this thread, and think what happens mod 4.
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..
It will be fantastic if you help me with proving this fact.
My previous comment was based on a misreading of the question. Ignore it.
However, the question as stated is plain wrong (unless I'm still misreading it):Try taking m=2 and n=3. Then . The only prime factor of z is 3, which is .We have: , where and are prime numbers. Prove that and are not the only prime factors of z.
Here's an approach that seems to work. Look at the polynomial , and let be a nonconstant polynomial in . Let be the quotient, and the remainder, when is divided by . I claim that if is nonzero then it consists entirely of odd powers of x. Reason: suppose not, let 2s be the highest even degree of a nonzero term in , and let 2t be the degree of . Then there is a nonzero term of degree 2(s+t) in their product (and there is no term of that degree in ). But there is no term of that degree in , so we have a contradiction.
Therefore consists entirely of terms of odd degree, and it follows that the constant term in r(x) must be –1.
Now let m, n be odd numbers, and suppose that is a repeated factor of . Notice that . Since both those powers of 2 are even, that is a polynomial in . So let , divide by and then put x=2. The above result says that the remainder will be r(2), and r(2) must be an odd number (since the constant term in r(x) will be the only odd term). In particular r(2) can never be 0.
Conclusion: , and similarly , can never be repeated factors of . And since their product can never be equal to , there must always be a third factor.
Well, you know I'm just trying to solve following problem found in the other thread: http://www.mathhelpforum.com/math-he...rs-155363.html
I try to prove that 2^mn - 1 (where m and n are two different odd primes) has at least 3 different prime factors. I already know about two of them that are quite obvious - 2^m -1 and 2^n -1. I can prove this statement when at least one of numbers 2^m - 1 and 2^n -1 is composite. I've got a problem when both are primes, I don't know how to solve it.
Or maybe I should try for example with Euler's theorem or Fermat's little theorem? I don't know... Please, help me with this problem.