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

Martin Sleziak
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user46220
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    umm, solve, how? There are two variables and one equation – hjpotter92 Mar 15 '13 at 00:07
  • Wolfram can't solve it... (fyi) – apnorton Mar 15 '13 at 00:08
  • @BackinaFlash It's a Diophantine-esque equation. They tend to have more variables than equations. – apnorton Mar 15 '13 at 00:09
  • If there is no solution you might check whether it fails to be solved mod some prime. – Lepidopterist Mar 15 '13 at 00:23
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    I would say that there is no solution (experimental evidence: there is none for $x \le 100000$). – George V. Williams Mar 15 '13 at 00:24
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    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 01:50
  • @AlexJBest, this is someone in your class? – Will Jagy Mar 15 '13 at 02:16
  • @WillJagy Yeah. – Alex J Best Mar 15 '13 at 02:19
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    @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 02:24
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    @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 02:28
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    @Alex, unfortunately as I do not know the assignment, I cannot rationalize taking it down. But it is wrong to cheat and ask others to do your own work. I think Will's suggestion of you submitting these pages to the professor/grader would be best. – davidlowryduda Mar 15 '13 at 06:56
  • I'd just like to say that OP should have been able to work out the proof for this based on the proof given from a previous answer to a question he asked ... http://math.stackexchange.com/questions/222093/show-that-the-equation-y2-x3-7-has-no-integral-solutions – Vincent Tjeng Mar 15 '13 at 16:05
  • @AlexJBest, perhaps too late for this week...I have successfully identified cheaters in the past, and can say two simple things. (A) it is worth doing so, instructors are interested, plus it calms one down a bit (B) the penalty for the cheater is generally not large. Oh, as far as deleting questions on MSE, it is mostly the cheaters who desire that, it hides the evidence. – Will Jagy Mar 15 '13 at 20:06

2 Answers2

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

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.

Will Jagy
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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 00:53
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    @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 01:06
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    nice! thanks :) – Vincent Tjeng Mar 15 '13 at 01:09
  • 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 01:58
  • @awllower, apparently not. The book I'm using gives as reference Cassels (1950), see http://link.springer.com/article/10.1007%2FBF02398279?LI=true I might expect the same information to be in Mordell's book Diophantine Equations. – Will Jagy Mar 15 '13 at 02:10
  • I see. Thanks for the information. – awllower Mar 15 '13 at 05:02
  • @awllower, I cannot help you with the rank as a question in its own right. I can say that I found the tables from Cassels (1950) repeated in H. E. Rose, A Course in Number Theory, pages 344-345 in the first edition. I can email you a scan of those two pages if you write to me, I don't think I will post the image here. – Will Jagy Mar 15 '13 at 20:12
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    @WillJagy Thanks very much for your munificence. – awllower Mar 16 '13 at 07:11
  • @awllower I just checked with Magma that the curve has rank 0, and trivial torsion subgroup, so there are no rational points at all (except the one at infinity). – Álvaro Lozano-Robledo Mar 18 '13 at 02:35
  • @ÁlvaroLozano-Robledo It is indeed a surprising result, in view of its similarity with $y^2=x^3-2$. In any case, thanks for sharing this information. – awllower Mar 18 '13 at 04:36
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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.

Vincent Tjeng
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