A counterexample may be sufficient, but since you were working with congruent arithmetic:
n^3 + 2n ~ 0mod2 (divisible by 2, no remainder, ~ means congruent)
n^3 + 2n = 2k
n^3 + 2n - 2k = 0
let k = n
n^3 + 2n - 2n = 0
n^3 = 0
Since n^3 cannot = 0 when n >=2, it's disproven.
Last edited by Yak; 07-28-2011 at 11:30 PM.
Reason: technical fix
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