Return to Revisions

Let $A_q=\{x^2:x\in\mathbb{F}_q^{*}\}$. Let $\pi^{\prime}$ be the permutation on $A_q$ defined by $$\pi^{\prime}(a_k)=a_{\pi(k)}.$$ Then $$ \sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{a\in A_q}(a\pi^{\prime}(a)) $$ so that if $\pi$ is the identity permutation $$ \sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{a\in A_q}(a^2). $$ Hence if if $\pi$ is the identity permutation and $q$ is a prime congruent to $3\bmod 4$, $$ \sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{a\in A_q}(a)=0. $$ Let $a\not=b$ and $\pi^{\prime}_{a,b}$ be the tranposition $\pi^{\prime}_{a,b}(a)=b$,$\pi^{\prime}_{a,b}(b)=a$ and $\pi^{\prime}_{a,b}(c)=c$, $c\not=a,b$. Hence if $q$ is a prime congruent to $3\bmod 4$, $$\sum_{m\in A_q}m\pi^{\prime}_{a,b}(m)=-(a+b)^2$$ where $a,b\in A_q$. Let $\pi^{\prime}_{a,b,c,d}$ be the product of two transpositions $\pi^{\prime}_{a,b}$, $\pi^{\prime}_{c,d}$, $a,b,c,d$ all distinct in $A_q$. Then if $q$ is a prime congruent to $3\bmod 4$, $$ \sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{m\in A_q}(m\pi^{\prime}_{a,b,c,d}(m))=-(a+b)^2-(c+d)^2. $$ So for $q$ a prime congruent to $3\bmod 4$, $$ \{\sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}:\pi\in S_{(q-1)/2}\}=\mathbb{F}_q $$ if every $k\in\mathbb{F}_q$ can be represented as $$ k=(a+b)^2+(c+d)^2 $$ where $a,b,c,d$ are distinct elements in $A_q$.