The equation $b^2=a(a^2-1)$ has no rational solutions except obvious ones I have problem with equation $b^2=a(a^2-1)$. How can I show that except $(a,b)=(1, 0), (-1,0),(0,0)$, the equation hasn't any other rational solutions ?

Editor's note. AFAICT the exercise is about showing that the rational points of the elliptic curve
$$y^2=x^3-x$$
are exactly the two-torsion points. This is probably known, but I am unfamiliar with this theory, so cannot point the OP to a resource, JL.
 A: The elliptic curve $b^2=a^3-a$ has a nontrivial rational solution iff $1$ is a congruent number. This is not the case, $1$ is not a congruent number. It is elementary to see that this would give a non-trivial integral solution to Fermat's equation $x^4+y^4=z^4$ (see exercise 3 in https://raw.github.com/williamstein/simuw12/master/day06/1.pdf‎).
A: $$gcd(a,a^2-1)=1$$Which implies the only four combinations are
$$a=b^2 \land a^2-1=1 \lor a=1 \land a^2-1=b^2\\a=\sqrt2,b=\sqrt{\sqrt{2}} \lor a=1,b=0$$
And the negative solutions
$$a=-b^2\land a^2-1=-1 \lor a=-1 \land a^2-1=-b^2\\a=b=0 \lor a=-1,b=0$$
A: Suppose a rational solution exists, namely; $b=p/q$, and $a=r/s$ where $gcd(p,q)=1=gcd(r,s)$ otherwise cancel and start afresh. We have, $(p/q)^2=(r/s)[(r/s)^2-1]$ so that 
$p^2=q^2*r(r^2-s^2)/s^3$, from here we must have at least $s^3|q^2$ similarly, we can show that $q^2|s^3$, so that $q^2=s^3$, now there are two cases; either $q=s$ [in which case we have $q^2=q^3$ or $q=s=1$, so that we are confined to integers whose solutions are the trivial cases provided above] or $gcd(q,s)=k$, so let $q=kq'$ and $s=ks'$, where $gcd(q',s')=1$ then, $k^2*q'^2=k^3*s'^3$ or $q'^2/s'^3=k$, which is impossible since $gcd(q',s')=1$, hence no solution exist.
