<s>[INST] 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 $(x,y)=(p,p)$ 
 [/INST] In the case that $a+b=a^2-ab+b^2$, we have $a^2+b^2=ab+a+b$, so $a^2+b^2=(a+1)b+a$. Without loss of generality (since the formula is symmetric in $a,b$) that $a\leq b$. Clearly, since $a+b$ is a prime, and $a,b=1$ isn't a solution, $a\neq b$. And, if $b>a+1$, then $a^2\geq a$ and $b^2>(a+1)b$, so $a^2+b^2>(a+1)b+a$, so there are no solutions.
Hence, any solutions to this case will be of the form $a, a+1$, for some natural $a$.
Now, we make that substitution to get $2a^2+2a+1=a^2+a+2a+1$, so $a^2=a$, so $a=0,1$. So, the only solution is $a=1,b=2$.  
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