How many solutions to prime = $a^3+b^3+c^3 - 3abc$ Let $a,b,c$ be integers.
Let $p$ be a given prime. 
How to find the number of solutions to $p = a^3+b^3+c^3 - 3abc$ ?
Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $a^3+b^3+c^3 - 3abc$ 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 ?

Andre's quadratic form $a^2 + b^2 + c^2 - bc - ca - ab$ is only positive semidefinite. It is 0 if $a=b=c.$ Meanwhile,
$$a^2 + b^2 + c^2 - bc - ca - ab \;  \geq \; \frac{3}{4} \; (b-c)^2,$$
$$a^2 + b^2 + c^2 - bc - ca - ab \;  \geq \; \frac{3}{4} \; (c-a)^2,$$
$$a^2 + b^2 + c^2 - bc - ca - ab \;  \geq \; \frac{3}{4} \; (a-b)^2.$$
So the quadratic form cannot be 1 unless all three letters are within 1 of each other. For example, we cannot have some negative and some positive. Indeed, all possible 1's are arrangements of $(n,n,n+1)$ or $(n,n+1,n+1).$ Taking $n$ as nonnegative, we see that we get the value of the cubic either $3n+1$ or $3n+2.$ So all primes not 3 are represented. Not sure about 3. 
EEEEDDDIITTT: Now that I think of it, cubes are $0,1,-1 \pmod 9.$ So, in order to get $a^3 + b^3 + c^3 - 3 a b c \equiv 0 \pmod 3,$ our choices are (A) $a,b,c$ are all divisible by 3, or (B) one of them is divisible by 3, the other two cubes give cancelling $\pm 1\pmod 9.$ In either case, the cubic is divisible by 9. So the prime 3 is not represented. 
