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

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 $x\equiv_4 3$ ....
