According to Fermat's Little theorem a^(p-1) mod(p) is 1. So a^k(p-1) mod(p)will also be 1 by splitting into k parts and apply modulus independently we get '1'. Am I missing something?
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math.stackexchange.com ? – Barmar May 07 '13 at 07:41
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I didn't get any response there – Alex May 07 '13 at 07:41
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1@Alex you posted your question there **ten minutes ago**. You generally should wait a bit more than **ten minutes** before deciding to cross post to a different site. And in any event this question is clearly off topic here. – AakashM May 07 '13 at 07:44
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@AakashM okay..but I thought there would be more users here – Alex May 07 '13 at 07:45
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That there are more users here doesn't somehow make it OK to ask off topic questions... – AakashM May 07 '13 at 07:46
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I will agree with alex, any other stack exchange forums are deserted compared to this forum. – Mercurial May 07 '13 at 07:49
2 Answers
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We know,
((a mod N) * (b mod N)) mod N = (a*b) mod N
a^(p-1) mod p = 1
Thus
( a^(p-1) * a^(p-1) * a^(p-1) * ... * a^(p-1) ) mod p = ( 1 * 1 * 1 * ... * 1) mod p = 1
Tada.
Mercurial
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You are right. In general, the equation holds as a^(k*phi(n)+b) is congruent with a^b modulo n where phi denotes the Euler-phi function, and a is relative prime to n.
elias
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