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).

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)