Does there exist a cubic polynomial $f(x)$ such that $f(x)\equiv0 \pmod p $ has no integer solutions if $p\equiv 3\pmod 4$? 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.
 A: If $p\equiv 1\pmod{3}$ and $a\in\mathbb{F}_p^*$ is not a cubic residue $\!\!\pmod{p}$, i.e.:
$$a^{\frac{p-1}{3}}\not\equiv 1\pmod{p},$$
there are no integer solutions to
$$ x^3 - a\equiv{0}\pmod{p}.$$
Just take $a\in\{2,3,4,5\}$ for $p=7$ or $a\in\{2,3,4,5,6,9,10,13,14,15,16,17\}$ for $p=19$.
In general, $f(x)$ has no roots in $\mathbb{F}_p$ iff it is irreducible over $\mathbb{F}_p$ - this depends only on the discriminant to be a quadratic residue or not and another function of the coefficients to be a cubic residue or not (we can drop the last condition if $p\equiv 2\pmod{3}$, since in such a case every element of the field is a cubic residue).

If you were looking for a "universal polynomial" $f(x)$ such that for every $p\equiv 3\pmod{4}$ 
$$f(x)\equiv 0\pmod{p}$$
has no solutions in $\mathbb{F}_p$, no such polynomial exists, since for the quadratic reciprocity law and the Dirichlet theorem the discriminant of $f(x)$ is a non-quadratic residue for some prime $q\equiv 3\pmod{4}$, hence $f(x)$ splits over $\mathbb{F}_q$ in virtue of the Stickelberger's criterion. Since $f(x)$ is a third-degree polynomial, this implies that $f(x)$ has a root in $\mathbb{F}_q$, contradiction.
A: 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 $3 \pmod 4$
A: Here is an analogous fact. If $$p\equiv 1\mod 3$$ where $p\neq m^2+27n^2$ for any integers $m$ and $n$, then $$x^3-2\equiv0\mod p$$ has no integer solutions by the cubic reciprocity law for $\left[\frac{2}{p}\right]_3$.
