1. legendre symbol proof

Hi,

Need help with this proof. (a/p) is legendre symbol not to be confused with 1 divided by p;

Prove

(1.2/p) + (2.3/p) + (3.4/p) + ..... + ((p-2)(p-1)/p) = -1

Thanks

2. no one?? this is must be a really tough one.

3. Originally Posted by htata123

(a/p) is legendre symbol not to be confused with 1 divided by p; Prove $\displaystyle \sum_{k=1}^{p-2}\left ( \frac{k(k+1)}{p} \right)= 0, \ \ p > 2.$
Originally Posted by htata123

no one?? this is must be a really tough one.
be patient! also your question is not hard, it's just false! each summand of your sum is either 1 or -1 and we have an odd number of them. so they can never add up to 0.

4. but then why is it a question on the text book? I'm sure there is a proof because, my professor told us it we had to use the fact (a.b/p) = (a/p).(b/p) and the fact that (a/p) = (a*/p) where a* is the multiplicative inverse of a.

5. Originally Posted by htata123
but then why is it a question on the text book? I'm sure there is a proof because, my professor told us it we had to use the fact (a.b/p) = (a/p).(b/p) and the fact that (a/p) = (a*/p) where a* is the multiplicative inverse of a.
i don't know why it's in your text book, but you can see it for yourself that the sum is not 0. for example what do you get for p = 3?

also, are you sure you've posted the question as it is in your text book?

6. I'm so sorry. It's actually supposed to equal -1. Apologies.

7. Originally Posted by htata123

I'm so sorry. It's actually supposed to equal -1. Apologies.
ok! first note that $\displaystyle \sum_{k=1}^{p-2}\left ( \frac{k(k+1)}{p} \right)=\sum_{k=1}^{p-1}\left ( \frac{k(k+1)}{p} \right)=S_1.$ for any $\displaystyle 0 \leq j \leq p-1,$ let $\displaystyle S_j=\sum_{k=1}^{p-1}\left ( \frac{k(k+j)}{p} \right).$ then $\displaystyle S_0=p-1$ and for any $\displaystyle 1 \leq j \leq p-1: \ S_j=S_1.$ why?

also note that for any integer $\displaystyle k: \ \ \sum_{j=0}^{p-1}\left ( \frac{k+j}{p} \right)= \sum_{j=1}^{p-1}\left ( \frac{j}{p} \right)=0.$ therefore: $\displaystyle (p \ - \ 1)(S_1+1)=\sum_{j=0}^{p-1}S_j=\sum_{k=1}^{p-1} \sum_{j=0}^{p-1}\left ( \frac{k(k+j)}{p} \right)=\sum_{k=1}^{p-1}\left ( \frac{k}{p} \right)\sum_{j=0}^{p-1}\left ( \frac{k+j}{p} \right)=0.$ thus $\displaystyle S_1=-1.$