= Integral solutions of x squared minus two equals y cubed
{title2=$x^2-2=y^3\ \Longrightarrow\ (x,y)=(\pm1,-1)$}
In the <ring of integers> $\mathbb Z[\sqrt{-6}]$, which has class number two and units $\pm1$, write $\alpha=x+(y+1)\sqrt{-6}$. Its norm is $(y+2)^3$. Parity modulo eight and cubic residues modulo nine show that $y$ is odd and $3$ does not divide $y+2$, so $\gcd(y+2,6)=1$. The ideals of $\alpha$ and its conjugate are coprime: a common prime cannot lie above two or three by their norms; at any other prime their difference forces divisibility of $y+1$, contradicting divisibility of $y+2$.
Unique factorization of ideals gives $(\alpha)=A^3$. The class group of order two forces $A$ principal, and its unit factor can be absorbed into a cube. Hence $\alpha=(u+v\sqrt{-6})^3$ with integers $u,v$, and $u^2+6v^2=y+2$. Comparing coefficients and eliminating $y$ yields
$$
9u^2=18v^2+24v+8-\frac1{3v-1}.
$$
Integrality requires $3v-1=\pm1$, so $v=0$ is the only integral possibility. Then $u=\pm1$, $y=-1$, and $x=\pm1$. Direct substitution verifies both solutions. The class-group step is what permits extracting a cube of an element from a cube of an ideal despite failure of unique factorization of elements.
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