Primitive elements with a nonsquare quadratic translate

The statement

For every odd prime power $q$, there exists $g\in\mathbb F_q^\times$ of multiplicative order $q-1$ such that

$$\chi_q(g^2+4)=-1,$$

where $\chi_q(0)=0$, and $\chi_q(u)=u^{(q-1)/2}\in\{1,-1\}$ for $u\ne0$. In particular, $g^2+4$ is not zero and is not a square.

This proves Conjecture 2 of Mabilat, arXiv:2510.09219v1, and its prime-field specialization proves Conjecture 1. The generator required here is a generator of the multiplicative group, not merely of the field extension. Neither conjecture excludes $q=3$ or imposes the restrictions on a parameter appearing in the preceding quiddity theorem.

1. Reduction to 24 fields by an established theorem

We use Booker, Cohen, Sutherland and Trudgian, Primitive values of quadratic polynomials in a finite field, Theorem 1 and equation (4), arXiv:1803.01435v2, published in Mathematics of Computation 88 (2019), 1903–1912, DOI 10.1090/mcom/3390.

Their theorem states that, for a finite field $\mathbb F_q$ and a quadratic

$$\begin{gathered} Q(X)=aX^2+bX+c,\\ a\ne0,\qquad b^2-4ac\ne0, \end{gathered}$$

there is a primitive element $g$ for which $Q(g)$ is also primitive, whenever $q$ is outside their displayed finite exception list. Its odd entries are exactly

$$\begin{aligned} E=\{&3,5,7,9,11,13,19,23,\\ &25,29,31,37,41,43,49,61,\\ &67,71,73,79,121,127,151,211\}. \end{aligned} \tag{1}$$

The even entries $2,4,16$ are irrelevant to the present statement. These are exceptions to the theorem's assertion uniformly over all coefficient triples; they are not assertions that the specific polynomial used here fails in those fields.

Apply the theorem with

$$Q(X)=X^2+4.$$

The leading coefficient is $1$, and its discriminant is $-16$, which is nonzero in every odd characteristic. Thus, when $q\notin E$, the theorem supplies primitive $g$ and primitive $g^2+4$.

Every primitive element $u$ of $\mathbb F_q^\times$, for odd $q$, is a nonsquare. Indeed, its order is the even number $q-1$, so $u^{(q-1)/2}\ne1$. The square of that element is $1$; since the field has odd characteristic, it follows that $u^{(q-1)/2}=-1$. This establishes the desired result for every $q\notin E$.

It remains to exhibit witnesses in the 24 fields in (1). The rest of the proof does exactly that; no extrapolation from a finite search is involved.

2. How to read the finite certificates

Mabilat already observes that $g=2$ works whenever $2$ is primitive in the proof of Proposition 6.9, and supplies witnesses for $q=9,25,49,41$ in the proof of Proposition 6.16. He also reports computational verification of Conjecture 1 for every prime from $3$ to $2{,}000{,}000$ in §6.2, referring to Appendix D. The following tables provide explicit certificates for the entire set $E$.

Put $n=q-1$. A nonzero field element $g$ has order $n$ if, for every prime divisor $\ell$ of $n$,

$$g^{n/\ell}\ne1. \tag{2}$$

To justify the criterion, the order $d$ of $g$ divides $n$ by Lagrange's theorem. If $d<n$, some prime $\ell$ divides $n/d$, and then $d\mid n/\ell$, contrary to (2).

Each row below gives $g$, an odd positive integer $k$ satisfying

$$g^2+4=g^k, \tag{3}$$

and all the values required by (2). The notation $\ell:v$ in the last column means $g^{n/\ell}=v$. All the listed values differ from $1$, and the listed $\ell$'s are precisely the prime divisors of $n$.

Consequently $g$ is primitive in every row. In particular, the entry for $\ell=2$ is $-1$, and (3), with odd $k$, gives the exact quadratic-character certificate

$$\begin{aligned} (g^2+4)^{(q-1)/2} &=\bigl(g^{(q-1)/2}\bigr)^k\\ &=(-1)^k=-1. \end{aligned} \tag{4}$$

Thus the tables certify both required properties, including the exclusion of zero as a purported nonsquare.

3. The twenty prime fields

In this table every equality is computed modulo the prime $q$.

$q$	$g$	$k$	Order tests
$3$	$2$	$1$	$2:-1$
$5$	$2$	$3$	$2:-1$
$7$	$3$	$3$	$2:-1$, $3:2$
$11$	$2$	$3$	$2:-1$, $5:4$
$13$	$2$	$3$	$2:-1$, $3:3$
$19$	$2$	$3$	$2:-1$, $3:7$
$23$	$7$	$1$	$2:-1$, $11:3$
$29$	$2$	$3$	$2:-1$, $7:16$
$31$	$3$	$11$	$2:-1$, $3:25$, $5:16$
$37$	$2$	$3$	$2:-1$, $3:26$
$41$	$7$	$13$	$2:-1$, $5:37$
$43$	$5$	$5$	$2:-1$, $3:36$, $7:16$
$61$	$2$	$3$	$2:-1$, $3:47$, $5:9$
$67$	$2$	$3$	$2:-1$, $3:37$, $11:64$
$71$	$7$	$23$	$2:-1$, $5:54$, $7:45$
$73$	$5$	$35$	$2:-1$, $3:8$
$79$	$7$	$25$	$2:-1$, $3:55$, $13:18$
$127$	$6$	$87$	$2:-1$, $3:107$, $7:64$
$151$	$7$	$87$	$2:-1$, $3:32$, $5:8$
$211$	$2$	$3$	$2:-1$, $3:196$, $5:107$, $7:171$

These are direct modular-power certificates, readily checked by repeated squaring. For example, modulo $31$,

$$\begin{aligned} 3^{15}&=-1,&3^{10}&=25,\\ 3^6&=16,&3^{11}&=13=3^2+4. \end{aligned}$$

Since $30=2\cdot3\cdot5$, these tests prove order $30$; the odd exponent $11$ proves the nonsquare property. The smallest field is covered as well: in $\mathbb F_3$, $g=2=-1$ has order $2$ and $g^2+4=2$.

4. The four quadratic extensions

Use the field models

$$\begin{gathered} \mathbb F_{p^2}=\mathbb F_p[T]/(T^2-d),\\ \alpha=T\bmod(T^2-d), \end{gathered}$$

with the following values of $p$ and $d$:

$$\begin{array}{c|c|c} q&p&d\\ \hline 9&3&2\\ 25&5&2\\ 49&7&3\\ 121&11&2 \end{array} \tag{5}$$

These are fields: the square sets in the four prime fields are

$$\begin{array}{c|l} p&\{a^2:a\in\mathbb F_p\}\\ \hline 3&\{0,1\}\\ 5&\{0,1,4\}\\ 7&\{0,1,2,4\}\\ 11&\{0,1,3,4,5,9\} \end{array}$$

In each case $d$ is absent, so $T^2-d$ has no root and is irreducible. Every element has a unique form $a+b\alpha$, and the arithmetic is explicitly

$$\begin{aligned} &(a+b\alpha)(c+e\alpha)\\ &\qquad=(ac+dbe)+(ae+bc)\alpha, \end{aligned} \tag{6}$$

with coefficients reduced modulo $p$. Any field of the same cardinality is isomorphic to this model over its prime field; multiplicative order and the equation $g^2+4=g^k$ are preserved by that isomorphism.

$q$	$g$	$k$	Order tests
$9$	$1+\alpha$	$3$	$2:-1$
$25$	$2+\alpha$	$3$	$2:-1$, $3:2+3\alpha$
$49$	$1+\alpha$	$11$	$2:-1$, $3:4$
$121$	$2+\alpha$	$21$	$2:-1$, $3:5+10\alpha$, $5:4$

The translated values in these four rows are:

$q$	$g^2+4$
$9$	$1+2\alpha$
$25$	$4\alpha$
$49$	$1+2\alpha$
$121$	$10+4\alpha$

For instance, in $\mathbb F_{25}$, where $\alpha^2=2$,

$$\begin{aligned} g^2&=1+4\alpha,\\ g^3&=4\alpha=g^2+4,\\ g^8&=2+3\alpha,\\ g^{12}&=-1. \end{aligned}$$

As $24=2^3\cdot3$, the last two equalities establish that $g$ has order $24$, while the first two give the odd-power certificate. The other rows follow from the same explicit multiplication rule (6).

There is also an independent short character check for these four rows. For $u=a+b\alpha$,

$$\begin{aligned} N_{\mathbb F_{p^2}/\mathbb F_p}(u)&=a^2-db^2,\\ u^{(p^2-1)/2}&=N(u)^{(p-1)/2}. \end{aligned}$$

For the four displayed values of $u=g^2+4$, their norms are

$$\begin{array}{c|c} q&N(u)\in\mathbb F_p\\ \hline 9&2\\ 25&3\\ 49&3\\ 121&2 \end{array}$$

Their Euler powers are $2,4,6,10$, respectively, which are exactly $-1$ in the corresponding prime fields. Hence the required quadratic character in each extension is exactly $-1$.

5. Conclusion

The published theorem handles every odd prime power outside (1), and the tables handle every element of (1). Thus every finite field of odd characteristic has a primitive element $g$ with $g^2+4$ nonsquare. Both Mabilat Conjectures 1 and 2 follow.