Let p be prime.

Assume that $\displaystyle \sum_{k=0}^pa^{2k-1} \equiv 0 mod p$.

Show that $\displaystyle p = 2$ or $\displaystyle p = 3$.

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

I was just wondering if this is a typo since $\displaystyle a^{2*0-1}=a^{-1}$ doesn't make sense when dealing with $\displaystyle modp$ 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 $\displaystyle a^{-1}$ is possible.

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

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)

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Let p be prime.

Assume that $\displaystyle \sum_{k=0}^pa^{2k-1} \equiv 0 mod p$.

Show that $\displaystyle p = 2$ or $\displaystyle p = 3$.