EDIT: NOT HELPING TOO MUCH.
What about p=31?
2^31 - 1 is prime.
Prove, that a composite number 2^p - 1 (where p is a prime) has at least 2 different prime factors.
How I tried to solve this problem:
I tried to write 2^p -1 as x^n (x, n are integers, n>1, just to prove that it can't be this way). I know that the only prime factors of 2^p - 1 can be numbers like 2kp+1 and I also tried this way. What is more, I tried with two cases- n odd and n even, separately. But I didn't manage to prove anything. Please, help. Thank you in advance.
We suppose that . It's clear that is odd.
If is even, then would be of the form . But has the form ; therefore is odd.*
The following identity holds, since is odd: . Note that the term is odd and thus relatively prime to . So must be equal to , namely .
Back to the original problem... Assume to the contrary that has only one prime factor, that is , where is prime. But the result implies that , contradiction.
* To see why: For any odd we have . Then if is even. But , so , contradiction. Sorry if this part is tedious...
@melese, maybe you have and idea how to solve this one:
z = (2^(mn) - 1)/[(2^m - 1)(2^n - 1)], where (2^m - 1) and (2^n - 1) are prime numbers.
Prove that (2^m - 1) and (2^n - 1) are not the only prime factors of z.
I tried to solve it writing z = (2^m - 1)^a * (2^n - 1)^b and proving that it is not correct. But I don't know how. I also noticed that it would be sufficient to prove that 2^(mn) - 1 can't be written as ((2^m - 1)^x * (2^n - 1)^y but I also have no idea how to prove it. I tried with comparing highest exponents, but nothing..
Thank you very much in advance!