TITLE: Does there exist a cubic polynomial $f(x)$ such that $f(x)\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$?
QUESTION [5 upvotes]: As we know that $f(x)=x^2+1\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$, does there exist a cubic polynomial $f(x)=ax^3+bx^2+cx+d~(a,b,c,d \in\mathbb Z,a\neq 0) $ such that $f(x)\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$?
I only know that $f(x)=0$ has no integer solutions.

REPLY [2 votes]: You want the polynomial to have no root modulo (at least) half of the primes.
If $G$ is the galois group of $f$ (a subgroup of $S_3$), the polynomial has a linear factor modulo $p$ iff the reciprocity symbol at $p$ is an element of the Galois group with a fixpoint. By Cebotarev's density theorem, you need $G' = \{g \in G \mid g $has a fixpoint$ \}$ to be less than half the size of $G$.
You can quickly enumerate the subgroups of $S_3$ and find that only a cyclic group of order $3$ can work.
Since such an extension is abelian, the primes where $f$ factors are those who land modulo $m$ into the kernel of a character $\chi : (\Bbb Z/m\Bbb Z)^* \to C_3$ for some $m$.
Since $(\Bbb Z / 2^k \Bbb Z)^*$ has no $3$-part, it is useless to have powers of $2$ in $m$ : if $\chi$ is such a character it can be defined modulo some odd $m$ (in fact the only things that may appear in $m$ are $3^2$ and $p^1$ for $p \equiv 1 \pmod 3$)
Then by Dirichlet's theorem there must be primes landing in $\ker \chi$ that are congruent to $3 \pmod 4$