TITLE: Prove that $2^{66}-1$ is not a prime number
QUESTION [9 upvotes]: I just started studying Discrete Mathematics and one of the exercises asks to prove that:
$$2^{66} - 1 \textrm{ is not a prime number}$$
The author suggests to use the well know relation: 
$$x^2 - 1 = (x + 1)(x - 1)$$
My proof goes as such (please forgive me — and do correct me — if I use any wrong symbol or term). Let's define the set $X$ as:
$$X=\{n\in\mathbb{N}\;|\;n>2\}$$
$\forall x\in X$ we define two numbers:
$$\begin{align}a & = x+1\\ b & = x-1\end{align}$$
Because $x>2$, then $a>1$ and $b>1$. Therefore we have
$$c=a\cdot b$$
which cannot be a prime number, because:
$$\frac{c}{a}=b \quad \textrm{or} \quad \frac{c}{b}={a}$$
This, in my view, is the proof of the theorem. 
Could you please let me know if my proof is correct, or if not, how do I prove the theorem?
Thank you very much!
Valerio
Edit: As cited in the comments/answers below, I didn't show the factorisation, so here we go:
$$2^{66}-1=(2^{33}+1)(2^{33}-1)$$
Now I understand I didn't need to use all the steps above, but I was trying to give a general proof, not just for the specific case.
Anyway, thank you all for your answers and showing interest in my question. I didn't expect it to create so much debate! :-)

REPLY [3 votes]: I'm surprised that there hasn't been a single mention of the term "Mersenne prime" or at least "Mersenne number" in the answers so far. I admit that aside from that detail, the answer I'm going to give is hardly different from the others.
It is a well-known fact that if $2^n - 1$ is prime (a Mersenne prime), then $n$ must itself be prime. And it is also well known that if $n$ is prime, then $2^n - 1$ is not necessarily prime.
In fact, it's much more likely to be composite. $2^{74207281} - 1$ is believed to be the $49$th Mersenne prime, and $74207281$ is itself the $4350601$st prime overall. Suppose they find a hundred Mersenne prime exponents between $37156667$ and $74207281$. That's unlikely, but it would still mean that less than $1\%$ of primes up to $74207281$ are Mersenne prime exponents.
The fact that composite $n$ means $2^n - 1$ is also composite has been known since before Marin Mersenne's time. To use the hint that was given to you, note that $$2^{2m} = (2^m)^2.$$ Since $66 = 2 \times 33$, it follows that $$2^{66} - 1 = (2^{33} - 1)(2^{33} + 1).$$ Clearly $1 < 2^{33} - 1 < 2^{33} + 1 < 2^{66} - 1$, meaning that $2^{66} - 1$ is divisible by positive integers other than $1$ and itself.