# Thread: a^p=b^p mod p ==> a^p=b^p mod p^2

1. ## 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).

Any thoughts?

2. 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 $k=1,...,p-1$

>, apbp, mod