Why does $x^2+41=y^3$ have no integer solutions?

I know how to find solutions for some of the Mordell's equations $(x^2=y^3+k)$ (using $\mathbb{Z}[\sqrt{-k}]$, and arguments of the sort). Still, I can't find a way to prove this particular equation has no integer solutions.

I know it doesn't because there are lists of solutions for $k \in [-100,100]$, like this one Numbers n such that Mordell's equation $y^2 = x^3 - n$ has no integral solutions and for what I've seen, the argument one uses to say a given Mordell equation has not solution is based on congruences, like in here.

Still, I can't seem to figure out why the one with $-41$ (meaning $x^2=y^3-41$) has no solutions. There's a theorem in the Apostol book (p191) that says that such an equation has no solution if $k$ has the form $k=(4n-1)^3-4m^2$, with $m$ and $n$ integers such that no prime $p\equiv -1 \pmod 4$ divides $m$. I've tried with this but can't find $m$ and $n$.

Any ideas?

  • $\begingroup$ It's a perfectly standard argument using quadratic number fields. $\endgroup$ – franz lemmermeyer May 11 '17 at 15:51

write the eqn as $X^2+7^2=Y^3+8=(Y+2)(Y^2-2Y+4)$ $X$ has to be even and hence $Y$ is odd. Thus we get $(Y^2-2Y+4)$ is congruent to $-1$ mod $4$. So it has a prime of the form $4k-1$ with an odd exponent in the prime factorization. $X^2+7^2$ can have only one prime of the form $4k-1$ namely $7$. So $7|gcd(Y+2,Y^2-2Y+4)$ but the $gcd$ is a factor $2^2+2.2+4=12$. Hence a contradiction.


$$x^2 + 49 = y^3 + 8$$ If $x$ were odd, then $x^2 + 49 \equiv 2 \pmod 8.$ Then $y$ would be even. However, this would give $y^3 + 8 \equiv 0 \pmod 8.$

So, in fact, $x$ is even and $y$ odd. Seems the other answer finishes $$ $$

  • $\begingroup$ Oh wow thank you. That is indeed insultingly easy haha, I guess sometimes you just don't see it. Thanks!! $\endgroup$ – Laura May 11 '17 at 23:03
  • $\begingroup$ @Laura - it is a good habit here, when some answer answers the question satisfyingly then to "accept" that answer so everybody knows the problem was solved. (Such behave gives the answerer and even yourself a bit more of "reputation"... :-) ) $\endgroup$ – Gottfried Helms May 13 '17 at 9:35

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