# N(p) number of solution to x^x =1 (mod p) Miklós Schweitzer 2010

Let $p$ a prime number and $N(p)$ the number of solution to $x^x \equiv 1$ (mod $p$) in $1\leq x \leq p$ . Prove that for sufficiently large $p$ there exist a constant $c < \frac{1}{2}$ such that $N(p) \leq p^c$

(Miklós Schweitzer 2010)

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does this work if u substitute 'a' with 'x' in this - uk.answers.yahoo.com/question/index?qid=20110604013314AAW75Z7 –  Bhargav Sep 10 '11 at 11:16
@Mahan:i am sure that your either of your question is going to get closed,as you posted the same at MO here its from my past experience, i suggest you just to post either at MO or at Math stack exchange,and wait for 2 days for your answer,i am sure that question gets closed ,i suggest that you close it before someone suggests you,as i had many bitter experiences posting in that way,thank you ,decision is left to you ,anyway –  Iyengar Sep 10 '11 at 11:41
and if you feel your Question to be a good one ,i mean if it is a Research level Question ,its better you post it at MO, –  Iyengar Sep 10 '11 at 11:43
buts its just a request,dont take it as a command,i neither have rights to say so,nor i am not a owner of MO,so i said just for your benefit,as some people just look at such small mistakes,and comment without answering the main question,which happened in my case many times –  Iyengar Sep 10 '11 at 11:48
@iyengar : thanks for your advice . I want to delete the MO one but I cant ! @anon : $kg^k$ mod $p-1$ is non-sense , I mean we cant work with it . –  Mahan Sep 10 '11 at 12:08