Find all prime numbers such that the square of the prime number can be written as the sum of cubes of two positive integers.

Question

Find all prime numbers such that the square of the prime number can be written as the sum of cubes of two positive integers.
I will need either

$x+y=1, x^2−xy+y^2=p^2$
$x+y=p^2, x^2−xy+y^2=1$
$x+y=x^2−xy+y^2=p$.

I will investigate only the last. That the first two are impossible should be easy to show. So it boils down to finding all the integer solutions to the second degree polynomial in two variables.

We can subtract $xy$ from both sides and factor the right hand side:$x+y−xy=(x−y)^2$. Now we can use calculus to figure out that the left hand side will be negative for the vast majority of combinations x,y but the right hand side is a square which is non-negative. I don't know what to do next.

You may like to read Broughan's theorem, see the condition given in Tito Piezas' answer. Not recommending this as a duplicate because there could be a simpler answer, but in any case the general answer is known. — Sarvesh Ravichandran Iyer, Jun 19 '22 at 16:08

score 1 · Accepted Answer · edited Jun 19 '22 at 16:16

We wish to find all solutions to $p^2=a^3+b^3$, where $a,b\in\mathbb{N}$ and $p$ is a prime.

Note that $p>2$, since $4=a^3+b^3$ is impossible. Hence $p$ is odd, so that $a, $b$ are of opposite parity.

Since $a^3+b^3=(a+b)(a^2−ab+b^2)$ and $a+b>1$, the only possibilities are $a+b=p^2$. If $a+b=p^2$, then $a^2−ab+b^2=1$. But then $4=4(a^2−ab+b^2)=(2a−b)2+3b^2$, so that $2a−b=±1$ and $b=1$. This leads to $a=b=1$, and $a+b=2≠p^2$.

Hence, $a+b=a^2−ab+b^2=p$. This gives $p=(a+b)^2–3ab=p^2–3ab$ , or $3ab=p(p−1)$ . Thus, $p=3$, or $p∣a$, or $p∣b$; we may assume $p∣a$, without loss of generality.

If $p>3$, then $p∣a$, so that $a^3+b^3>a^3≥p^3>p^2$. So $p=3$, $ab=p−1=2$, and ${a,b}={1,2}.$

The only solution is $1^3+2^3=3^2$.

Find all prime numbers such that the square of the prime number can be written as the sum of cubes of two positive integers.

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