Wondering if this is a typo?

**Th3sandm4n** · March 29th 2009, 07:22 PM

Let p be prime.
Assume that $\sum_{k=0}^pa^{2k-1} \equiv 0 mod p$ .
Show that $p = 2$ or $p = 3$ .

I can prove this for $\sum_{k=1}^pa^{2k-1} \equiv 0 mod p$ (notice the summation goes from 1 to p instead of 0 to p).

I was just wondering if this is a typo since $a^{2*0-1}=a^{-1}$ doesn't make sense when dealing with $modp$ This was a handwritten one, so maybe my teacher just made the error of starting the summation at 0 instead of 1?

As of now I'm assuming it was an error since I can prove it if it starts at 1

EDIT : Just read about Modular Inverses, so I guess $a^{-1}$ is possible.

So my original plan of saying if it's even then $p = 2$ works, and odd $p = 3$ works...well, doesn't work if you start at $k=0$ . And my intuition is also telling me that that wasn't even correct. The way I was doing it makes the odds not work since i get LHS some multiple of two $\not \equiv 0 mod 3$ and LHS has remainder 1 $\not\equiv 0 mod 2$ .

It's 2am now though, and I've been doing these problems for a few hours and need rest

. Hopefully I can think better in the morning,and any help would be appreciated.

To restate the problem in case it was lost in all my typing and thinking (which is probably illogical at this hour)
__________
Let p be prime.
Assume that $\sum_{k=0}^pa^{2k-1} \equiv 0 mod p$ .
Show that $p = 2$ or $p = 3$ .

**Media_Man** · April 30th 2009, 01:01 PM

In the wording of the problem, are you supposed to consider "for all a" being all natural numbers???

Consider the case where p=2. The scary-looking $\sum_{k=0}^pa^{2k-1}$ is actually equal to $a^{-1}+a+a^3$ . And $\equiv 0 \bmod p$ just means "is even." Now, $0^{-1}$ does not exist $\bmod p$ for any $p$ . So If $a$ is even, $p=2$ does not work, and if $3|a$ , then $p=3$ does not work.

In modular arithmetic, $a^{-1}$ exists $\bmod p$ if and only if $a$ and $p$ are relatively prime.

Therefore, we need some restrictions on a. Either way, this problem needs to be clarified before any proof is pursued.

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Not a typo most likely, so now don't know how to solve.

This is a typo

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