Integer solutions for $x^3=y^2+5$

Question

I have been able to solve the equation $x^3=y^2+a$ for integers where $1 \leq a \leq 4$ by splitting in $\Bbb Z [i\sqrt a]$. However as a natural continuation I would like to know whether the equation $x^3=y^2+5$ can be solved using elementary methods. Seems this case is much tougher though .I would like some hints on how to proceed.

Answer 1

Working mod 4 we see that $y$ must be even and $x=1\bmod4$. Now we have $$y^2+4=x^3-1=(x-1)(x^2+x+1)$$ But $x^2+x+1=3\bmod4$, so $x^2+x+1$ is odd. It is also positive, so it must be at least 3. Hence it must have a prime factor $p=3\bmod4$. So $y^2+4=0\bmod p$, in other words -4 is a quadratic residue of $p$. But that implies that $p=1\bmod 4$. Contradiction.

Answer 2

Write $$ (x-1)(x^2+x+1) = y^2+4$$

If $x\equiv_4 0$ then $x-1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x-1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!

If $x\equiv_4 1$ then $x^2+x+1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x^2+x+1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!

If $x\equiv_4 2$ then $x^2+x+1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x^2+x+1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!

We are left with the case $x\equiv_4 3$ ....