# How to approach proof of "For an integer n, if $n > 7$, then $n^2-8n+12$ is composite"?

Mathematics Asked by lswift on December 2, 2020

For the first part of this question, I was asked to find the either/or version and the contrapositive of this statement, which I found as follows:

i) either $$n leq 7$$, or $$n^2-8n+12$$ is composite

ii) if $$n^2-8n+12$$ is not composite, then $$n leq 7$$

Then we are asked to prove the statement.

I’ve examined all of our class notes and found mention of proof via factorising "to be covered in more detail later" (but no detail later included – perhaps it is considered too obvious?).

As this is such a simple question, judging by the two marks available for it, I’m concerned that asking another student will cause them to inadvertently tell me the answer.

I am finding this question hard to google around because findable simple proof examples do not seem to include quadratic expressions.

What would the right steps be to approach a proof like this? And/or is there an online resource that shows some similar examples with working/notes? I don’t mind handing in an incorrect answer if I’m able to make an attempt, but I’m just so uncertain of the right starting point that I cannot make a start.

$$n^2-8n+12=(n-2)(n-6)$$ Now if $$n>7$$ then both factors are at least $$2$$, which means $$n^2-8n-12$$ is composite. This settles the matter.

Correct answer by Parcly Taxel on December 2, 2020

$$n^2-8n+12=(n-4)^2-2^2=(n-2)(n-6)$$

Thus, if $$n>7$$, then both of the factors $$n-2$$ and $$n-6$$ are greater than $$1$$ and $$n^2-8n+12=(n-2)(n-6)$$ is a nontrivial factorization of $$n^2-8n+12$$.

Answered by Stinking Bishop on December 2, 2020