Example 3. Prove that n(n² + 5) is an integral multiple of 6 for all integers n, that is, there is an integer u such that n(n² + 5) = 6u.

Proof: We begin by proving the desired result for all the integers greater than or equal to 0 by mathematical induction.

When n = 0, n(n² + 5) is 0. Sinceis a multiple of 6, the result holds for n = 0.

We now assume it true for n = k, and seek to derive from this its truth for n = k + 1. Hence we assume that

with r an integer. We then wish to show that

with s an integer. Simplifying the difference between the left-hand sides of (2) and (1), we obtain

Since k and k + 1 are consecutive integers, one of them is even. Then their product k(k + 1) is even, and may be written as 2t, with t an integer. Now (3) becomes

Transposing, we have

Using (1), we can substitute 6r for k(k² + 5). Hence

Letting s be the integer r + t + l, we establish (2), which is the desired result when n = k + 1. This completes the induction and proves the statement for 

Now let n be a negative integer, that is, let n = -m, with m a positive integer. The previous part of the proof shows that m(m² + 5) is of the form 6q, with q an integer. Then
 

a multiple of 6. The proof is now complete.
 

Saturday, April 11, 1998