Prove that there are infinitely many prime numbers $p\equiv1\pmod3$ such that there is no integer $x$ such that $p\mid (x^3+3)$.
We know that $P(x)=x^3+3$ is irreducible in $Q[x] $; thus, the density of its prime divisors is $1/3$. on the other hand $x^3$ produces all remaiders modulo any $3k+2$ prime number. so if there were finitely many such $3k+1$ prime numbers, the density of its prime divisors would be $1$. But I want an elementary solution!
Well, all primes $p$ that can be expressed as $$ p= 7 x^2 + 3xy + 9 y^2 $$ have no cube root of $3.$
The primes $$ r = u^2 + uv + 61 v^2 $$ do have a cube root of $3.$
Theorem of Jacobi. Cox gives this as an exercise in his Primes of the Form $x^2 + n y^2.$ In the first(1989) edition this is exercise 4.15 on pages 90 and 91. He compares this to Theorem 4.15 on page 80, which does the cube root of $2.$ Oh, Cox uses the equivalent version $4p = s^2 + 243 t^2$
Ireland and Rosen also make this an exercise, second edition (1990), it is exercise 23 on page 135. They also write this as $4p = C^2 + 243 B^2$ for there to be cube roots of $3$
Why not: Theorem 9.12 in Cox is:
Theorem 9.12 Let $ax^2 + bxy + c y^2$ be a primitive positive definite quadratic form of discriminant $D < 0,$ and let $S$ be the set of primes represented by $ax^2 + bxy + c y^2.$ Then the Dirichlet density $\delta(S) $ exists and is given by the formulas
$\delta(S) = \frac{1}{2h(D)} $ if $ax^2 + bxy + c y^2$ is properly equivalent to its opposite; otherwise $\delta(S) = > \frac{1}{h(D)} $
In particular, $ax^2 + bxy + c y^2$ represents infinitely many prime numbers.
In particular, $h(-243) = 3,$ the principal form is $x^2 + xy + 61 y^2$ and is equivalent to its opposite, so the principal form represents $1/6$ of primes. The other two classes are $7 x^2 + 3xy + 9 y^2$ and $7 x^2 - 3xy + 9 y^2.$ These are opposites and not equivalent, but they represent the same primes, that being $1/3$ of all primes.
