All Questions
10
questions
1
vote
1
answer
250
views
Showing that $\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$, where $\left(\frac{x}{p}\right)$ is the Legendre symbol
The question is,
Show that $$\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$$ where the operation $\left(\frac{x}{p}\right) = \pm 1$ if $x$ is a quadratic residue/non-residue and $0$ ...
- 1,166
5
votes
3
answers
843
views
Prove the congruence $ \sum_{r=1}^{p-1}{(r|p) * r } \equiv 0 \pmod p.$
Prove that if $p$ is prime and $p\equiv 1 \pmod4$, then $$ \sum_{r=1}^{p-1}{(r|p) * r } \equiv 0 \pmod p.$$
( $(r|p)$ is a Legendre Symbol )
I know that $\sum_{1 \le r \le p}{(\frac{r}{p})} = 0$, but ...
- 105
2
votes
1
answer
1k
views
Sum of Legendre symbols: $\sum _{n=1}^p \left(\frac{an+b}{p}\right)=0$ [duplicate]
If $(a,p)=1$ and $p$ is an odd prime, prove the Legendre symbol sum
$$\sum _{n=1}^ p \left(\frac{an+b}{p}\right)=0.$$ Where $b$ is any integer.
I know the fact that $\sum_{a=1}^p \left( \frac{a}{p} \...
- 129
2
votes
1
answer
271
views
Why is $\sum_{x=1}^{p-1}\left(\frac{x}{p}\right)=\left(\frac{0}{p}\right)$?
For $p$ an odd prime, why does $$\sum_{x=1}^{p-1}\left(\frac{x}{p}\right)=\left(\frac{0}{p}\right)$$
where $\left(\frac{x}{p}\right)$ is the Legendre symbol.
I'm not sure if I have given enough ...
- 2,297
5
votes
2
answers
4k
views
Proving summations involving the Legendre symbol
In the following, let $(\frac{a}{p})$ denote the Legendre symbol. Then
Show that $$\sum _{a=1}^{p-2} \left(\frac{a(a+1)}{p}\right)=-1$$ for an odd prime $p$.
I was thinking of factoring out $a^2$, ...
- 435
5
votes
1
answer
2k
views
Sums of Primitive Roots and Quadratic Residues when $p \equiv 3\pmod 4$
Define
$$R_{p}=\{ r \mid r: \text{primitive root of p}, 1 \le r \le p \}$$
and also
$$Q_{p}=\{ a \mid a: \text{quadratic residue of p}, 1 \le a \le p \}$$
$$Q_p^c=\{a \mid a: \text{...
- 1,502
6
votes
2
answers
2k
views
How can I prove these summations for the legendre symbol
How can I prove for the Legendre symbol that:
$$\sum_{a=1}^{p-1}{\left(\frac{a(a+1)}{p}\right)}= -1 = \sum_{b=1}^{p-1}{\left(\frac{(1+b)}{p}\right)}$$
- 61
5
votes
1
answer
3k
views
sum of the product of consecutive legendre symbols is -1
How do I prove the formula $\newcommand{\jaco}[2]{\left(\frac{#1}{#2}\right)}\sum\limits_{a=1}^{p-2} \jaco{a(a+1)}p = -1$ where a varies from 1 to p-2 and p is a prime
I got as far as $\jaco{p-a}p = \...
- 653
5
votes
1
answer
2k
views
Legendre symbol: Showing that $\sum_{m=0}^{p-1} \left(\frac{am+b}{p}\right)=0$
I have a question about Legendre symbol.
Let $a$, $b$ be integers. Let $p$ be a prime not dividing $a$. Show that the Legendre symbol verifies:
$$\sum_{m=0}^{p-1} \left(\frac{am+b}{p}\right)=0.$$
I ...
- 686