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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?

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  • $\begingroup$ It's a perfectly standard argument using quadratic number fields. $\endgroup$ May 11, 2017 at 15:51

2 Answers 2

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Write the equation 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 $y^2-2y+4\equiv-1\bmod 4$. 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 $7\mid\gcd(y+2,y^2-2y+4)$, but the $\gcd$ is a factor of $2^2+2\cdot2+4=12$. Hence a contradiction.

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$$x^2 + 49 = y^3 + 8$$ 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 $$ $$

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