On prime numbers let $q$ be a prime 
let $p = 2^q -1 $
is p must be prime always for any prime q ? 
is this is true always ? 
or it is false for some prime q ? 
if it is false , give an example to show that there is a prime q such that 
$2^q -1$ is not a prime 
thanx 
 A: It is a theorem that if $p=2^q-1$ is prime then $q$ is necessarily prime. However, the converse is not true. One counterexample to the converse is: $2^{11}-1 = 89\times 23$. 
A: Such primes $q$ seem to be in fact very rare. The first example is that $2^{11}-1$ is not prime. I leave it to you to find its non-trivial factors.
You may want to read about Mersenne primes.
A: Such a number is called a Mersenne prime. In fact, we only know 48 such Mersenne primes, so most numbers in this form are not prime.
A: Here's some example computations to suggest a way you might try to prove that $q$ must be prime:
$$x^2 - 1 = (x - 1)(x + 1)$$
So:
$$x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1)$$
And so:
$$2^6 - 1 = 63 = (2^3 - 1)(2^3 + 1) = 7 \cdot 9$$
Or:
$$x^3 - 1 = (x - 1)(x^2 + x + 1)$$
So:
$$x^{15} - 1 = (x^5)^3 - 1 = (x^5 - 1)(x^{10} + x^5 + 1)$$
And so:
$$2^{15} - 1 = 32767 = (2^5 - 1)(x^{10} + 2^5 + 1) = 31 \cdot 1057$$
A: $q=2^p-1$
$q$ is a $prime$. There its factors are $1$ and $q$.
$1.q=2^p-1$
If $p$ wasn't $prime$ and was of the form $a_1a_2a_3...a_n$,
$2^{a_1a_2...a_n}-1= (2^{a_2a_3...a_n})^{a_1}-1^{a_1}= A.B$
And $A,B \ge 2$
Which contradicts the fact that $q$ is prime. Therefore, $p$ has to be prime.
But when we take $p$ .
$2^p-1=(2-1)(2^{p-1}+2^{p-2}....1)$
Note that $(2^{p-1}+2^{p-2}....1)$ can/cannot be a $prime$.
