# Prove the Mordell's equation $x^2+41=y^3$ has no integer solutions

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?

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

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