1
$\begingroup$

Let. $p$ be an odd prime. Consider the following sum of Legendre Symbols:

$(\frac{1}{p})(\frac{2}{p}) + (\frac{2}{p})(\frac{3}{p}) + \cdots + (\frac{p-2}{p})(\frac{p-1}{p})$.

Show that this sum is equal to $-1$. Using the algebra of the Legendre symbol i can show that this sum is the same as

$\sum_{i = 2}^{\frac{p-1}{2}}(\frac{i-1}{p})(\frac{i}{p}) + (\frac{\frac{p-1}{2}}{p})(\frac{\frac{p+1}{2}}{p})$.

I can also show that this last term is equal to 1 if $p \equiv 1 \mod 4$ and $-1$ if $p \equiv 3 \mod 4$ via Guass' Lemma. I'm really more interested in a hint than a full solution but any help would be greatly appreciated.

$\endgroup$
1
$\begingroup$

With the assumptions $\left(\frac{0}{p}\right)=0$, $p\equiv 1\pmod{2}$, by exploiting the multiplicativity of the Legendre symbol we have

$$ \sum_{k=1}^{p-2}\left(\frac{k}{p}\right)\left(\frac{k+1}{p}\right)=\sum_{k=1}^{p-2}\left(\frac{k^2+k}{p}\right)=\sum_{k=1}^{p-2}\left(\frac{1+k^{-1}}{p}\right)$$ where $k^{-1}$ stands for the inverse of $k$ in $\mathbb{F}_p^*$. Now it is enough to consider how that map $k\mapsto 1+k^{-1}$ acts on $\{1,2,\ldots,p-1\}$ and recall that $$ \sum_{k=1}^{p-1}\left(\frac{k}{p}\right)=0, $$ i.e. that in $\mathbb{F}_p^*$ there are as many quadratic residues as quadratic non-residues.

$\endgroup$
  • $\begingroup$ I'm little confused by the second inequality where we go from $k^2 + k$ to $1 + k^{-1}$. How do we know equality is preserved. $\endgroup$ – Sean Haight Oct 4 '16 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.