Prove that for every natural $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$ Prove that for every natural number $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$.
I've noticed that $(n^2 + n) = n(n+1)$ so these are two successive numbers hence one of them can be divided by two.
I suppose that I should prove that $(n^2 + n)(n^2 + 2)$ can be divided by $3$ but I don't know how to do that.
 A: Either $n$ is divisible by $3$, or $n^2+2=(n+1)(n-1)+3$ is.
A: HINT: 
$$\begin{align} & (n^2+n)(n^2+2) \\
&=n(n+1)(n^2+2) \\
&=n(n+1)(n^2-1+3)\\
&=\color{red}{n(n+1)(n-1)} \cdot (n+1)+3\cdot\color{blue}{n(n+1)}\end{align}$$
Get the clue?
A: You can also use the brute force method (but don't let the name fool you; its actually pretty easy). This method is great because fundamentally, it requires no real cleverness; you don't have to work out tricky ways of factorizing things or anything like that.
The idea, in short, is to work modulo 6 the whole way through.
Begin by noting that if we're working modulo $6$, the sequence $\mathop{\lambda}_{n:\mathbb{N}}n^2$ looks like this:
$$(0,1,4,3,4,1; 0,1,4,\ldots)$$
So $\mathop{\lambda}_{n:\mathbb{N}}n^2+n$ equals
$$(0,2,0,0,2,0;\ldots)$$
And $\mathop{\lambda}_{n:\mathbb{N}}n^2+2$ equals
$$(2,3,0,5,0,3; \ldots)$$
So $\mathop{\lambda}_{n:\mathbb{N}}(n^2+n)(n^2+2)$ equals
$$(0,0,0,0,0,0; \ldots)$$
In other words, we have that $$6 \mid (n^2+n)(n^2+2)$$ for all $n \in \mathbb{N}$.
So we're done. I should add that the problem with this method is it provides very little insight into why the theorem is actually true. But still, you get a proof essentially for free, without really having to think. That's got to count for something, right? For what its worth, this is similar to proving identities in the language of Boolean algebra by brute force, using truth tables.
