Prove that the equation $y^2=x^3-73$ has no integer solutions

Question

Prove that there are no integers $x,y$ such that $y^2=x^3-73$. Thank you.

Answer 1

Equations of the form $$ y^2 = x^3 + k$$ are known as Bachet equations. I will quote the statement of Theorem 4.2 from Richard Mollin's Algebraic Number Theory:

Let $F=\mathbb{Q}(\sqrt{k})$ be a complex quadratic field with radicand $k< -1$ such that $k \neq 1 \pmod 4,$ and $h_{\mathfrak{D}_F} \neq 0\pmod 3.$ Then there are no solutions to the Batchet equation in integers $x,y$ except in the following cases: there exists an integer $u$ such that $$(k,x,y) = (\pm 1-3u^2, 4u^2 \mp 1, \epsilon \cdot u(3 \mp 8u^2) ),$$

where the $\pm$ signs correspond to the $\mp$ signs and $\epsilon = \pm 1$ is allowed in either case.