I tried with comparing highest exponents

  • Thread starter helgamauer
  • Start date
  • Tags
    Exponents
In summary, the conversation discusses how to prove that (2^m - 1) and (2^n - 1) are not the only prime factors of z, where z = (2^(mn) - 1)/[(2^m - 1)(2^n - 1)] and (2^m - 1) and (2^n - 1) are prime numbers. The speaker mentions trying to solve it by writing z as (2^m - 1)^a * (2^n - 1)^b and proving it is not correct, but is unsure of how to do so. They also suggest proving that 2^(mn) - 1 cannot be written as ((2^m -
  • #1
helgamauer
8
0
We have:
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.
 
Physics news on Phys.org
  • #2


* And i meant (2^m -1) and (2^n -1) are two different ODD primes. May anybody help?
 

Similar threads

Replies
8
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
41
Views
3K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
11
Views
2K
Replies
5
Views
1K
Back
Top