interesting math problem

[gumpzilla]

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The prime decompositions of different integers m and n involve the same primes. The integers m+1 and n+1 also have this property. Is the number of such pairs (m,n) finite or infinite?

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Here's a related question. What are all the pairs of numbers (k,k+1) such that one is a power of 2 and the other is a power of 3.

Examples are (1,2), (2,3), (3,4), (8,9). Are there any others. Anyone know? Does anyone know of techniques to answer questions like this? Is it some standard number theory, or is it really difficult?

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There's a trick I remember for proving that in base 10, all possible natural numbers will eventually occur as the leading digits in some power of 2. If I remember correctly, the trick involved looking at the circle map that you get by considering the fractional part of the log base 10 of successive powers of 2. Since this is like adding log_10 of 2 over and over again, and that number is irrational, you can make an argument about the distribution of points being dense in the circle and then you're done.

I thought for a minute a similar trick might work for this. But now I realize it probably won't, because for each iteration around the circle, you need a closer and closer approach of the two points in order for them to be different by 1; it's not a fixed epsilon that stays constant no matter how many times you traverse the circle. So so much for that.

[Siegmund]

Almost surely infinite, but I don't see any easy proof of it.

Notice that all Proth primes work (k*2^n+1). So far, people have always been able to find more than one k for any n, and more than one n for any k, and just about everybody believes there are infinitely many of them - but I believe that, as with the Mersennes, it's an open question to formally prove there are infinitely many of them.

[Fly]

I actually do have a simple solution which I will post later. I didn't have any luck trying to solve it so I posted it to find how difficult it is.

[blah_blah]

[ QUOTE ]
[ QUOTE ]
The prime decompositions of different integers m and n involve the same primes. The integers m+1 and n+1 also have this property. Is the number of such pairs (m,n) finite or infinite?

[/ QUOTE ]

Here's a related question. What are all the pairs of numbers (k,k+1) such that one is a power of 2 and the other is a power of 3.

Examples are (1,2), (2,3), (3,4), (8,9). Are there any others. Anyone know? Does anyone know of techniques to answer questions like this? Is it some standard number theory, or is it really difficult?

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http://mathworld.wolfram.com/CatalansConjecture.html

[thylacine]

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Here's a related question. What are all the pairs of numbers (k,k+1) such that one is a power of 2 and the other is a power of 3.

Examples are (1,2), (2,3), (3,4), (8,9). Are there any others. Anyone know? Does anyone know of techniques to answer questions like this? Is it some standard number theory, or is it really difficult?

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http://mathworld.wolfram.com/CatalansConjecture.html

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Wow. I guess that means it is really difficult. Let me ask if you know any information about the following similar questions.

(1) Let P be a finite set of primes.

(1a) Are there only finitely many solutions to the equation a+b=c where positive integers a,b,c are relatively prime and have all there prime factors in the set P?

(1b) Is there a method for finding all such solutions?

(2) Let n be a positive integer. Let X_n be the set of all products of powers of numbers of the form z and 1-z where z^n=1.

(2b) Is there a method for finding all solutions to a+b=c where a,b,c are all elements of X_n?

(2a) Are there only finitely many solutions to this equation, up to scaling.

Any info appreciated. [img]/images/graemlins/smile.gif[/img] I'll be happy with answers just to (2a,b). [img]/images/graemlins/wink.gif[/img]

[blah_blah]

Honestly I'm not really a number theorist by trade, although I've taken a couple of courses in analytic number theory in the past (in particular, I don't know anything about the questions you posted).

There is almost certainly a 'simple' proof for the cases 2^x - 3^y = \pm 1 of the Catalan Conjecture (which, as noted in that article, has been proven), although 'simple' in this case doesn't preclude the use of semi-sophisticated tools from algebraic number theory.

[bigpooch]

Obviously, not many even looked at my post!

In my previous post, the requirement that q needed to be
prime was unimportant.

Let p>=2 be any integer.

m = (2^p-2); m+1 = (2^p-1); thus, m is even

n+1 = (m+1)^2 = (2^p-1)^2

n = (m+1)^2 - 1 = (m+2)m = (2^p)m

so n has the same prime factors as m (since m is even).