= No integer solutions to y cubed equals x squared plus thirty
{title2=$y^3=x^2+30$}
The equation has no <integer> solutions. Work in the <ring of integers> $\mathbb Z[\theta]$ for $\theta=\sqrt{-30}$. If a rational prime divides both $y$ and thirty, it divides $x$, and the valuation of $x^2+30$ is one because thirty is squarefree, contradicting the valuation of a cube. Thus $\gcd(y,30)=1$. The two <principal ideals> $(x+\theta),(x-\theta)$ are coprime: any common <prime ideal> lies above a prime dividing both $N(2\theta)=120$ and $y$. Their product is $(y)^3$, so unique <prime ideal> factorization gives $(x+\theta)=A^3$. The <Ideal class group of Q of square root minus thirty> has exponent two, forcing $A$ to be principal. The only units are $\pm1$ and both are cubes, so $x+\theta=(a+b\theta)^3$ for <integers> $a,b$. Comparison of the coefficient of $\theta$ gives $1=3b(a^2-10b^2)$, impossible modulo three.
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