Solve $y^2= x^3 − 33$ in integers This is not homework, could someone provide a nice clear proof as I have been struggling with this for some time.

Solve the equation $y^2= x^3 − 33$; $x, y \in \mathbb{Z}$


$$  y^2 + 25 = x^3 - 8.   $$
Note $y^2 + 25$ can not be divisible by any prime $q \equiv 3 \pmod 4,$ therefore not by any number $m \equiv 3 \pmod 4.$
If $y$ were odd, $y^2 + 25 \equiv 2 \pmod 4,$ impossible for $x^3 - 8.$ Therefore $y$ is even and $x$ is odd.
$$  y^2 + 25 = (x - 2)(x^2 + 2 x + 4).  $$
As we need $x-2 \equiv 1 \pmod 4,$ we know $x \equiv 3 \pmod 4.$ However, then $(x^2 + 2 x + 4) \equiv 1 + 2 + 4 \equiv 3 \pmod 4.  $ This is prohibited from dividing $y^2 + 25,$ so we have arrived at a contradiction of the equation in integers.
Done.
