$ and $a,b$ coprime. Note that, since $y$ and $z$ are odd, one of $a,b$ is odd and the other even.

The main proof

We want to show that the equation $$x^4+4y^4=z^2$$ has no non-trivial integer solutions. The strategy of the proof – Fermat’s method of descent – is to show that, if it does, then there is another triple of integers $p,q,r$ with $p^4+4q^4=r^2$ having $0<r<z$.

If $x,y$ have a common prime factor $w$ then $(\frac x w)^4+4(\frac y w)^4$ = $(\frac{z}{w^2})^2$, and we are done. Otherwise $x,y$ are coprime. By the classification above we have that $x$ is odd and $$2y^2=2ab,\qquad x^2=b^2-a^2,\qquad z^2=a^2+b^2$$ for $a<b$ with $a,b$ coprime. Since $y^2=ab$, it follows that $a$ and $b$ are both square, say $a=c^2$ and $b=d^2$. Therefore $x^2=d^4-c^4$, which we rewrite as $$x^2+c^4=d^4.$$Using the classification again, recalling that $x$ is odd, we have$$c^2=2ef,\qquad x=f^2-e^2,\qquad d^2=e^2+f^2$$ for some coprime $e,f$. Since $c^2=2ef$ we know that one of $e,f$ is a square and the other is twice a square, so $e,f$ are equal in some order to $p^2,2q^2$ for some $p,q$, and since $d^2=e^2+f^2$ it follows that $$p^4+4q^4=d^2$$ where $d<d^2=b<z$, as required.