Let n be an integer. If 3 does not divide (n^2 -1), then 3 divides n.
I have no idea where to even start. Please help!
I have no idea where to even start. Please help!
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Prove the contrapositive; I suspect it's much easier!
If 3 does not divide n, then 3 divides (n^2 - 1).
Generally speaking, 3 can only not divide n in 2 ways...
If n = 3m + 1 OR n = 3m + 2.
Sticking n = 3m + 1 into (n^2 - 1) gives...
9m^2 + 6m = 3(3m^2 + 2), so 3 does divide this!
Sticking n = 3m + 2 into (n^2 -1) gives...
9m^2 + 12m + 3 = 3(3m^2 + 4m +1), so 3 divides this too.
Hence we've proved the claim! :-)
If 3 does not divide n, then 3 divides (n^2 - 1).
Generally speaking, 3 can only not divide n in 2 ways...
If n = 3m + 1 OR n = 3m + 2.
Sticking n = 3m + 1 into (n^2 - 1) gives...
9m^2 + 6m = 3(3m^2 + 2), so 3 does divide this!
Sticking n = 3m + 2 into (n^2 -1) gives...
9m^2 + 12m + 3 = 3(3m^2 + 4m +1), so 3 divides this too.
Hence we've proved the claim! :-)