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Fairly new to Discrete Mathematics and I'm stumped on this one. So we're asked to prove:

If $n$ is an even integer greater than 2, then $2^n - 1$ is not a prime.

What I can come up with is that since $n > 2$, we know that $n$ is not prime since the only even $n$ happens to be $2$. We can write $n = 2k$ and so we rewrite $$2^n - 1 = 2^{2k}-1 = (2^k)^2 - 1 = (2^k-1)(2^k+1)$$

Up to here, am I even remotely correct? I'm not sure what else to say to take it from here to fully prove this. I also apologize for how I worded it, as I'm still trying to understand how to explain my proofs.

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    Hint: what does it mean for an integer to be prime? – 2017-02-09
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    Hint: $2^k-1>1$ if $n>2$ – 2017-02-09
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    try $n=4$ and $n=6$ – 2017-02-09
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    @vadim123 For an integer to be prime, it can only be divisible by 1 and itself. Am I able to say that $2^n-1$ is divisible by $2^k-1$, thus makes it not a prime? – 2017-02-09
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    @ProjectDefy, only if $2^k-1$ isn't $1$. – 2017-02-13

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You are doing fine. Now that you have shown a factorization of $2^n-1$ the only thing that can go wrong is that one of the factors is $1$. So if $n \gt 2, \ldots$

A more general statement is that if $n$ is composite, $2^n-1$ is never prime. The reasoning is the same. If $n=ab$ with $a,b \gt 1$ then $2^n-1$ is divisible by $2^a-1$ and $2^b-1$. You can just do the division or search this site for the proof. Your problem is the $a=2$ case of this.

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    In order to have a number be prime, it needs to be factored by atleast $1$. And since we know that $n>2$, it never is divisible by $1$? – 2017-02-09
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    @ProjectDefy It's always divisible by 1. Your way of writing it shows that it's also divisible by something else smaller than itself. – 2017-02-09
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    Ah, the if $n$ is composite, then $2^n-1$ is never prime makes much more sense now. I didn't think of that. I think I got it now. Thank you! – 2017-02-09
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    "In order to have a number be prime, it needs to be factored by at least $1$": Whoa, reality check! A positive integer $n$ is prime if and only if its only factors are $1$ and $n$, – 2017-02-09
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    @ProjectDefy: you are trying to prove a number composite, which means that it has a factor that is neither $1$ nor the number. If your $n=ab$ and neither $a$ nor $b$ is $1$, $n$ is composite. – 2017-02-09
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General advice: if you have a statement of the problem, make sure that you used all the points of the statement. Specificially, in this problem you should ask yourself a question: where did you used that $n>2$? Why your proof doesn't work for $n=2$? Then you'll see that to complete the proof you should in fact say why both $2^k-1$ and $2^k+1$ are not equal to $1$.

Also note that $2^{2k}-1$ is always divisible by $3$, this is another way to approach this problem.

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    That's some pretty good advice. So would something like: Since we know that $n$ is even and also is greater than $2$, we have to explicitly say that $n > 2$. – 2017-02-09
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    Well, if you never say in the proof the phrase like "Since $n>2$ we have…" while you have $n>2$ in the statement, it is suspicious and can point that something is wrong with the proof (though in some cases it can be just that the statement of the problem is excessive). – 2017-02-09
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    Well, here you say something like this though, I didn't read carefully. But you never use the consequence ($n$ is not prime). – 2017-02-09