legendre symbol proof

**htata123** · April 18th 2009, 03:27 PM

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

**htata123** · April 18th 2009, 05:21 PM

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

**NonCommAlg** · April 18th 2009, 07:10 PM

Originally Posted by htata123

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

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.

**htata123** · April 18th 2009, 07:19 PM

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.

**NonCommAlg** · April 18th 2009, 07:24 PM

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?

**htata123** · April 18th 2009, 07:32 PM

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

**NonCommAlg** · April 18th 2009, 10:13 PM

Originally Posted by htata123

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

ok! first note that $\sum_{k=1}^{p-2}\left ( \frac{k(k+1)}{p} \right)=\sum_{k=1}^{p-1}\left ( \frac{k(k+1)}{p} \right)=S_1.$ for any $0 \leq j \leq p-1,$ let $S_j=\sum_{k=1}^{p-1}\left ( \frac{k(k+j)}{p} \right).$ then $S_0=p-1$ and for any $1 \leq j \leq p-1: \ S_j=S_1.$ why?

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

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