A diophantine equation of degree 3

Question

Find the integer solutions of $y^2+6=x^3$. I guess it does not have integer solutions but I cannot prove it. By $\pmod 8$, I can know that $y$ is odd and $x\equiv7 \pmod 8$. Then what else can I do?

Answer 1

This paper talks a lot about Mordell's equations. I haven't read through all of it, but on Page 2, Theorem 2.3, there is a proof of how $y^2=x^3-6$ has no integral solutions. Here's an outline of the proof:

1. Prove $y$ is odd and $x \equiv 7 \pmod 8$.
2. Get $y^2-2=x^3-8=(x-2)(x^2+2x+4)$. Note that the latter factor is always positive and is $\equiv 3 \pmod 8$.
3. Therefore, $x^2+2x+4$ must have a prime factor that is $\equiv \pm3 \pmod 8$ because otherwise, all of its prime factors are $\equiv \pm1 \pmod 8$, making it $\equiv \pm1 \pmod 8$.
4. Let the prime be $p \equiv \pm3 \pmod 8$. (Note that $p$ is an odd prime.) Thus, $p$ divides $y^2-2$, so $y^2 \equiv 2 \pmod p$.
5. Thus, $2$ is congruent to a square $\pmod p$ where $p$ is an odd prime. Using quadratic residues, we know $p \equiv \pm1 \pmod 8$, so we have a contradiction with the fact that $p \equiv \pm3 \pmod 8$.