Question

let (a,57)=1 then a^18 congruent to 1 in modulo 57.

Answer 1

Little Fermat gives you

a^18 = 1 ( mod 19 )
a^2 = 1 ( mod 3 )

Answer 2

second congruence implies
a^18 = 1 ( mod 3 )

Answer 3

That means a^18 - 1 is divisible by both 3 and 19

Answer 4

Since 3 and 19 are primes we can conclude that a^18 - 1 is also divisible by their product 3*19