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Math Help - congruence property modulo p^2

  1. #1
    Junior Member
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    congruence property modulo p^2

    I have attached the problem, and my work so far. Thanks in advance!
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  2. #2
    Super Member fardeen_gen's Avatar
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    Jun 2008
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    Assume that such b doesn't exist. Then,

    f(k)=(kp+a)^{p-1}=(p-1)kpa^{p-2}+a^{p-1}=-kpa^{p-2}+a^{p-1}=1(\mbox{mod}\ p^2) for all integers k,

    i.e. f(k_{1})-f(k_{2})=(k_{2}-k_{1})pa^{p-2}=0(\mbox{mod}\ p^2) for all integers k_{1}, k_{2}, which is clearly impossible \rightarrow contradiction!
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