Can two Mersenne numbers share a factor?

[chris2be8]

Thanks everyone. I thought the the proof for base 2 would extend to other bases, I just wanted confirmation I hadn't made a mistake. Now I'm convinced.

Thanks again.

Chris K

[R.D. Silverman]

Originally Posted by LaurV

in fact missed the part with "except the case when their number of digits share a common factor". As long as I know, 55 and 5 share a common factor (5), ?

"number of digits share a common factor" is gibberish. 5 and 55 do indeed
share a common factor. But what has this to do with "their number of digits
share a common factor"?

Take my earlier hint for (b^q-1)/(b-1) and (b^p-1)/(b-1) with p,q, prime.
Look at the proof for b = 2. Generalize.

[science_man_88]

Originally Posted by R.D. Silverman

"number of digits share a common factor" is gibberish. 5 and 55 do indeed
share a common factor. But what has this to do with "their number of digits
share a common factor"?

Take my earlier hint for (b^q-1)/(b-1) and (b^p-1)/(b-1) with p,q, prime.
Look at the proof for b = 2. Generalize.

it has everything to do with it! 4^5-1 has 5 digits 4^55-1 has 55 digits the number of digits is the exponent in this case so the "number of digits share a common factor" can be said to be the same as saying "powers share a common factor"

[science_man_88]

Originally Posted by science_man_88

it has everything to do with it! 4^5-1 has 5 digits 4^55-1 has 55 digits the number of digits is the exponent in this case so the "number of digits share a common factor" can be said to be the same as saying "powers share a common factor"

this is in base 4 notation.

[R.D. Silverman]

Originally Posted by science_man_88

it has everything to do with it! 4^5-1 has 5 digits

More horsesh*t.

What have you been smoking???? 4^5 - 1 = 2^10 - 1 = 1023. This has
4 digits!!

4^55-1 has 55 digits the number of digits is the exponent in this case so the "number of digits share a common factor" can be said to be the same as saying "powers share a common factor"

More totally stupid crap. 4^55-1 definitely does NOT have 55 digits. It has
34.

The ONLY b for which b^55-1 has 55 digits is b = 10. Why do you
think that they are called DIGITS???????

[science_man_88]

Originally Posted by R.D. Silverman

More horsesh*t.

What have you been smoking???? 4^5 - 1 = 2^10 - 1 = 1023. This has
4 digits!!



More totally stupid crap. 4^55-1 definitely does NOT have 55 digits. It has
34.

The ONLY b for which b^55-1 has 55 digits is b = 10. Why do you
think that they are called DIGITS???????

digits base 4:

Quaternary is the base-4 numeral system. It uses the digits 0, 1, 2 and 3 to represent any real number.

hence although it might be technically wrong it has been used in this sense before

[science_man_88]

Originally Posted by science_man_88

digits base 4:

hence although it might be technically wrong it has been used in this sense before

what I mean is 4^5-1 = (33333)₄ which has 5 digits in base 4. and (4^55-1) = (3333333333333333333333333333333333333333333333333333333)₄ which has 55 digits in base 4. so 55/5 = power2/power1 = numberofdigits2/numberofdigits1 .

[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 time I heard it in highschool in 82. How should we call it? Figure?

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. They are all of the form zzzz....zzzz with z=b-1. For example 10^3-1 is 999 in base 10. When you divide by b-1 you get the repunit. For the rest, read again post 22 onward.

Originally Posted by R.D. Silverman

"number of digits share a common factor" is gibberish. 5 and 55 do indeed
share a common factor. But what has this to do with "their number of digits
share a common factor"?

Take my earlier hint for (b^q-1)/(b-1) and (b^p-1)/(b-1) with p,q, prime.
Look at the proof for b = 2. Generalize.

We just did. Thank you. Not all of us are born mathematicians, we feel better working with polinomes and bits (digits) and not with modulus and functions. If you want to continue the discussion, then read last 20 posts carefully and point what is wrong. But read it without skipping paragraphs. Including the examples.

For me, the case is closed. Two repunits, in the SAME base b can never share a common factor, unless their SIZES share a common factor. By size we understand the number of units in the repunit, that is the number of digits, or figures, or whatever you want to call it, necessary to represent the number in base b, that is the same as x and y in the expression of numbers as (b^x-1)/(b-1) and (b^x-1)/(b-1), that "size" as should be returned by a (virtual) operator sizeof() in C programming language (now well, this is more gibberish :D)

[axn]

Originally Posted by LaurV

For me, the case is closed. Two repunits, in the SAME base b can never share a common factor, unless their SIZES share a common factor. By size we understand the number of units in the repunit, that is the number of digits, or figures, or whatever you want to call it, necessary to represent the number in base b, that is the same as x and y in the expression of numbers as (b^x-1)/(b-1) and (b^x-1)/(b-1), that "size" as should be returned by a (virtual) operator sizeof() in C programming language (now well, this is more gibberish :D)

Please note the restriction that the base b itself should not be a power.

[wblipp]

Originally Posted by R.D. Silverman

What have you been smoking???? 4^5 - 1 = 2^10 - 1 = 1023. This has
4 digits!!

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.

[LaurV]

Originally Posted by axn

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

Originally Posted by axn

Please note the restriction that the base b itself should not be a power.

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...