Can Mersenne Numbers Be Multiples of Each Other?

[helgamauer]

Hello. I've got three questions.
1.Can a Mersenne number be a multiple of another Mersenne number? Prove. (I mean is that possible that 2^a - 1 | 2^b - 1 where a, b are prime numbers)
2.Can a Mersenne number be equal to X^n ? (X and n are integer). Prove. (This Mersenne number is 2^d - 1 where d is a prime number)
3.Is it possible that (2^t - 1)^a * (2^u - 1)^b is equal to 2^tu - 1? (t, u, 2^t -1, 2^u -1 are prime numbers).

 

[Petek]

Even if it's not homework, you're expected to show how you've tried to solve the problems.

 

[robert Ihnot]

#1) Anyway, one example is (2^6-1)/(2^3-1) =63/7=9. That's a start!

 

[Petek]

#1) Anyway, one example is (2^6-1)/(2^3-1) =63/7=9. That's a start!

Robert,

In the initial post, the exponents are supposed to be prime.

 

[robert Ihnot]

O.K. #1 is not possible because in the division of 2^p-1 by 2^j-1, for p>j, We begin by looking at the Euclidean algorithm p=kj+r, for r<j Now we take the case of (1+2++2^(n-1)/(1+2+2^2++2^(j-1)=1+(2^j)(1+2+2^2+++2^(p-1-j+1)/(1+2+++2^j)...etc.

This finally results in the form 1+2^j +2^(2j)+++2^((k-2)j +(2^(J-1))(R). r being 1 or less . Should R be 1, then we have the form (2^p-1)/(2^j-1) =1 +2^j +2^2j+++2^(p-(k-1)j)=(2^p-1)/(2^j-1). Otherwise we have a fractional remainder. So that they do not divide evenly

This proves that p=kj if whole number division takes place. So they can not both be prime exponents.

 

[helgamauer]

Oh.. Would it be possible for you to write it in latex or annex a photo of those equations? I can't fully understand it this way.. Please, I would really appreciate.

 

[robert Ihnot]

It probably can be written up in a simpler way.

 

[helgamauer]

Then could you please write it more simply? Please, it is very important for me.

 

[robert Ihnot]

\frac{2^p-1}{2^j-1} we look at this fraction, which is equal to: 1+2^j+2^(2j)++2^(n-1)J +(2^(nj))(2^r-1)/(2^j-1).
In this case we are using p=nj+r, where r<j by the Eucledian algorithm.

In the event that r=0, the last term drops out and multiplying bothsides by 2^j-1 makes both sides equal to 2^p-1. However in this case p=nj, so both are not primes unless n=1.

In the even that r is not 0, we have a final term \frac{(2^{nj})(2^r-1)}{(2^j-1)}

Now the term 2^(nj) is all powers of 2 and has nothing in common with 2^j-1. So that we are left with \frac{2^r-1}{2^j-1}, where r<j , so the final term is not an integer.

So the only cases that divide evenly are the ones where p is divisible by j.

 

[helgamauer]

Well... You are great! Thank you so much.
I don't want to be importunate, but do you have any ideas how can I solve two remaining problems?
2.Prove that a composite Mersenne number ((2^d - 1) is composite, d is prime) has at least 2 prime divisors.
3.Is it possible that (2^t - 1)^a * (2^u - 1)^b is equal to 2^tu - 1? (t, u, 2^t -1, 2^u -1 are prime numbers). / In other words - can (2^t - 1) and (2^u - 1) be the only divisors of 2^tu - 1 ?

 

[helgamauer]

And one more.. Is from #1 easily derivable that If (m,n) = 1 then (2^m - 1 , 2^n - 1) = 1 ?