= No integral solutions to y cubed equals x squared plus twenty-two
{title2=$y^3=x^2+22$}
The equation $y^3=x^2+22$ has no integer solutions. Factoring it in $\mathbb Z[\sqrt{-22}]$, the two conjugate factors are coprime. <Unique factorization of ideals in a number field> and the absence of three-torsion in the class group force $x+\sqrt{-22}$ to be a cube up to a unit. Writing it as $(a+b\sqrt{-22})^3$ makes its $\sqrt{-22}$ coefficient
$$
b(3a^2-22b^2)=1,
$$
which has no integer solution.
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