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

Solution 1:

$$ y^2 + 25 = x^3 - 8. $$

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.

Solution 2:

The equation is a Mordell Curve, and the number of solutions can be found at this link. Inspired by the proof here, which works for certain types of Mordell Curves, we can proceed as follows.

Firstly, if $x$ is even, then

$$y^2 \equiv -33 \equiv 3 \mod 4$$

This is not possible; therefore, $x$ must be odd.

We re-write the equation as follows to allow the RHS to be factorized.

$$y^2+25=x^3-8=(x-2)(x^2+2x+4)$$

Note that $x^2+2x+4=(x+1)^2+3>0$, and since $x$ is odd,

$$(x+1)^2+3\equiv 3 \mod 4$$

Therefore, $x^2+2x+4$ must have a prime factor $p \equiv 3 \mod 4$, and thus $p|(y^2+25)$.

But this implies that $$y^2\equiv -25 \mod p$$

and the above equation has no solutions since $25$ is a quadratic residue for any such $p$, and since $p \equiv 3 \mod 4$, the negative of a residue modulo p is a non-residue.

Thus, the original equation admits no solutions too.