test for n = 1
1^3 + 2(1) = 3 , which is a multiple of 3
assume it is true for n = k
that is, k^3 + 3k is a multiple of 3, or it is divisible by 3
show that it is also true for n = k+1
that is, show that (k+1)^3 + 3k is a multiple of 3
let's take the difference
(k+1)^3 + 2(k+1) - (k^3 + 2k)
= k^3 + 3k^2 + 3k + 1 + 2k + 2 - k^3 - 2k
= 3k^2 + 3k + 3
= 3(k^2 + k + 1)
which is divisible by 3, (since 3 is a factor)
so n^3 + 3n is always a multiple of 3
the property I used is the following:
if 2 numbers are divisible by the same number, then their difference is divisible by that same number
e.g. 91 and 49 are both divisible by 7
then 91-49 or 42 is also divisible by 7
-- try it for other numbers.
since we knew the second number, k^3 + 2k , was divisible by 3 and the result was divisible by 3, then the first number, (k+1)^3 + 2(k+1) has to be divisible by 3
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