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.