Natural number solutions of $x^3+y^3=p^2$, $x$ and $y$ are integers, $p$ is prime number.

Question

Natural number solutions of $x^3+y^3=p^2$, $x$ and $y$ are integers, $p$ is prime number.

I have found $(1,2,3)$ is a solution and there seem to be no other solution.

Can anyone prove it?

Answer 1

Hint: Since $p$ is a prime number, we must have : $x+y = p, x ^2 - xy + y^2 = p$ or $x + y = 1, x^2 - xy + y^2 = p^2$ or $x+y = p^2, x^2-xy+y^2 = 1$. Either case is quite simple enough to handle. But for the last case which is the case the OP pointed out as a missing case in the comment.We claim that this case can't happen.For if it were, then since $p^2 > 0$, $x,y$ can't be both negative. Thus if both are $>0$, then let's say if $x > y$, then $x^2-xy+y^2 = x(x-y)+y^2 \ge x+y^2 > x+y \implies 1 > p^2$, contradiction. If $xy < 0$, then $x^2-xy+y^2 > x^2+y^2 > x+y^2 > x+y \implies x^2 -xy+y^2 > x+y \implies 1 > p^2$ contradiction again. Thus this case is ruled out.