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I am studying the sums of Legendre symbol and I have a question about it.

Let $p$ is a prime number, $p >7$, $p\equiv 7 \pmod 8$. Find $$\sum_{x=4}^{p-1} \left(\frac{x(1-p-x)}{p}\right).$$ I had some basic background working with Legendre symbols but not with sums of series. Any help is appreciated.

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The corresponding complete sum, where $x$ runs over all integers from $0$ to $p-1$, is \begin{align*} S &= \sum_{x=0}^{p-1} \left(\frac{x(1-x)}{p}\right) \\ &= \sum_{x=1}^{p-1} \left(\frac{x(1-x)/x^2}{p}\right) \\ &= \sum_{x=1}^{p-1} \left(\frac{(1/x)-1}{p}\right) \\ &= \sum_{y=1}^{p-1} \left(\frac{y-1}{p}\right) \\ &= \sum_{y=0}^{p-1} \left(\frac{y-1}{p}\right) - \left(\frac{-1}{p}\right) \\ &= - \left(\frac{-1}{p}\right) \\ &= 1 \end{align*} (where all divisions are in $\mathbb F_p$, and where $y=1/x$). It follows that the original incomplete sum is $$ 1 - \left(\frac{2(1-2)}{p}\right) - \left(\frac{3(1-3)}{p}\right) = 1 + \left(\frac{2}{p}\right) + \left(\frac{6}{p}\right) = 2 + \left(\frac{3}{p}\right). $$ Notice that the value of $(3/p)$ is determined by the residues class of $p$ modulo $3$.

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  • $\begingroup$ and you simplified $(2/p)$ because it is given by $p\bmod 8$ $\endgroup$ – reuns Dec 20 '19 at 14:23
  • $\begingroup$ @reuns: right, $p\equiv 7\pmod 8$ implies $(2/p)=1$. $\endgroup$ – W-t-P Dec 20 '19 at 14:25
  • $\begingroup$ @129492: Is the computation clear? If so, please, consider accepting the answer. $\endgroup$ – W-t-P Dec 21 '19 at 19:16
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Let $S$ be your sum. Then $S+p-4$ is the number of solutions in $\mathbb{F}_p^2$ of $y^2=x(1-x)$ with $x \neq 0,1,2,3$.

There is exactly one solution if $x=0$ or $x=1$, for $x=2$ the equation is about $y^2=-2$ which is impossible (since $8|p-7$), and for $x=3$ the equation is $y^2=-6$. Since $2$ is a square in $\mathbb{F}_p$, this equation has the same number of solutions as $y^2=-3$. By quadratic reciprocity, we can see that $\left(\frac{p}{3}\right)=\left(\frac{-3}{p}\right)$.

As a consequence, $S+p-1+\left(\frac{p}{3}\right)$ is the number of solutions in $\mathbb{F}_p^2$ of $y^2=x(1-x)$.

Define, for a point $P=(x,y) \neq O=(0,0)$, $t(P)=\frac{y}{x} \in \mathbb{F}_p^{\times}$.

Define, for a $t \in \mathbb{F}_p^{\times}$, $x(t)=\frac{1}{t^2+1}$, $y(t)=tx$ and $Q(t)=(x(t),y(t))$. One easily checks that $t(Q(t))=t$ and $Q(t(P))=P$ for a nonzero point on the curve. Thus the equation has $p$ solutions in total.

Therefore, $S+p-1+\left(\frac{p}{3}\right)=p$, ie $S=1-\left(\frac{p}{3}\right)$.

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