Algebra problem (28)

Problem

Prove that n3+5n {n=1,2,3,...} is a multiple of 6

Solution

Suppose that the above is true for some value of n.
We prove that in this case, it is true as well for n+1.
n + 1:
(n+1)3 + 5(n+1) = n3+3n2+8n+6
Subtract n3+5n, which results in
3n2+3n+6=3(n2+n+2).

This is a multiple of 3.
Also 3n2+3n+6 is even for all n, so a multiple of 2.
3n2+3n+6 is a multiple of 6.

So far we have proved that if n3+5n is a multiple of 6 for some n,
it is also a multiple of 6 for (n+1).

For n = 1 we get:
n3+5n = 6, which is true.
So it is true for n=2,3,4...etc.

This method is called "induction".