Can two Mersenne numbers share a factor?

[only_human]

Originally Posted by LaurV

I saw that in your post #19 too, but I intentionally skipped, because it did not seem to be important at that time.

Now, if you insist on it, THIS become interesting! Because in my "proof" I never used the fact that the base is or is not a power. So, if ANY restriction applies to the case when the base itself is a (prime or not) power, then something must be wrong in that proof of mine... Let me think about it...

I've been thinking about it too but am lost at the moment.

In a paper "Primes of the Form (bⁿ + 1)/(b - 1)" by Harvey Dubner, I read:

For certain special forms of b, Q has algebraic factors for all n. If b = c^t is a perfect power where t is greater than 2 and not a power of 2 then Q has algebraic factors and is almost always composite. There are rare cases when Q may be prime for small n but again Q(b; n) can only be prime when n is prime.

Q(b;n) is notation for (bⁿ + 1)/(b - 1).

I can see how this is significant when looking for prime or composite repunits themselves but don't see how the fact that the repunit has an algebraic factorization matters when looking for common factors between two different repunits to the same base but with relatively prime exponents.

[R.D. Silverman]

Originally Posted by LaurV

Digit is digit, in any base. It can be a hexadecimal digit, a quaternary digit, octal, heptal, etc. Even the word "bit" as we have it in English is an acronym for BInarry digiT, defined 50 years ago, or more. First

No, it is not. 'digit', when unmodified means base 10.
In binary we either use 'binary digit', or 'bit'. In ternary we use 'trinary
digit' or 'trit'. In hex, we use 'hex digit'. Note the modifiers

Well, we go away from the subject. Let's take the constructive part. I don't want to create a flame war here.

Any b^x-1 has x DIGITS in base b. So b^y-1 has y DIGITS.

No, it does NOT have y digits. It has y 'base b digits'. Learn to use the
correct definitions.

We just did. Thank you. Not all of us are born mathematicians, we feel better
working with polinomes (sic) and bits (digits) and not with modulus and functions.

Unfortunately for you this sub-forum is for mathematics. If you don't like
this fact, then you are welcome to leave.

[R.D. Silverman]

Originally Posted by wblipp

Bob,

You probably skipped over the details of LaurV's post. Since these questions are being asked not by mathematicians but by people interested in "Generalized Rep Units," he chose to discuss the problem in that framework. As you probably know, people interested in this problem typically started being interested in primes and factors of numbers that are strings of the digit "1", which they called "Rep Units." Later they found this generalized to (a^n-1)/(a-1), which is a string of 1's if written in base a. So in the context of his discourse, when he talks about "the number of digits" he is talking about what a more mathematically inclined audience would call the exponent.

I expect to see what you call "a more mathematically inclined audience "
in this sub-forum because it is SPECIFICALLY FOR THE DISCUSSION OF
MATHEMATICS.

And I have given them the strongest possible hint in answer to their
questions. --> Follow the proof for base 2.

[R.D. Silverman]

Originally Posted by only_human

I've been thinking about it too but am lost at the moment.

In a paper "Primes of the Form (bⁿ + 1)/(b - 1)" by Harvey Dubner, I read:Q(b;n) is notation for (bⁿ + 1)/(b - 1).

I doubt it. b^n + 1 is not divisible by b-1.

I can see how this is significant when looking for prime or composite repunits themselves but don't see how the fact that the repunit has an algebraic factorization matters when looking for common factors between two different repunits to the same base but with relatively prime exponents.

Go READ the Cunningham book. Learn the definitions of 'primitive factor',
'algebraic factor', and 'intrinsic prime factor'.

One needs to PROVE that algebraic factors of b^n-1 are irrelevant to
a discussion of whether b^n-1 might have a common prime factor with
b^m-1 when (m,n) = 1. If k divides n then b^k-1 | b^n-1.
(I assume that you can prove this much)

One then needs to show that (b^k-1, b^m-1) = 1 when (n,m) = 1.

[wblipp]

Originally Posted by R.D. Silverman

Go READ the Cunningham book. Learn the definitions of 'primitive factor', 'algebraic factor', and 'intrinsic prime factor'

A change at AMS seems to have broken links to the full text of the book - links that used to go there now go to an AMS front page. If anyone knows the link to the full text, it would be helpful to post it here - it is indeed excellent reading on this topic, but presently hard to find.

[R.D. Silverman]

Originally Posted by wblipp

A change at AMS seems to have broken links to the full text of the book - links that used to go there now go to an AMS front page. If anyone knows the link to the full text, it would be helpful to post it here - it is indeed excellent reading on this topic, but presently hard to find.

It is on Sam Wagstaff's web site.

[wblipp]

Originally Posted by R.D. Silverman

It is on Sam Wagstaff's web site.

Are you sure? It used to be. But the closest I can find now is this

http://homes.cerias.purdue.edu/~ssw/...ird/index.html

and the link to the full text, which is what you want them to read, leads to the AMS home page. So once again I repeat, it would be helpful to post a link here.

[R.D. Silverman]

Originally Posted by wblipp

Are you sure? It used to be. But the closest I can find now is this

http://homes.cerias.purdue.edu/~ssw/...ird/index.html

and the link to the full text, which is what you want them to read, leads to the AMS home page. So once again I repeat, it would be helpful to post a link here.

Indeed! I am surprised. It did indeed used to be directly available on Sam's
website.

It seems that it no longer is. I don't know when the change happened.

[only_human]

Originally Posted by R.D. Silverman

I doubt it. b^n + 1 is not divisible by b-1.



Go READ the Cunningham book. Learn the definitions of 'primitive factor',
'algebraic factor', and 'intrinsic prime factor'.

One needs to PROVE that algebraic factors of b^n-1 are irrelevant to
a discussion of whether b^n-1 might have a common prime factor with
b^m-1 when (m,n) = 1. If k divides n then b^k-1 | b^n-1.
(I assume that you can prove this much)

One then needs to show that (b^k-1, b^m-1) = 1 when (n,m) = 1.

I doubt it. b^n + 1 is not divisible by b-1.

Thank you, that was careless of me; I was reading about Aurifeuillian Factorization late at night and mentally slipped a cog or three.

I will look into these things that you mentioned.

I was looking for the Cunningham Book last night but ran up against the same dead end on the AMS website. I think I might have a copy somewhere on an offline drive from back when the AMS link was good.

Google found me a link to the book while searching for: conm22 pdf cunningham
[PDF] Factorizations of b^n+-1, b=2,3,5,6,7,10,11,12 up to high powers

[only_human]

Originally Posted by R.D. Silverman

Hint: Look at the proof for GCD(2^a-1, 2^b-1)

Originally Posted by axn

Try to generalize LaurV's approach to arbitrary (non-power) base.

Ok, I think that I might be looking using the Euclidean Algorithm in base b to prove that gcd(bⁿ − 1,b^m − 1) = b^gcd(n,m) − 1

I'll be thinking about this as I do errands

[R.D. Silverman]

Originally Posted by only_human

Ok, I think that I might be looking using the Euclidean Algorithm in base b to prove that gcd(bⁿ − 1,b^m − 1) = b^gcd(n,m) − 1

I'll be thinking about this as I do errands

Hint: GCD(A,B) = GCD(A-B, B). use descent.