# summations in number theory

prove that $\sum_{j=1}^{p-1} (\frac jp) = 0$ and that $\sum_{k=1}^{p-2} (\frac{k(k-1)}{p}) = -1$ where $(\frac ap)$ is the Legendre symbol.

you should be able to use the latter relation to show that there are always either two consecutive quadratic residues or two consecutive nonresidues mod p

• The answer here might prove useful here. – Old John Dec 18 '13 at 22:02
• I have editied your question, but in the second sum you have $k=1$ ... should that be $k-1$ or $k+1$ maybe? – Old John Dec 18 '13 at 22:08
• @OldJohn: OP can decide, but we need $k+1$. – André Nicolas Dec 18 '13 at 22:40
• @AndréNicolas exactly - I was trying to get OP to make a decision, but the last editor seems to have made the decision for them. – Old John Dec 18 '13 at 22:42
• it should have been k+1. thank you for all the help! its much appreciated – david Dec 18 '13 at 23:09

The first result is an immediate consequence of the fact that if $p$ is an odd prime, it has exactly as many QR as NR in the interval $[1,p-1]$.
For the second, I take it we want to find $\sum_1^{p-2}\left(\frac{k(k+1)}{p}\right)$.
Let $k^\ast$ be (the least positive residue of) the inverse of $k$ modulo $p$. Then $$\left(\frac{k(k+1)}{p}\right)=\left(\frac{k(k+kk\ast)}{p}\right)=\left(\frac{k^2(1+k^\ast)}{p}\right)=\left(\frac{1+k^\ast}{p}\right).$$
As $k$ travels through the numbers $1$ to $p-2$, the numbers $1+k^\ast$ travel through the integers from $2$ to $p-1$. So our sum is the sum of all the Legendre symbols $\left(\frac{z}{p}\right)$ except $\left(\frac{1}{p}\right)$. This Legendre symbol is $1$. By the first result, the sum of all the Legendre symbols is $0$. This yields the second result.
Remark: We leave to you to show that from the second result that any prime $\gt 3$ has $2$ consecutive QR or two consecutive NR. (There are simpler ways to show that.)