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

**carabidus** · November 1st 2009, 04:10 PM

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

Any thoughts?

**PaulRS** · November 2nd 2009, 03:45 AM

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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