Factorization of huge integer I have to factorize the integer $n = 2^{214313833}-1$. Obviously this is not a prime, because $214313833 = 9623 \cdot 22271$, so $n_1=2^{9623}-1$, $n_2=2^{22271}-1$ are divisor of $n$, though $n\neq n_1 \cdot n_2$.
How can I get at least one prime factor of $n$? Are the smaller numbers still handleable by a normal computer (or am I too naive?)? Is there any other trick?

Let $p$ be one odd prime factor of $2^{22271}-1$
$$2^{22271}-1\equiv 0 \pmod{p}\implies 2^{22271}\equiv 1 \pmod{p}$$
Since $22271$ is prime, the order of $2$ modulo $p$ is $22271$ and the order must divide $\phi(p)$ 
$$22271 \mid \phi(p)\\22271\mid (p-1)\\p = 22271k+1$$ 
$k$ has to be even for $p$ to be odd 
$$p = 44542l + 1$$
Now you may plugin different values for $l$ and check divisibility for those that produce primes. 
