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Math Help - really cant do

  1. #1
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    really cant do

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  2. #2
    Newbie Klaus's Avatar
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    Hello,
    For question (a): Each number n in the sum is lower than p. So, (n,p)=1.
    Hence, we can apply Fermat's Little Theorem:
    n^{p-1}\equiv 1\text{ mod }p
    Let's multiply both sides by n.
    n^{p}\equiv n\text{ mod }p
    Thus
    \sum_{n=1}^{p-1} n^p\equiv \sum_{n=1}^{p-1} n\equiv \frac{(p-1)(p)}{2}\text{ mod }p
    Isn't it ?
    Last edited by Klaus; June 23rd 2008 at 06:57 AM.
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  3. #3
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    For (a) we can strengthen the problem. 1^k + 2^k + ... + (p-1)^k \equiv 0 (\bmod p) if (p-1)\not | k and \equiv p if (p-1)| k.

    For (b) note 1+...+p = p\cdot \frac{p-1}{2}. Let a=\frac{p-1}{2}.
    Let r be remainder of (p-1)! mod pa.
    Thus, (p-1)!\equiv r(\bmod pa) thus (p-1)!\equiv r(\bmod p) and (p-1)!\equiv r(\bmod a).

    By Wilson's theorem it follows that r\equiv -1(\bmod p).
    Also r\equiv 0(\bmod a) because a divides (p-1)!.
    If we let r=p-1 it satisfies both conditions.
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  4. #4
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    For the second problem let y be the order of a mod b. Let a^x\equiv 1(\bmod b). Write x=qy+r where 0\leq r < y. Then, a^x \equiv 1 (\bmod b)\implies a^{qy+r}\equiv 1(\bmod b)\implies a^r\equiv 1(\bmod b). Since y is the order of a it means a^r\not \equiv 1 if 0<r<y. But since 0\leq r < y and a^r\equiv 1(\bmod b) it follows that r=0. And thus x=qy+r = qy\implies y|x.
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