Integral solution of $y^2=x^3-51$

Question

I have the following equation: $y^2 = x^3 -51$. I want to show that this equation has no integral solutions.

I found out that $x$ must be odd and that $\gcd(y,51)= 1$. But how can we finally show that there is no integral solution? (By contradiction I guess, but how exactly?)¨

Thanks for any help!

Answer 1

Since $\mathbb{Q}(\sqrt{-51})$ has class number $2$, it is easy to solve this. Firstly, note that $y$ is even and $x$ is odd.

Consider $$(y-\sqrt{-51})(y+\sqrt{-51}) = x^3$$ any prime ideal divisor of two terms on left must lie above $2$ or $3$ or $17$, since $x$ cannot be even nor multiple of $3$ nor $17$, the two terms are coprime.

Therefore $(y+\sqrt{-51})$ is the cube of an ideal, which must be principal from class number, so $$y+\sqrt{-51} = \left(\dfrac{a+b\sqrt{-51}}{2}\right)^3$$ for integers $a,b$, which implies $$3a^2b-51b^3 = 8$$ this has no integer solutions, so neither does the original.