# How many solutions to prime = $2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$

Let $a,b,c$ be integers, no sign restriction. Let $p$ be a given prime. How to find the number of solutions to $p = 2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ ?

Note, from Heron's formula, $$2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4 = (a + b + c) ( b + c - a) ( c + a - b) (a + b - c).$$

Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ below $w$. How does the function $f(w)$ behave ? How fast does it grow ? Are those primes of type $A$ $mod$ $B$ for some integers $A$ and $B$ ? How to deal with this ?

Can this be solved without computing the class number ?

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What do these questions have to do with class number? –  Old John Feb 11 '13 at 1:41
@OldJohn, I have no idea. –  Will Jagy Feb 11 '13 at 2:27

It seems the only prime representable this way is $p=3$, which has eight such representations by using $a,b,c=\pm 1$ (independent sign choices). [Note that $-3$, which is prime, has no such representations.] I ran a check of all choices of $a,b,c$ between $-100$ and $100$ just to make sure of this.
It wouldn't be hard to prove this, since three of the four factors must be $\pm 1$ and the fourth $\pm p$ for some prime $p$. By looking at all 32 ways of setting three of the four factors to $\pm 1$, and solving each such system, one gets (at most) 32 cases to look at, and for each one computes the value of the fourth factor.