Cubic polynomials over finite fields whose roots are quadratic residues or non-residues

Question

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?

Answer 1

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.)

Answer 2

Yes, it does. Here is a stronger claim:

Theorem 1. Let $F$ be a field of characteristic $\neq2$. Let $c\in F$. Let $r\in F$ be a nonzero square, and let $n_{1},n_{2}\in F$ be such that the polynomial $\left( X-r\right) \left( X-n_{1}\right) \left( X-n_{2} \right) \in F\left[ X\right] $ equals $X^{3}+X^{2}+\dfrac{1}{4}X+c$. Then, $n_{1}$ and $n_{2}$ are squares.

Proof of Theorem 1. We have \begin{equation} \left( X-r\right) \left( X-n_{1}\right) \left( X-n_{2}\right) =X^{3}+X^{2}+\dfrac{1}{4}X+c \label{darij1.pf.t1.1} \tag{1} \end{equation} in $F\left[ X\right] $. Comparing coefficients before $X^{2}$ in this equality, we thus obtain $-\left( r+n_{1}+n_{2}\right) =1$. In other words, $r+n_{1}+n_{2}=-1$.

Comparing coefficients before $X$ in the equality \eqref{darij1.pf.t1.1}, we obtain $rn_{1}+rn_{2}+n_{1}n_{2}=\dfrac{1}{4}$. Hence, $4\left(rn_1 + rn_2 + n_1n_2\right) = 1$.

Now, \begin{align*} & \underbrace{\left( n_{1}-n_{2}+r\right) ^{2}}_{=n_{1}^{2}+n_{2}^{2} +r^{2}-2n_{1}n_{2}+2n_{1}r-2n_{2}r}-4n_{1}r\\ & =n_{1}^{2}+n_{2}^{2}+r^{2}-2n_{1}n_{2}+2n_{1}r-2n_{2}r-4n_{1}r\\ & =n_{1}^{2}+n_{2}^{2}+r^{2}-2n_{1}n_{2}-2n_{1}r-2n_{2}r\\ & = \left( \underbrace{r + n_1 + n_2}_{= -1} \right)^2 - \underbrace{ 4\left(rn_1 + rn_2 + n_1n_2 \right) }_{= 1} = \left(-1\right)^2 - 1 = 0 . \end{align*} Hence, $\left( n_{1}-n_{2}+r\right) ^{2}=4n_{1}r$. Solving this for $n_{1}$, we obtain \begin{equation} n_{1}=\dfrac{\left( n_{1}-n_{2}+r\right) ^{2}}{4r} \end{equation} (since $r\neq0$ and $4\neq0$). Thus, $n_{1}$ is a square (since $\left( n_{1}-n_{2}+r\right) ^{2}$, $4$ and $r$ are squares). The same argument (but with the roles of $n_{1}$ and $n_{2}$ swapped) yields that $n_{2}$ is a square. This proves Theorem 1. $\blacksquare$

A few remarks:

1. We cannot drop the assumption that $r$ is nonzero; otherwise, we would get a counterexample by taking $r = 0$, $n_1 = -1/2$, $n_2 = -1/2$ and $c = 0$ (whenever $-1/2$ is a non-square).

2. The above proof of Theorem 1 looks unmotivated; I have obtained it by neating up a more straightforward argument, which you can find in revision 3 of this answer.

3. The formula $n_{1}=\dfrac{\left( n_{1}-n_{2}+r\right) ^{2}}{4r}$ reminds me of Vieta jumping -- could it be something we get from it?

Answer 3

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.