<s>[INST] Rational solutions of $y^2 = x^3 - x$ I believe that the only rational solutions of $$y^2 = x^3 - x$$ are the obvious ones $(-1,0)$, $(0,0)$, $(1,0)$, and that this was proved by Fermat using the method of descent.
Can anyone outline a proof, either Fermat’s proof or a different one; or, failing that, point to a suitable reference, preferably available online?
 [/INST] 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.
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