a^p=b^p mod p ==> a^p=b^p mod p^2 - Math Help Forum

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#1

$Old$ November 1st, 2009, 06:10 PM

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$Default$ a^p=b^p mod p ==> a^p=b^p mod p^2

Show that a^p=b^p mod p ==> a^p=b^p mod p^2 (where p is prime).

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$Old$ November 2nd, 2009, 05:45 AM

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First note that $a^p\equiv{b^p}(\bmod.p)$ implies $a\equiv{b}(\bmod.p)$ by Fermat's little Theorem.
Now, try using the binomial Theorem and note that $\binom{p}{k}\equiv{0}(\bmod.p)$ for $k=1,...,p-1$

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