# Prove that if a^p is congruent to b^p (mod p) and a and b are integers not divisible by the prime p, then a is congruent to b (mod p).

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### 1 Answer

It is given that a and b are not divisible by p. We have to prove that if a^p is congruent to b^p(mod p), then a is congruent to b(mod p).

a^p is congruent to b^p(mod m), implies (a^p - b^p) = k*p, where k is an integer.

a^p - b^p can be expressed as a product (a - b)[C(p, 0)*a^(p-1) - C(p, 1)*a^(p-2)*b + C(p, 2)*a^(p-3)*b^2 -... - C(p, p)*b^(p-1)]

As a and b are not divisible by p, none of the terms in the expansion above except (a - b) can be divisible by p.

But a^p - b^p = k*p, implies a - b is divisible by p

As a - b is divisible by p, a is congruent to b(mod p)

**This proves that a is congruent to b(mod p) if a^p is congruent to b^p(mod p)**