Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$ \sum_{\substack{d\mid n\\d>0}} d^{\frac{p-1}{2}}\equiv0\pmod{p} $$

I've tried using the fact that the sum divides evenly into p to prove it directly. But I just can't seem to figure out a concrete proof.

share|improve this question

2 Answers 2

Change the sum to Euler Product form: $$ \sum_{d\mid n} d^{\frac{p-1}{2}}=\prod_{\substack{q\mid n\\q\in\mathbb{P}}} (1+q^{\frac{p-1}{2}}) $$

By Euler's Criterion, If $q$ is a QNR, $q^{\frac{p-1}{2}}\equiv-1\pmod{p}$. Since $n$ is a QNR and quadratic character is multiplicative, there must be at least one such $q$.
$$ \Rightarrow \sum_{d\mid n} d^{\frac{p-1}{2}}=0 $$

share|improve this answer

Since $n$ is not a perfect square, the divisors of $n$ can be divided into pairs $\{a,b\}$, where $ab=n$.

Note that $ab=n$ implies that one of $a$ and $b$ is a QR, and the other is an NR. (The product of two QR is a QR, as is the product of two NR.)

If follows that one of $a^{(p-1)/2}$ or $b^{(p-1)/2}$ is congruent to $1$ modulo $p$, and the other is congruent to $-1$. So their sum is congruent to $0$ modulo $p$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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