Verified answer

How do you think you'd prove it? Write down what's given, and just work it out. It's pretty straightforward.

If 6 divides n+1. What does that mean? It means that there is an integer k such that 6k = n+1.

We need to show something about an expression involving n^2, so obviously you need to solve that for n, which is easy. We get n = 6k - 1

So what is n^2 +11? No problem figuring that out now, since we have an expression for n, and can just use simple algebra.

It's (6n - 1)^2 + 11= 36k^2 -12k + 1 + 11

= 36k^2 -12k + 12

Since we want to show this is a multiple of 12, we're obviously almost there. Just factor out a 12 to get

= 12 (3k^2 - k + 1)

= 12 times an integer.

That's what it means to be divisible by 12.

So n^2 + 11 is divisible by 12.

Well let's see. 6 divide n + 1, so that means n + 1 = 6k for some integer k.

So n = 6k - 1

Then n^2 = 36k^2 - 12k + 1

And n^2 + 11 = 36k^2 - 12k + 12

What do you notice about all 3 terms?

n+1 = 6k

(n + 1)^2 = n^2 + 2n + 1 = 36 k^2

n^2 + 2(n+1) - 1 = 36 k^2

n^2 + 12k - 1 = 36k^2

n^2 - 1 = 12 (3k^2 - k)

n^2 + 11 = 12(3k^2 - k + 1)

Edit: Alternative easier proof

n + 1 = 6k

n = 6k - 1

n^2 = 36k^2 - 12k + 1

n^2 + 11 = 36k^2 - 12k + 12 = 12 (3k^2 - k + 1)