Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Assume that the prime is $p$.

$p^2 = a^3+b^3 = (a+b) (a^2-ab+b^2)$

Case 1:

$a+b=1$ and $a^2-ab+b^2 = p^2$

Given that $a,b\geq1$, $a+b\geq2$ which contradicts the case.

Case 2:

$a+b=p^2$ and $a^2-ab+b^2 = 1\implies (a-b)^2 = 1-ab$

But $(a-b)^2\geq 0$ and $a-b = 1-ab<1$

As $(a-b)$ is an integer, only possible value of $(a-b)^2$ is 0 which implies $a=b$

So, $2a^2-a^2 = 1 \implies a =b= 1 \implies p^2 = 2$ which also doesn't have any integer solution.

Case 3:

$a+b=p\implies a=p-b$ 

And $a^2-ab+b^2=p\implies (p-b)^2-(p-b)b+b^2=p\implies  p^2-3pb+3b^2=p$

$\implies p(3b+1-p) = 3b^2 \implies p\mid 3b^2$

Either, $p\mid 3 \implies p=3$ which is $p^2=9$.

9 can be expressed as $a^3+b^3$ as $1^3+2^3$ and $2^3+1^3$ and these are the only ways to do that.

So, $(a,b,p) = (1,2,3)$ and $(2,1,3)$ work.

Or, $p\mid b$. Similarly, $p\mid a$.

Assume that $a = pA$ and $b = pB$

So, $p^2 = a^3+b^3 = p^3(A^3+B^3) \implies p(A^3+B^3) = 1$

But, that is not possible for any value of $p$.

So, the only solutions are 

$(a,b,p) = (1,2,3)$ and $(2,1,3)$

Key points:

• Factorize $a^3+b^3$.
• $p$ can not be the multiplication of two integers greater than 1. So, there are 3 cases.
• The cases are $a+b=1,~a+b=p^2$ and $a+b=p$
• $a+b\geq 2$ eliminates the first case. For the second case, $a^2-ab+b^2$ can be bounded.
• Solve the third case using $a=p-b$ and expanding $a$ in the other equation.