# Show that n(n + 2)(n + 4) is either divisible by 16, or is an odd number.

I understand more or less how to do this problem, however, I am having trouble actually showing that n can be divisible by 16.

Here's what I have done so far

If n is an odd integer, then n = 2k + 1, where k is any integer.

2k + 1 * (2k + 3) * (2k + 5) must be an odd number (due to multiply odd numbers)

If n is an even integer, then n = 2k, where k is any integer.

2k * (2k + 2) * (2k + 4)

How do I show that 2k * (2k + 2) * (2k + 4) is a multiple of 16?

$$2k \cdot (2k+2) \cdot (2k+4) = 2\cdot 2 \cdot 2 \cdot \underbrace{k(k+1)(k+2)}_{\text{focus here}}.$$ Now, notice that the final three terms are three consecutive integers, so one of them must be even. So, factor the two out of that one, and get $$2^4=16$$ out front.
• Yes, you'll have to figure out how to write it up. The terms either go odd-even-odd, in which case $k+1=2m$ or they go even-odd-even in which case $k=2m$. This allows you to move the $2$ correctly in the proof. – Randall Mar 8 '19 at 3:13
Your approach is fine and, as noted by Randall, will lead to a solution. The other way to handle the problem is to note that if $$n$$ is even, either $$n \equiv 2 \pmod{4}$$ or $$n \equiv 0 \pmod {4}$$, and in either case the result will follow.