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This is not homework, could someone provide a nice clear proof as I have been struggling with this for some time.

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

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umm, solve, how? There are two variables and one equation – hjpotter92 Mar 15 '13 at 0:07
I would say that there is no solution (experimental evidence: there is none for $x \le 100000$). – George V. Williams Mar 15 '13 at 0:24
It's on my assignment due for tomorrow also (along with a couple of other of this users questions) and it really irks me that the same user keeps doing this every week. – Alex J Best Mar 15 '13 at 1:50
@AlexJBest, can you print out screen captures of the questions involved, with some of the answers, and give those to the person grading the assignments? I'm afraid I was not sufficiently suspicious this time. – Will Jagy Mar 15 '13 at 2:24
@WillJagy, that sounds like a good idea. Sorry to have caused a bother, being non-suspicious is best I think! – Alex J Best Mar 15 '13 at 2:28
up vote 13 down vote accepted

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

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.


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How do we see that $y^2+25$ can not be divisible by any such prime $q$ quickly? It took me some time. – Vincent Tjeng Mar 15 '13 at 0:53
@VincentTjeng, if $q \equiv 3 \pmod 4$ then $(-1|q)=-1$ Legendre symbol. If $a^2 + b^2 \equiv 0 \pmod q,$ and, say, $a \neq 0 \pmod q,$ then $b^2 \equiv -a^2 \pmod q,$ and $ (b/a)^2 \equiv -1 \pmod q,$ contradiction. So actually $a \equiv 0 \pmod q,$ then $b^2 \equiv 0 \pmod q$ gives also $b \equiv 0 \pmod q.$ – Will Jagy Mar 15 '13 at 1:06
nice! thanks :) – Vincent Tjeng Mar 15 '13 at 1:09
Just an aside, are there any solutions of this equation in rationals? I am quite interested in knowing its rank. And thanks for this good answer! – awllower Mar 15 '13 at 1:58
@WillJagy Thanks very much for your munificence. – awllower Mar 16 '13 at 7:11

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.


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.

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