TITLE: Cubic polynomials over finite fields whose roots are quadratic residues or non-residues
QUESTION [11 upvotes]: For a cubic polynomial $f(x)=x^3+x^2+\frac{1}{4}x+c$ over $\mathbb{F}_q$, where $q$ is a odd prime power, I find that for a lot of $q$, there does not exist $c\in\mathbb{F}_q$ such that $f$ has three distinct roots in $\mathbb{F}_q$, one of which is a quadratic residue and the other two are non-residues. I have not found any counter examples, so my question is, does it hold for any $q$? If so, what forms of cubic polynomials have such property?

REPLY [4 votes]: I have an indirect proof using a result from Leonard's paper in 1969 "On Factoring Quartics (mod $p$)":

Lemma. Let $f(x)=x^4+a_2x^2+a_1x+a_0$ be a quartic polynomial over $\mathbb{F}_q$ having four distinct roots in its splitting field, where $q$ is an odd prime power and $a_1\ne 0$, then $$g(x)=x^3+8a_2x^2+(16a_2^2-64a_0)x-64a_1^2$$has one root being a non-zero quadratic residue in $\mathbb{F}_q$, and the other two roots being quadratic non-residues in $\mathbb{F}_q$, if and only if $f(x)=h_1(x)h_2(x)$, where $h_1,h_2$ are irreducible quadratics over $\mathbb{F}_q$.

Assume that among the roots of $g(x)=x^3+x^2+\frac{1}{4}x+c$, there are one non-zero quadratic residue and two non-residues, then $-c$ must be a square in $\mathbb{F}_q$. It is easy to see that $g(x)$ is a resolvent cubic polynomial of $f(x)=x^4+\frac{1}{8}x^2+\frac{1}{8}\sqrt{-c}x$. This contradicts the above lemma.
Furthermore, for $f(x)=x^4+a_2x^2+a_1x+a_0$ , let $a_0=0$, then $b=\frac{a^2}{4}$ for its solvent cubics $g(x)=x^3+ax^2+bx+c$. Thus $b=\frac{a^2}{4}$ is a sufficient condition for the property.