Sum of two cubes equal to prime square If $a,b\in \mathbb{N}$ find all primes $p$ such that $a^3+b^3=p^2$
My approach-
$a^3+b^3=(a+b)(a^2-ab+b^2)=p^2$ suppose $a+b=x$ and $a^2-ab+b^2=y$ then there are two cases- $(x,y)=(p^2,1),(p,p)$
Now I am struggling for case 02 where $(x,y)=(p,p)$ 

In the case $(x,y)=(p,p)$, we have $a+b=(a+b)^2-3ab=p$, hence $p^2\equiv p\pmod 3$.
If $p\neq 3$, then $p\equiv 1\pmod 3$, hence
$$ab=\frac{p-1}3p$$
Then $p|(ab)$ but since $0<a,b<a+b=p$ this is impossible.
