Past exam of the number theory course of the University of Oxford / 2008 / B / 8

For integral solutions of x squared minus two equals y cubed, the equation implies $y\ge -1$, so $y+2>0$. If $y$ were even, then $x$ would be even and $x^2-2$ would be two or six modulo eight, whereas $y^3$ would be zero modulo eight. Thus $y$ and $x$ are odd. Also $3\nmid x$: otherwise $x^2-2\equiv 7(\mathrm{mod}9)$, which is not a cubic residue modulo nine. Reducing modulo three then gives $y\equiv 2(\mathrm{mod}3)$. ConsequentlyPut $\eta =\sqrt{-6}$ and $\alpha =x+(y+1)\eta$. Its field norm is $(y+2{)}^3$. The principal ideals $(\alpha )$ and $(\bar{\alpha })$ are coprime. Indeed a common prime ideal has underlying rational prime $p$ dividing $y+2$. The displayed gcd excludes two and three, so both two and $\eta$ are invertible modulo that prime ideal. The difference $\alpha -\bar{\alpha }=2(y+1)\eta$ therefore implies $p\mid y+1$, contradicting $p\mid y+2$.

Their product is $(y+2{)}^3$, so unique factorization of ideals in a number field and coprimality imply $(\alpha )=A^3$ for an integral ideal $A$. In the given ideal class group of order two, $[A{]}^3=1$ implies $[A]=1$. Hence $A=(u+v\eta )$ with integers $u,v$. Equality of principal ideals gives $\alpha =\epsilon (u+v\eta {)}^3$. The only units are $\pm 1$, and both are cubes, so the sign can be absorbed into $u,v$. Taking norms, and using $y+2>0$, givesThe coefficient of $\eta$ gives $y+1=3u^{2}v-6v^3$. Eliminating $y$ with the field norm identity yieldsThe denominator is nonzero for integral $v$. Polynomial division makes the integrality obstruction especially clear:If $u$ is integral, the reciprocal must be an integer, forcing $3v-1=\pm 1$. The only integral possibility is $v=0$. Then $u^2=1$, the field norm gives $y=-1$, and the real coefficient in the cube gives $x=u^3=\pm 1$. Both pairs directly satisfy the original equation. Therefore all integral solutions are