If a is congruent to b (mod3n) prove that a^3 is congruent to b^3 (mod9n) Solution a is congruent to b mod(3n) So, 3n divides (b-a) To Prove: a 3 is congruent to b 3 mod(9n) So, we need to prove 9n divides (b 3 - a 3 ) Now, (b 3 - a 3 ) = (b - a)( b 2 + ab + a 2 ) ------------(1) since in R.H.S 3n divides (b-a), So we only need to show that 3 divides ( b 2 + ab + a 2 ) We can write (b-a) = 3nq , for some integer q or b = 3nq + a So, ( b 2 + ab + a 2 ) = [ (3nq+a) 2 + a(3nq+a) + a 2 ] = 9n 2 q 2 + 6nq + 3nq 2 + 3a 2 = 3 (3n 2 q 2 + 2nq + nq 2 + a 2 ) So, 3 divides ( b 2 + ab + a 2 ) Hence proved :) .