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let p be a prime number, f(x) a polynomial with integer terms. assuming the congruence f(x) = 0 (mod p) has k different roots (x1, x2, ... , xk) mod p, and p does NOT divide: f`(x1), f`(x2), ... f`(xk) Prove: the congruence f(x) = 0 (mod p^a) has k different roots mod p^a, for each a >=1. 2. let p be a prime number >= 3. prove that p divides A, where: 1 + 1/2 + 1/3 + ... + 1/(p-1) = A / (p-1)!
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