I have a problem that I haven't quite been able to solve.
Prove for any n, n^3 mod 6 = n mod 6
I know that a mod n = b mod n iff n|(a-b) but I can't quite get there.
This is as far as I have gotten:
Rewrite n^3 mod 6 = n^3 = 6q + r and n mod 6 = n = 6t + s for some integers q, t, s, r where r >= s.
Then subtract both equations so:
n^3 - n = (6q +r) - (6t +s) = 6(q-t) + (r-s)
If r = s then r-s = 0 and 6|(n^3 -n)
After that point I get stuck. Is this the right way to prove this or am I making it too complicated?
Any help would be appreciated. Thanks.