5

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 $a^2$, but…

Show that $$\sum _{a=1}^{(p-1)/2} \left(\frac{a}{p}\right)=0$$ for a prime $p \equiv 1 \pmod 4$.

Martin Sleziak
  • 56,060
user2214
  • 445
  • Are those quadratic residues? – Beni Bogosel Apr 08 '15 at 22:32
  • Yes they are quadratic residues @BeniBogosel – user2214 Apr 08 '15 at 22:35
  • Hint: Let $a^\ast$ be the inverse of $a$. Then $a(a+1)\equiv a(a+aa^\ast)\pmod{p}$, so $(a(a+1)/p)=(a^2((1+a^\ast)/p)=((1+a^\ast)/p)$. – André Nicolas Apr 08 '15 at 22:41
  • It seems interesting can you be more explicit @AndréNicolas How does one approach these ones? – a1bcdef Apr 09 '15 at 18:28
  • 3
    There is already an answer to the first question, on the hint side, but my answer would be just an expansion, so I prefer to omit it. For the second, we need to use the fact the prime is of the shape $4k+1$, and therefore $-1$ is a QR. Thus for any $a$ with $1\le a\le \frac{p-1}{2}$ we have $(a/p)=((p-a))/p)$. So the sum of the Legendre symbols from $1$ to $p-1$ is twice the sum of the Legendre symbols from $1$ to $\frac{p-1}{2}$. But the sum of the Legendre symbols from $1$ to $p-1$ is $0$ (that was the hint given). – André Nicolas Apr 09 '15 at 20:23
  • You can find your first question here: http://math.stackexchange.com/questions/333704/sum-of-the-product-of-consecutive-legendre-symbols-is-1 (and in other posts linked to that one). – Martin Sleziak Apr 16 '15 at 05:14
  • Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Apr 16 '15 at 05:17

2 Answers2

1

The first question is answered in several other posts:

For the second one, first you can notice that $$\newcommand\jaco[2]{\left(\frac{#1}{#2}\right)}\sum_{a=1}^{p-1} \jaco ap = 0,$$ since this sum contains the same number of $1$'s and $(-1)$'s.

Using the fact that $$\jaco{p-a}p = \jaco{-a}p = \jaco{-1}p \jaco ap \overset{(*)}= \jaco ap$$ you can divide the above sum into two sums which are equal to each other and therefore they are both zero.

(Can you say why the equation denoted by $(*)$ holds?)

Community
  • 1
Martin Sleziak
  • 56,060
0

Some hints:

  1. $a(a+1)$ is a quadratic residue if and only if $\frac{a+1}{a}=1+a^{-1}$ is quadratic residue and then how many quadratic residu of the form $1+x$ with $x$ invertible are there?
  2. How many quadratic residues are there in $\Bbb Z^*$ and how many non quadratic residue are there?
Elaqqad
  • 13,983