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Thread: Problem with Congruences

  1. #1
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    Problem with Congruences

    Let p is prime number.
    Prove a^p \equiv b^p\ (\mod p) \Rightarrow a^p \equiv b^p\ (\mod p^2). How to do it?

    It is clear that: a^p-b^p \equiv 0\ (\mod p)

    a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+\ldots + b^{p-1}), a^{p-1}\equiv 1 (\mod p), b^{p-1}\equiv 1 (\mod p)

    Is it a^{p-i}b^{i-1}\equiv 1 (\mod p) and why?

    Thank you.
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  2. #2
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    Re: Problem with Congruences

    Since $a^p\equiv a\,(mod\, p)$, $a=b+qp$ for some q. Thus
    $$a^p-b^p=\sum_{k=1}^p{{p}\choose{k}}b^{p-k}(qp)^k=pb^{p-1}qp+\sum_{k=2}^p{{p}\choose{k}}b^{p-k}(qp)^k$$
    We're done.
    Worth remembering: $p$ divides each binomial coefficient ${{p}\choose{k}}\text{ for }1\leq k<p$.
    Thanks from feferon11
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