I am a bit confused with the question: For what prime $p$, $\left(\frac{-21}{p}\right) = 1$? I did something like that: $$\left(\frac{-21}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\left(\frac{7}{p}\right).$$ But I don’t have any ideas what to do next. Thanks for any help.

Given $a,b\in\mathbb Z^+$, and let $F_{a,b}:\mathbb N\to\mathbb N$ be a function such that $F_{a,b}(0)=0$, $F_{a,b}(1)=1$ and $F_{a,b}(n+1)=a\cdot F_{a,b}(n)+b\cdot F_{a,b}(n-1)$. $F_{1,1}$ correspond to the Fibonacci function. Conjecture for which I would like to see a proof: If $a,b$ are co-prime and $p>2$ is a prime that doesn’t divide $b$, then $F_{a,b}\left(p-\left(\frac{a^2+4b}{p}\right)\right)\equiv 0\pmod p$, where $\left(\frac{a^2+4b}{p}\right)$ is […]

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 I don’t know what to do with the multiplication by r.

If $n\in\Bbb Z^+$ is not a square, prove exist infinitely many primes $p$ such that $\left(\frac{n}{p}\right)=-1$. Note that if $p\nmid n$ and $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, then $\left(\frac{n}{p}\right)=\left(\frac{p_1}{p}\right)^{\alpha_1}\left(\frac{p_2}{p}\right)^{\alpha_2}\cdots \left(\frac{p_k}{p}\right)^{\alpha_k}$ $=\left(\frac{p_1}{p}\right)^{\alpha_1\pmod {2}}\left(\frac{p_2}{p}\right)^{\alpha_2\pmod{2}}\cdots \left(\frac{p_k}{p}\right)^{\alpha_k\pmod{2}}$ So i.e. prove exist infinitely many primes $p$ such that $\left(\frac{q_1}{p}\right)\left(\frac{q_2}{p}\right)\cdots \left(\frac{q_n}{p}\right)=-1$, no matter what primes $q_i$ you take.

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 context for this to necessarily be true, but I read it in lecture notes and can’t understand why it is true.

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 $\jaco{p-a}p = \jaco{-1}p \jaco ap$ so I can reduce the sum but that does not seem of much help.

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 know that $\displaystyle\sum_{m=0}^{p-1} \left(\frac{m}{p}\right)=0$, but how do I connect this with the previous formula? Any help is appreciated.

$$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}$$ how did they get the exponent. May be from Gauss lemma, but how. Suppose we have a = 2 and p = 11. Then n = 3 (6,8,10), but not $$15 = (11^2-1)/8$$ n is a way to compute Legendre symbols from Gauss lemma: $$\left(\frac{a}{p}\right) = (-1)^n$$

This question already has an answer here: Legendre symbol: Showing that $\sum_{m=0}^{p-1} \left(\frac{am+b}{p}\right)=0$ 1 answer

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