Square-Product-Square Primes

If $\omega =-\tfrac12 + \tfrac12i\sqrt{3}$ is a primitive cube root of unity, then the ring $R = \mathbb{Z}[\omega]$ is an Euclidean domain with distance function $N(u + v\omega) \; = \; |u + v\omega|^2 \; = \; u^2 - uv + v^2 \hspace{2cm} u,v \in \mathbb{Z}$ If $a,b \in \mathbb{N}$ are such that $a^2 + ab + b^2 = p > 3$ is prime, then $a - b\omega \in R$ is such that $N(a - b\omega)=p$ , and hence $a - b\omega$ is irreducible in $R$ . Similarly, $a - b\omega^2\in R$ is irreducible.

If $c,d \in \mathbb{N}$ are such that $c^2 - cd + d^2 = p$ , then $(c + d\omega)(c + d\omega^2) \; = \; (a - b\omega)(a - b\omega^2)$ and hence either $c + d\omega = u(a - b\omega)$ or $c + d\omega = u(a - b\omega^2)$ for some unit $u \in R$ . The units of $R$ are $\pm1,\pm\omega,\pm\omega^2$ . Checking the twelve possibilities for $c +d\omega$ for solutions where both $c$ and $d$ are positive, we deduce that we must have $(c,d) = (a,a+b)$ , $(b,a+b)$ , $(a+b,a)$ or $(a+b,b)$ . If $a=b$ , then $p = 3a^2$ , and hence $3$ divides $p$ , which is not possible, since $p>3$ . Thus we deduce that $a \neq b$ , and hence there are four distinct solutions for $(c,d)$ . Thus the sum of their values is $a + (a+b) + b + (a+b) + (a+b) + a + (a+b) + b \; = \; 6(a+b)$ making the result $\boxed{\mathrm{True}}$ .