Counting Polynomials with Distinct Zeros in Finite Fields

TL;DR

This paper derives explicit formulas for counting degree n polynomials over finite fields with a fixed polynomial part that have exactly k distinct roots, and applies these results to analyze the spectra of certain Wenger graphs.

Contribution

It provides new combinatorial formulas for counting polynomials with a specified number of roots when the polynomial degree difference is small.

Findings

Explicit formulas for N_k(u(x),ll) when n-ll=1, 2, 3

Application to spectrum analysis of jumped Wenger graphs

Enhanced understanding of polynomial root distributions over finite fields

Abstract

Let $\mathbb{F}_q$ be a finite field with $q=p^e$ elements, where $p$ is a prime and $e\geq 1$ is an integer. Let $\ell<n$ be two positive integers. Fix a monic polynomial $u(x)=x^n +u_{n-1}x^{n-1}+\cdots +u_{\ell+1}x^{\ell+1} \in \mathbb{F}_q[x]$ of degree $n$ and consider all degree $n$ monic polynomials of the form $$f(x) = u(x) + v_\ell(x), \ v_\ell(x)=a_\ell x^\ell+a_{\ell-1}x^{\ell-1}+\cdots+a_1x+a_0\in \mathbb{F}_q[x].$$ For integer $0\leq k \leq {\rm min}\{n,q\}$, let $N_k(u(x),\ell)$ denote the total number of $v_\ell(x)$ such that $u(x)+v_\ell(x)$ has exactly $k$ distinct roots in $\mathbb{F}_q$, i.e. $$N_k(u(x),\ell)=|\{f(x)=u(x)+v_l(x)\ |\ f(x)\ {\rm has\ exactly}\ k\ {\rm distinct\ zeros\ in}\ \mathbb{F}_q\}|.$$ In this paper, we obtain explicit combinatorial formulae for $N_k(u(x),\ell)$ when $n-\ell$ is small, namely when $n-\ell= 1, 2, 3$. As an application, we define two kinds of Wenger graphs called jumped Wenger graphs and obtain their explicit spectrum.

Equations

$$f(x)=u(x)+v_{\ell }(x),v_{\ell }(x)=a_{\ell }{x}^{\ell }+a_{\ell -1}{x}^{\ell -1}+\cdots +a_{1}x+a_0\in {\mathbb{F}}_q[x].$$

$$N_k(u(x),\ell )=|\{f(x)=u(x)+v_l(x)|f(x)\mathrm{has}\mathrm{exactly}k\mathrm{distinct}\mathrm{zeros}\mathrm{in}{\mathbb{F}}_q\}|.$$

$$f(x)=u(x)+v_{\ell }(x),v_{\ell }(x)=a_{\ell }{x}^{\ell }+a_{\ell -1}{x}^{\ell -1}+\cdots +a_{1}x+a_0\in {\mathbb{F}}_q[x].$$

$$N_k(u(x),\ell )=|\{f(x)=u(x)+v_l(x)|f(x)\mathrm{has}\mathrm{exactly}k\mathrm{distinct}\mathrm{zeros}\mathrm{in}{\mathbb{F}}_q\}|.$$

$$|S_j|=\sum _{|I|=j}|S_I|-\sum _{|I|=j+1}|S_I|-\cdots +(-1{)}^{n-j}\sum _{|I|=n}|S_I|,\mathrm{where}j=0.$$

$$N_{k}(x^{n},n-1)=q^{n-k}\left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{n-k}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{-i}.$$

$$S_I(J)=\left\{g(x)\in {\mathbb{F}}_q[x]|g(\alpha )=0\mathrm{for}\mathrm{all}\alpha \in I\right\}.$$

$$\sum _{I\subseteq {\mathbb{F}}_q-J,|I|=i}|S_I(J)|=\left(\genfrac{}{}{0ex}{}{q-k}{i}\right)q^{n-k-i}.$$

$$\begin{array}[]{ll}N_{k}(x^{n},n-1)&=\sum\limits_{|J|=k}\sum\limits_{I\subset\mathbb{F}_{q}-J}(-1)^{|I|}|S_{I}(J)|\\ &=\sum\limits_{i=0}^{n-k}(-1)^{i}\sum\limits_{|J|=k}\sum\limits_{I\subset\mathbb{F}_{q}-J,|I|=i}|S_{I}(J)|\\ &=q^{n-k}\left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{n-k}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{-i}\end{array}.$$

$$N_{k}(x^{n},n-1)=q^{n-q}(q-1)\sum\limits_{r=k}^{q-1}q^{r-k}\left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{r-k}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{-i}.$$

$$\begin{array}[]{l}\sum\limits_{r=k}^{q-1}q^{r-k}\sum\limits\limits_{i=0}^{r-k}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{-i}\\ \hskip 113.81102pt=1+q\sum\limits_{i=0}^{1}(-1)^{i}{q-k\choose i}q^{-i}+\cdots+q^{q-1-k}\sum\limits_{i=0}^{q-1-k}(-1)^{i}{q-k\choose i}q^{-i}\\ \hskip 113.81102pt=\frac{1}{1-q}\sum\limits_{i=0}^{q-k-1}(-1)^{i}{q-k\choose i}(1-q^{q-k-i})\\ \hskip 113.81102pt=(q-1)^{q-k-1},\end{array}$$

$$N_k(x^n,n-1)=\left(\genfrac{}{}{0ex}{}{q}{k}\right)q^{n-q}(q-1{)}^{q-k}.$$

$$M(n,b)=\frac{1}{q}\left(\begin{array}[]{c}q\\ n\end{array}\right).$$

$$M(n,b)=\frac{1}{q}\left(\begin{array}[]{c}q\\ n\end{array}\right)+(-1)^{n+\frac{n}{p}}\frac{v(b)}{q}\left(\begin{array}[]{c}q/p\\ n/p\end{array}\right),$$

$$N_{k}(x^{n}-bx^{n-1},n-2)=q^{n-k-1}\left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{n-k}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{-i}.$$

$$\begin{array}[]{rl}N_{k}(x^{n}-bx^{n-1},n-2)=&q^{n-k-1}\left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{n-k}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{-i}\\ &+(-1)^{\frac{n}{p}+n}\frac{v(b)}{q}\left(\begin{array}[]{c}n\\ k\end{array}\right)\left(\begin{array}[]{c}q/p\\ n/p\end{array}\right)\end{array}.$$

$$g(x)=x^{n-k}+d_1{x}^{n-k-1}+\cdots +d_{n-k-1}x+d_{n-k}\in {\mathbb{F}}_q[x]$$

$$\begin{array}[]{ll}f(x)&=x^{n}-bx^{n-1}+a_{2}x^{n-2}+\cdots+a_{n}\\ &=(x-c_{1})(x-c_{2})\cdots(x-c_{k})g(x)\\ &=(x^{k}+b_{1}x^{k-1}+\cdots+b_{k-1}x+b_{k})(x^{n-k}+d_{1}x^{n-k-1}+\cdots+d_{n-k-1}x+d_{n-k}).\end{array}$$

$$\left\{\begin{array}[]{ccc}b_{1}+d_{1}&=&-b\\ b_{2}+b_{1}d_{1}+d_{2}&=&a_{2}\\ b_{3}+b_{2}d_{1}+b_{1}d_{2}+d_{3}&=&a_{3}\\ \cdots&\cdots&\cdots\\ b_{k}d_{n-k}=a_{n}\end{array}\right.$$

$$\begin{array}[]{ll}S_{I}(J)=&\left\{g(x)=x^{n-k}+d_{1}x^{n-k-1}+\cdots+d_{n-k}\in\mathbb{F}_{q}[x]|d_{1}=b-(c_{1}+c_{2}+\cdots+c_{k}),\right.\\ &\left.g(\alpha)=0\ {\rm for}\ {\rm all}\ \alpha\in I\right\}.\end{array}$$

$$\sum _{I\subseteq {\mathbb{F}}_q-J,|I|=i}|S_I(J)|=\left(\genfrac{}{}{0ex}{}{q-k}{i}\right)q^{n-k-1-i}.$$

$$\sum\limits_{|J|=k}\sum\limits_{I\subseteq\mathbb{F}_{q}-J,|I|=n-k}|S_{I}(J)|=\left\{\begin{array}[]{l}\left(\begin{array}[]{c}n\\ k\end{array}\right)\frac{1}{q}\left(\begin{array}[]{c}q\\ n\end{array}\right),p\nmid n,\\ \left(\begin{array}[]{c}n\\ k\end{array}\right)\left[\frac{1}{q}\left(\begin{array}[]{c}q\\ n\end{array}\right)+(-1)^{n+\frac{n}{p}}\frac{v(b)}{q}\left(\begin{array}[]{c}q/p\\ n/p\end{array}\right)\right],p\mid n,\end{array}\right.$$

$$N_k(x^n-b{x}^{n-1},n-2)=\sum _{|J|=k}\sum _{I\subset {\mathbb{F}}_q-J}(-1{)}^{|I|}|S_I(J)|=\sum _{i=0}^{n-k}(-1{)}^i\sum _{|J|=k}\sum _{I\subset {\mathbb{F}}_q-J,|I|=i}|S_I(J)|.$$

$$N_{k}(x^{n}-bx^{n-1},n-2)=\left\{\begin{array}[]{l}\left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{n-k-1}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{n-k-1-i}+\\ (-1)^{n-k}\left(\begin{array}[]{c}n\\ k\end{array}\right)\frac{1}{q}\left(\begin{array}[]{c}q\\ n\end{array}\right),p\nmid n,\\ \left(\begin{array}[]{c}q\\ k\end{array}\right)\sum\limits\limits_{i=0}^{n-k-1}(-1)^{i}\left(\begin{array}[]{c}q-k\\ i\end{array}\right)q^{n-k-1-i}+\\ (-1)^{n-k}\left(\begin{array}[]{c}n\\ k\end{array}\right)\left[\frac{1}{q}\left(\begin{array}[]{c}q\\ n\end{array}\right)+(-1)^{n+\frac{n}{p}}\frac{v(b)}{q}\left(\begin{array}[]{c}q/p\\ n/p\end{array}\right)\right],p\mid n.\end{array}\right.$$

$$N_k(x^n-b{x}^{n-1},n-2)=q^{n-q-1}\left(\genfrac{}{}{0ex}{}{q}{k}\right)(q-1{)}^{q-k}.$$

$$N_{k}(x^{q}-bx^{q-1},q-2)=\left\{\begin{array}[]{ll}0&k=q,\\ \frac{1}{q}{q\choose k}((q-1)^{q-k}-(-1)^{q-k})&k\leq q-1.\end{array}\right.$$

$$N_{k}(x^{q}-bx^{q-1},q-2)=\left\{\begin{array}[]{ll}1&k=q,\\ 0&k=q-1,\\ \frac{q-1}{q}{q\choose k}((q-1)^{q-k-1}+(-1)^{q-k})&k\leq q-2.\end{array}\right.$$

$$\Bigg{\{}\begin{array}[]{lll}a_{1}x_{1}^{2}+\cdots+a_{n}x_{n}^{2}&=&a_{0},\\ b_{1}x_{1}+\cdots+b_{n}x_{n}&=&b_{0},\end{array}$$

$$N(n,a_{0},b_{0})=\{\begin{array}[]{ll}q^{n-2}&{\rm if}\ n\ {\rm even},\\ q^{n-2}+q^{(n-3)/2}(q-1)\chi((-1)^{(n-1)/2}ab)&{\rm if}\ n\ {\rm odd},\end{array}$$

Full text

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For $0\leq|I|<n-k$, it is obvious to obtain that the cardinality of the set $S_{I}(J)$ is $q^{n-k-|I|}$. It follows that for $0\leq i<n-k$,

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For $|I|>n-k$, it is clear that $|S_{I}(J)|=0$. By the inclusion-exclusion principle, we deduce that

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∎

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Since

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we have

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If $k=q$, then $h(x)=0$. Then $N_{k}(x^{n},n-1)=q^{n-q}$. Hence for $n\geq q$,

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**3  **The case $n-\ell=2$

In the special case that $k=n$, this is the counting version for the $n$-subset sum problem over $\mathbb{F}_{q}$, which is already handled in [12]. We state this result as a lemma and will use it in our proof.

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If $p|n$, then

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where $v(b)=-1$ if $b\neq 0$, and $v(b)=q-1$ if $b=0$.

In terms of our earlier notations, we have $M(n,b)=N_{n}(x^{n}-bx^{n-1},n-2)$.

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(ii) If $p\mid n$, then

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such that

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Comparing the coefficients, we have

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For fixed $b$ and fixed $k$-subset $J=\{c_{1},\cdots,c_{k}\}\subset\mathbb{F}_{q}$, the coefficient $d_{1}=-(b+c_{1}+c_{2}+\cdots+c_{k})$ is then fixed. The other coefficients $\{d_{2},\cdots,d_{n-k}\}$ of the polynomial $g(x)$ are free since $\{a_{2},\cdots,a_{n}\}$ are free.

For a subset $I\subseteq\mathbb{F}_{q}-J$, define the set

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For $0\leq|I|<n-k$, the above argument shows that the cardinality of the set $S_{I}(J)$ is $q^{n-k-1-|I|}$. It follows that for $0\leq i<n-k$,

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When $i=n-k$, $f(x)$ is forced to have $n$ distinct roots in $\mathbb{F}_{q}$ with sum equal to $b$. Then by Lemma 3.1, we deduce

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For $|I|>n-k$, it is clear that $|S_{I}(J)|=0$. By the inclusion-exclusion principle, we deduce that

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Hence,

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This Theorem is proved from the fact \left(\begin{array}[]{c}q\\ k\end{array}\right)\left(\begin{array}[]{c}q-k\\ n-k\end{array}\right)=\left(\begin{array}[]{c}n\\ k\end{array}\right)\left(\begin{array}[]{c}q\\ n\end{array}\right).

∎

$1)$* If $n>q$, then $n-1>q-1$. Similar arguments to those used in the Remark 2.1 show that*

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$2)$* If $n=q$, then $f(x)=x^{q}-x-bx^{q-1}+a_{2}x^{q-2}+\cdots+a_{q-2}x^{2}+(a_{q-1}+1)x+a_{q}$. It is easy to get the following conclusions:*

When $b\neq 0$,

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When $b=0$,

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**4  **The case $n-\ell=3$

In this section, we always assume that $q$ is an odd number. Let $\mathbb{F}_{q}^{n}$ be the Cartesian product of $n$ copies of $\mathbb{F}_{q}$. For convenience, we firstly state some results on the number of common solutions in $\mathbb{F}_{q}^{n}$ of the equations

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where $a_{0},b_{0},b_{1},\cdots,b_{n}\in\mathbb{F}_{q}$, $a_{1},\cdots,a_{n}\in\mathbb{F}_{q}^{*}$, $b_{i}\neq 0$ for at least one $i$, $1\leq i\leq n,$ see Exercises 6.31-6.34 in [16].

$i)$* For $b\neq 0,\ c=0,$*

[TABLE]

where $\chi$ is the quadratic character of $\mathbb{F}_{q}.$

$ii)$* For $b\neq 0,\ c\neq 0,$*

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$iii)$* For $b=c=0,$*

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where $v$ is as in Lemma 3.1.

$iv)$* For $b=0,\ c\neq 0$, $N(n,a_{0},b_{0})=q^{n-2}$.*

Now, we begin to recall a sieve for distinct coordinate counting, see [13]. Let $X$ be a subset of $\mathbb{F}_{q}^{n}$. Motivated by diverse applications in coding theory and graph theory, it is very interesting to count the number of elements in the set

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In [13], J. Li and D. Wan discovered the new sieving formula about $|\overline{X}|$. Let $S_{n}$ be the symmetric group on $\{1,2,\cdots,n\}$. For a given permutation $\tau=(i_{1}i_{2}\cdots i_{t_{1}})\cdots(l_{1}l_{2}\cdots l_{t_{s}})$ with $t_{i}\geq 1,\ 1\leq i\leq s$, define

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Now the symmetric group $S_{n}$ acts on $\mathbb{F}_{q}^{n}$ by permuting coordinates. That is, for given $\tau\in S_{n}$ and $x=(x_{1},\cdots,x_{n})\in\mathbb{F}_{q}^{n}$, we have

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$X$ is called symmetric if any $x\in X$ and any $\tau\in S_{n}$, $\tau\circ x\in X$. Furthermore, If $X$ satisfies the ”strongly symmetric” condition, that is, for any $\tau$ and $\sigma$ in $S_{n}$, one has $|X_{\tau}|=|X_{\sigma}|$ provided $l(\tau)$ and $l(\sigma)$, then we call X a strongly symmetric set. Let $C_{n}$ be the set of conjugacy classes of $S_{n}$. If $X$ is symmetric, then

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where $C(\tau)$ is the number of permutations conjugate to $\tau$ and $l(\tau)$ is the number of cycles including the trivial cycle.

A permutation $\tau\in S_{n}$ is said to be of type $(c_{1},c_{2},\cdots,c_{n})$ if $\tau$ has exactly $c_{i}$ cycles of length $i$. We denote by $N(c_{1},c_{2},\cdots,c_{n})$ the number of permutations in $S_{k}$ of type $(c_{1},c_{2},\cdots,c_{n})$ and we have (see [19]),

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Since two permutations are conjugate if and only if they have the same type, we have $C(\tau)=N(c_{1},c_{2},\cdots,c_{n}).$

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Then

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where

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and

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Then we get the following exponential generating function

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For given generating function $f(x)$, denote by $[x^{i}]f(x)$ the coefficient of $x^{i}$ in the formal power series expansion of $f(x)$.

$1)$ If $t_{i}=-\sqrt{q}$ for $p\nmid i$ and $t_{i}=-q$ for $p|i$, then we have

\begin{array}[]{l}C_{n}(-\sqrt{q},\cdots,-\sqrt{q},-q,-\sqrt{q},\cdots,-\sqrt{q},-q,\cdots)\\ \hskip 113.81102pt=[\frac{u^{n}}{n!}]e^{-\sqrt{q}(u+\frac{u^{2}}{2}+\frac{u^{3}}{3}+\cdots)+\frac{-q+\sqrt{q}}{p}(u^{p}+\frac{u^{2p}}{2}+\frac{u^{3p}}{3}+\cdots)}\\ \hskip 113.81102pt=[\frac{u^{n}}{n!}]e^{\sqrt{q}\ln(1-u)+\frac{q-\sqrt{q}}{p}\ln(1-u^{p})}\\ \hskip 113.81102pt=[\frac{u^{n}}{n!}](1-u)^{\sqrt{q}}(1-u^{p})^{\frac{q-\sqrt{q}}{p}}\\ \hskip 113.81102pt=[\frac{u^{n}}{n!}](\sum\limits_{i\geq 0}(-1)^{i}\left(\begin{array}[]{c}\sqrt{q}\\ i\end{array}\right)u^{i})(\sum\limits_{j\geq 0}(-1)^{j}\left(\begin{array}[]{c}\frac{q-\sqrt{q}}{p}\\ j\end{array}\right)u^{pj})\\ \hskip 113.81102pt=n!\sum\limits_{i+pj=n,0\leq i\leq\sqrt{q}}(-1)^{n}\left(\begin{array}[]{c}\sqrt{q}\\ i\end{array}\right)\left(\begin{array}[]{c}\frac{q-\sqrt{q}}{p}\\ j\end{array}\right).\end{array}

Similarly, if $t_{i}=\sqrt{q}$ for $p\nmid i$ and $t_{i}=-q$ for $p|i$, then we have

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Thus, this lemma is proved from

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∎

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$i)$* For $p\nmid n$,*

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$ii)$* For $p|n$,*

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For a type $(c_{1},c_{2},\cdots,c_{n})$ permutation $\tau$, we have $\sum\limits\limits_{i=1}^{n}ic_{i}=n$ and $l(\tau)=\sum\limits\limits_{i=1}^{n}c_{i}$. Denote by $r$ the number of the cycles of $\tau$ such that the length of it is divisible by $p$, and denote by $s$ the number of the cycles of $\tau$ such that the length of it is not divisible by $p$. Note that $r+s=l(\tau)$ and $\sum\limits_{p\nmid i}ic_{i}\equiv n\pmod{p}$.

$i)$ Since $p\nmid n$, we have $s\geq 1$. If $s=1$, then $|X_{\tau}|=q^{l(\tau)-1}$. If $s\geq 2$, then by $i)$ of Lemma 4.1, we have

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Since $q=p^{2e}$, by Lemma 4.3 we have

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Therefore, we have

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$ii)$ Since $p|n$, we have $s\neq 1$. If $s=0$, then $|X_{\tau}|=q^{l(\tau)}$. Denote by $CP_{n}$ the conjugacy classes in $C_{n}$ whose every cycle length is divisible by $p$, and denote by $p(n,i)$ the number of permutations in $S_{n}$ of $i$ cycles with the length of its each cycle divisible by $p$. If $s>0$, then by $iii)$ of Lemma 4.1 and 4.3 we have

\begin{array}[]{rl}M(n,0,0)=&\frac{1}{n!}\sum\limits\limits_{\tau\notin CP_{n}}(-1)^{n-l(\tau)}C(\tau)|X_{\tau}|+\frac{1}{n!}\sum\limits\limits_{\tau\in CP_{n}}(-1)^{n-l(\tau)}C(\tau)|X_{\tau}|\\ =&\frac{1}{n!}\sum\limits\limits_{\tau\notin CP_{n}}(-1)^{n-l(\tau)}C(\tau)|X_{\tau}|+\frac{1}{n!}\sum\limits\limits_{i=1}^{n}(-1)^{n-i}p(n,i)q^{i}\\ =&\frac{1}{n!}\sum\limits\limits_{i=1}^{n}(-1)^{n-i}(c(n,i)-p(n,i))q^{i-2}+\frac{1}{n!}\sum\limits\limits_{i=1}^{n}(-1)^{n-i}p(n,i)q^{i}\\ &+\frac{1}{n!}(q-1)q^{-1}\sum\limits\limits_{s>0\ {\rm is\ even}}(-1)^{n-l(\tau)}C(\tau)\prod\limits_{p|i}q^{c_{i}}\prod\limits_{p\nmid i}\sqrt{q}^{c_{i}}\\ =&\frac{1}{n!}\sum\limits_{i=1}^{n}(-1)^{n-i}c(n,i)q^{i-2}+\frac{1}{n!}\frac{q-1}{q^{2}}\sum\limits\limits_{i=1}^{n}(-1)^{n-i}p(n,i)q^{i}\\ &+\frac{1}{n!}(q-1)q^{-1}\sum\limits\limits_{s\ {\rm is\ even}}(-1)^{n-l(\tau)}C(\tau)\prod\limits_{p|i}q^{c_{i}}\prod\limits_{p\nmid i}\sqrt{q}^{c_{i}}\end{array}.

Recall that \sum\limits_{i=1}^{n}(-1)^{n-i}p(n,i)q^{i}=(-1)^{n+\frac{n}{p}}n!\left(\begin{array}[]{c}q/p\\ n/p\end{array}\right) (See Lemma 3.1 in [13]) and $p$ is an odd prime number. Therefore,

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∎

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Then $M^{\prime}(n,0,0)=M(n,0,0)$

Let $X$ be the set of all solutions of the equations (4.3) in $\mathbb{F}_{q}^{n}$. Put

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Now the symmetric group $S_{n-1}$ acts on $\mathbb{F}_{q}^{n}$ by permuting the first $n-1$ coordinates. The similar arguments of the $|\overline{X}|$ show that

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where $C(\tau)$ is the number of permutations conjugate to $\tau\in S_{n-1}$ and $l(\tau)$ is the number of cycles including the trivial cycle. Denote by $M_{1}(n,0,0)$ the number of $n$-subsets $\{x_{1},x_{2},\cdots,x_{n}|x_{i}\neq x_{j},\ \forall 1\leq i\neq j\leq n-1\}$ of $\mathbb{F}_{q}$ whose elements are the common solutions of the equations (4.3). The similar proof of Theorem 4.1 shows the following theorem:

$i)$* For $p\nmid n$ ,*

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$ii)$* For $p|n$,*

[TABLE]

where

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and

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(i) If $p\nmid n$, then

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where $D(n-1)=\alpha(n-1)+(-1)^{n-1}\beta(n-1)$ and $D(n)=\alpha(n)-(-1)^{n}\beta(n)$.

(ii) If $p|n$, then

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where $P(n-1)=\alpha(n-1)-(-1)^{n-1}\beta(n-1)$ and $P(n)=\alpha(n)+(-1)^{n}\beta(n)$.

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This result follows from the similar proof of Theorem $\ref{thm:a1=0}.$ ∎

$2)$* For $n\geq q$, the formulae of $N_{k}(x^{n},n-3)$ can be obtained from the similar arguments of Remark 2.1.*

$i)$* If $n>q+1$, then $N_{k}(x^{n},n-3)=q^{n-q-2}{q\choose k}(q-1)^{q-k}.$*

$ii)$* If $n=q+1$, then*

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$ii)$* If $n=q$, then*

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5 The spectrum of some jumped Wenger graphs

In [22], Wenger introduced a family of $p$-regular bipartite graphs and then Lazebnik and Ustimenko arrived at a family of bipartite graphs using a construction based on a certain Lie algebra for a prime power $q$ in [8]. Later, Lazebnik and Viglione gave an equivalent representation of these graphs in [10]. Then another useful representation of these graphs was given in [20], on which we concentrate in this section. All graph theory notions can be found in [1].

Let $m\geq 1$ be a positive integer and $g_{k}(x,y)\in\mathbb{F}_{q}[x,y]$ for $2\leq k\leq m+1$. Let $\mathfrak{P}=\mathbb{F}_{q}^{m+1}$ and $\mathfrak{L}=\mathbb{F}_{q}^{m+1}$ be two copies of the $(m+1)-$dimensional vector space over $\mathbb{F}_{q}$, which are called the point set and the line set respectively. If $a\in\mathbb{F}_{q}^{m+1}$, then we write $(a)\in\mathfrak{P}$ and $[a]\in\mathfrak{L}$. Denote $\mathfrak{G}=G_{q}(g_{2},\cdots,g_{m+1})=(V,E)$ by the bipartite graph with vertex set $V=\mathfrak{B}\bigcup\mathfrak{L}$ and the edge set $E$ is defined as follows: there is an edge from a point $P=(p_{1},p_{2},\cdots,p_{m+1})\in\mathfrak{B}$ to a line $L=[l_{1},l_{2},\cdots,l_{m+1}]\in\mathfrak{L}$, denoted by $P~{}L$, if the following $m$ equalities hold:

[TABLE]

If $g_{k}(x,y)=x^{k-1}y,\ k=2,\cdots,m+1$, then the graph is just the original Wenger graph denoted by $W_{m}(q)$ in [5]. The spectrum, the diameter and the automorphism group of $W_{m}(q)$ were studied in $\cite[cite]{[\@@bibref{}{CL14}{}{}]},\ \cite[cite]{[\@@bibref{}{LU93}{}{}]},\ \cite[cite]{[\@@bibref{}{LU95}{}{}]}$ and [21]. In [2], a new class of bipartite graphs called linearized Wenger graphs was introduced. These graphs were defined by $\eqref{def:edge}$ together with $g_{k}(x,y)=x^{p^{k-2}},\ k=2,\cdots,m+1$, which denoted by $L_{m}(q)$. When $m\geq e$, the spectrum of $L_{m}(q)$ was explicitly determined using results on linearized polynomials over finite fields. The diameter and girth of $L_{m}(q)$ also were obtained. Furthermore, the spectrum of a general class of graphs, which defined by $g_{k}(x,y)=f_{k}(x)y$ and the mapping $\sigma:\ \mathbb{F}_{q}\rightarrow\mathbb{F}_{q}^{m+1};\ u\rightarrow(1,f_{2}(u),\cdots,f_{m+1}(u))$ is injective, was studied. The eigenvalues of such a graph were determined and their multiplicities were reduced to counting certain polynomials with a given number of roots over $\mathbb{F}_{q}$. That is, for all prime power $q$ and positive integer $m$, the eigenvalues of $\mathfrak{G}$, counted with multiplicities, are

[TABLE]

where $F_{\omega}(u)=\omega_{1}+\omega_{2}f_{2}(u)+\cdots+\omega_{m+1}f_{m+1}(u)$ and $N_{F_{\omega}}=|\{u\in\mathbb{F}_{q}:\ F_{\omega}(u)=0\}|$. For $0\leq i\leq q$, the multiplicity of $\pm\sqrt{qi}$ is

[TABLE]

In this section, we use our previous results to get the spectrum of a general class of graphs defined by $\eqref{def:edge}$ together with polynomials $g_{k}(x,y)=f_{k}(x)y,\in\mathbb{F}_{q}[x,y]$ , where $f_{k}(x)=x^{k-1},\ 2\leq k\leq m,f_{m+1}(x)=x^{m+1}$ and $f_{k}(x)=x^{k-1},\ 2\leq k\leq m,f_{m+1}(x)=x^{m+2}$. These graphs are denoted by $JW_{m}^{1}(q)$ and $JW_{m}^{2}(q)$ respectively.

Theorem 5.1**.**

For all prime power $q$ and $1\leq m+1\leq q-1$, the distinct eigenvalues of $JW_{m}^{1}(q)$ are

[TABLE]

The multiplicity of the eigenvalue $\pm\sqrt{iq}$ is

[TABLE]

and

[TABLE]

Proof.

Let $(\omega_{1},\omega_{2},\cdots,\omega_{m+1})\in\mathbb{F}_{q}^{m+1}$ and $f(X)=\omega_{1}+\omega_{2}X+\cdots\omega_{m}X^{m-1}+\omega_{m+1}X^{m+1}$.

In the case of $f=0$, $|\{u\in\mathbb{F}_{q}|f(u)=0\}|=q$. Thus, $JW_{m}^{1}(q)$ has $\pm q$ as its eigenvalues.

For any $0\leq i\leq m-1$, there exists a polynomial $f$ over $\mathbb{F}_{q}$ of degree at most $m+1\leq q-1$, which has exactly $i$ distinct roots in $\mathbb{F}_{q}$.

For $i=m,m+1$, it is easy to compute $N_{1}(m+1,i)>0$ by Theorem 3.1. Then there exists a polynomial of degree $m+1$ which has exactly $i$ distinct roots in $\mathbb{F}_{q}$. Thus by Theorem 2.2 in [2], $JW_{m}^{1}(q)$ has $\pm\sqrt{iq},\ 0\leq i\leq m+1$ as its eigenvalues, and by Theorem 2.1 and Theorem 3.1, we obtain that the multiplicity of the eigenvalue $\pm\sqrt{iq}$ is

[TABLE]

and

[TABLE]

∎

Similarly, we have the following theorem about the spectrum of $JW_{m}^{2}(q)$ by Theorem 4.1.

Theorem 5.2**.**

Suppose that $p$ is an odd prime number and $q=p^{2e}$. If $1\leq m+2\leq q-1$, then we have

$1)$**

[TABLE]

are the distinct eigenvalues of $JW_{m}^{2}(q)$. The multiplicity of the eigenvalue $\pm\sqrt{iq}$ is

[TABLE]

$2)$* If $m\leq i\leq m+2$, then $\pm\sqrt{iq}$ are the distinct eigenvalues of $JW_{m}^{2}(q)$ if and only if $N_{i}(x^{m+2},m-1)>0$. Furthermore, the multiplicity of the eigenvalue $\pm\sqrt{iq}$ is*

[TABLE]