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Old 07-28-2011, 09:53 PM
sjh sjh is offline
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Join Date: Oct 2009
Posts: 580
Quote:
Originally Posted by Yak View Post
A counterexample may be sufficient, but since you were working with modular 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.
I do not understand the transition from step 1 to 2 but I do not doubt it.

Once that is accomplished the rest of your proof is simple, clear and well done.

My career has not involved mod/etc except tangentially but for some reason I'v always enjoyed geometric proofs.
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