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- February 1st 2010, 04:30 PM #1

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- Feb 2010
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## n^3 mod 6 = n mod 6

Hi,

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.

- February 1st 2010, 04:35 PM #2