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?



Cubic polynomials over finite fields whose roots are quadratic residues or non-residues
EDIT: Following a clever observation of user44191 in the comments:


If $f(x)$ is a monic polynomial, and $c$ a number, then the polynomial $xf(x)^2+c$ has a similar property to your example (the case $f(x)=x+1/2$). Indeed, we have $x = \frac{-c}{f(x)^2}$ so 


* If $-c$ is a nonzero square then all rational roots are squares.
* if $-c$ is a nonsquare then all rational roots are not square, but their ratios are square.
* If $-c$ is zero then one root is zero and the rest are double (this doesn't really fit the pattern).


This produces polynomials of odd degree. For even degree examples, we can do $f(x)^2+cx$. This gives $x =\frac{ f(x)^2}{-c}$ so we have the same thing except if $-c$ is zero than all roots are double, and there is a special case if $f(0)=0$.


So we have many examples of polynomials of this type.


(See the edit history for an earlier argument, special to the case of degree 3 polynomials, if desired. This was inspired by darij grinberg's answer, and that earlier answer inspired user44191's comment, so both of them are partially responsible for this solution.)


