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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?

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3 Answers 3

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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.)

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  • $\begingroup$ Nice! Do you think something like this can work for higher degrees? $\endgroup$ Dec 24, 2019 at 17:09
  • $\begingroup$ @darijgrinberg Thanks! Probably nothing so simple. Unless we make the Galois group small, it seems that the weakest condition we can look for (other than the product of the roots being a square) is the product of any two roots being a square, and that seems unlikely to have a simple way to force for polynomials of degree greater than 3. But I don't really have a reason to say that, it's just a hunch. $\endgroup$
    – Will Sawin
    Dec 24, 2019 at 22:31
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    $\begingroup$ The condition that $b = a^2/4$ is equivalent to the cubic being a square multiplied by $x$ plus a constant. Is that entirely a coincidence? $\endgroup$
    – user44191
    Dec 25, 2019 at 23:23
  • $\begingroup$ @user44191 Not a coincidence! In fact that's the key observation - see my new answer. $\endgroup$
    – Will Sawin
    Dec 26, 2019 at 0:17
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    $\begingroup$ A more general (possibly the most general?) version would be $c f(x)^2 - x g(x)^2$. Is it clear that these are the only examples? $\endgroup$
    – user44191
    Dec 26, 2019 at 2:39
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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?

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  • $\begingroup$ Shouldn't there be a more conceptual argument using Frobenius elements of a splitting field? (Maybe of the polynomial $X^6+X^4+1/4X^2+c$) I thought about it for a bit, but didn't see it. $\endgroup$ Dec 24, 2019 at 14:50
  • $\begingroup$ @AchimKrause: I'm loathe to sacrifice the generality, seeing that the result holds for any field of characteristic $\neq 2$. (And I think Frobenius elements would sacrifice this generality.) But I do think I'm missing the forest of the trees. I suspect the discriminant of the cubic should have a say here. $\endgroup$ Dec 24, 2019 at 15:36
  • $\begingroup$ The same argument shows, if $r$ is not a square, then $n_1$ and $n_2$ are not squares. $\endgroup$
    – Will Sawin
    Dec 24, 2019 at 16:34
  • $\begingroup$ @WillSawin: Even better, it shows that the mutual products of the roots are always squares. $\endgroup$ Dec 24, 2019 at 16:35
  • $\begingroup$ Thank you, from your proof we can see that any cubic $x^3+ax^2+bx+c$ such that $a^2-4b=0$, this still holds. $\endgroup$
    – user150240
    Dec 25, 2019 at 12:37
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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.

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