I've been working on this problem for the past two and a half hours. The problem is: Prove by induction that n^3 + 2n is divisible by 3 for non-negative integer values of n. No matter how I try, I cannot manipulate the expression of the inductive step so that it is divisible by 3. I have done the base step and assumed that P(k): k^3 + 2k is divisible by 3. Now I need to show that this implies that P(k+1): (k+1)^3 + 2(k+1) is divisible by 3. I have tried manipulating P(k+1) in probably every possible way. The closest I have got to solving the problem, is getting part of the expression to be k^3 + 2k (which is divisible by 3 by assumption) and a constant 3. (k+1)^3 + 2(k+1) = k^3 + 2k^2 + 2k + 1 + 2k + 2 = k^3 + 2k^2 + 4k + 3 I also tried P(k+1) - P(k) and see if that would give an expression that is factorizable by 3. (Since the difference of two integers a and b is divisible by x, if and only if a and b are divisible by x.) P(k+1) - P(k) = [(k+1)^3 + 2(k+1)] - [k^3 + 2k] = [k^3 + 2k^2 + 2k + 1 + 2k + 2] - [k^3 + 2k] = 2k^2 + 2k + 3 which does not lead to a desirable expression. I think I've exhausted all the options that I know of. Can anyone help me please?