Let $p$ be an odd prime number. Prove that $$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$$ where $\left( \frac{a}{p}\right)$ is the Legendre symbol.

This seems to be a tricky one! I've tried using the property $\left( \frac{ab}{p} \right)=\left( \frac{a}{p}\right) \left( \frac{b}{p} \right)$ and prime factoring all the non-primes but to no avail. I had a quick thought of induction haha, but that was silly. I tried factoring the common Legendre symbols like $\left( \frac{3}{p}\right) \left[ \left( \frac{2}{p} \right) + \left( \frac{4}{p} \right) \right]$ but that didn't bring anything either. And I've been looking for pairwise cancellation with $1$ term leftover, but that does not seem to work.

Can you help?

  • 2
    $\begingroup$ If $p=3$, you only have $\left(\frac23\right)=1$, which is false.Am I missing something? $\endgroup$ – chubakueno Nov 5 '13 at 4:54
  • $\begingroup$ I apologize, the sum should be $-1$ and not $1$. Edited! $\endgroup$ – Numbersandsoon Nov 5 '13 at 15:16

Let $a^\ast$ be the inverse of $a$ modulo $p$. Then $$\left(\frac{a(a+1)}{p}\right)=\left(\frac{a(a+aa^\ast)}{p}\right)=\left(\frac{a^2(1+a^\ast)}{p}\right)=\left(\frac{1+a^\ast}{p}\right).$$

As $a$ ranges from $1$ to $p-2$, the number $1+a^\ast$ ranges, modulo $p$, through the integers from $2$ to $p-1$. But the sum from $1$ to $p-1$ of the Legendre symbols is $0$, so our sum is $-1$.

  • $\begingroup$ Ahhhh, whenever we have congruences modulo some prime, always consider modular inverses!!! And as you said, the sum should be -1 (now edited). Thank you, André Nicolas!! $\endgroup$ – Numbersandsoon Nov 5 '13 at 14:58
  • $\begingroup$ You are welcome. I have removed the now irrelevant last sentence. $\endgroup$ – André Nicolas Nov 5 '13 at 15:04

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.