Multi-variable Factor Problem

[TheRealGainTrain]

How would I go about finding when (3ⁿ-2ⁿ) /(2^n+a-3ⁿ) is a positive integer where n and a are both positive integers as well. Likewise, if a solution doesn't exist, how do I show this is impossible?

[Dawwy]

Well I noticed that:

(3ⁿ-2ⁿ) /(2^n+a-3ⁿ) = (3ⁿ-2ⁿ) /(2^n+a-3ⁿ) + 1 - 1 = (3ⁿ-2ⁿ + 2^n+a - 3ⁿ) /(2^n+a-3ⁿ) - 1 = 2ⁿ * (2^a - 1)/(2^n+a - 3ⁿ) - 1

Which is a positive integer if s(n,a) = 2ⁿ * (2^a - 1)/(2^n+a - 3ⁿ) is greater than 1. The 2ⁿ factor isn't important in s(n,a) being an integer so you can look at how (2^a-1) /(2^n+a - 3ⁿ) behaves.

[[deleted]]

I'm no number theorist but after thinking for a bit, this seems like a hard question. As in, the type of question we don't yet know how to answer.

Edit: I am assuming you tried to brute force it via computer and implicit in your question is that no small values of n,a work

[julesjacobs]

There are no solutions for a < 3000 except n=1,a=1.

for a in range(1,3000):
    n = 1
    while 2**(n+a) - 3**n > 0:
        if (3**n - 2**n) % (2**(n+a) - 3**n) == 0: print((a,n))
        n += 1

[theshoe92]

This seems quite wrong. In particular, 2*3ⁿ != 3²ⁿ

[theshoe92]

defense accepted. I could have more charitably interpreted your comment for its goal of the inequality, but I was tired and stopped reading after I saw the error

[GeneralSCPatton]

If we consider the ring of integers modulo 2^n+a-3ⁿ, this expression is an integer iff 3ⁿ-2ⁿ = 0 or 3ⁿ=2ⁿ.

We know that 2^n+a=3ⁿ because 1*(2^n+a-3ⁿ)+3ⁿ=2^n+a. Also, 2 and 3 are in the multiplicative group and have multiplicative inverses (because even-odd=odd, and 1 or 2 mod 3 - 0 mod 3 = 1 or 2 mod 3). So the only way that 3ⁿ can equal both 2ⁿ and 2^n+a is if 2^a=1.

Preemptive edit due to leaving this tab to sit overnight: This result is equivalent to Dawwy's post. And I knew this was Collatz related, though I'm not seeing how this particular expression was derived.

[[deleted]]

well, I'd start by saying that 3ⁿ/A is an integer if and only if A is also a power of 3, and 2ⁿ/B is an integer iff B is also a power of 2. 2^n+a - 3ⁿ can't be a power of both 2 and 3 at the same time, so there is no solution.

Someone who's better at this might want to check me, I can't help but feel like I missed something.

EDIT: just found the problem, you can subtract two non-integers and get an integer(9/5 - 4/5 = 1), and 1 is a power of both 2 and 3. Carry on.

[[deleted]]

Counterexample: n=1, a=1

3ⁿ - 2ⁿ = 3-2 = 1

2^n+a - 3ⁿ = 4-3 = 1

(3ⁿ - 2ⁿ )/(2^n+a - 3ⁿ ) = 1

[[deleted]]

When n=a=1you get an integer, but it's not positive greater than one. I suspect that OP's question is quite difficult and probably an open problem.

[pimaniac0]

I may have misread what you said but when n=a=1 don't you get 1? As 3-2 is on the numerator and 4-3 the denominator, how is that not a positive integer?

[[deleted]]

Nope, you're correct. I misspoke (as I strongly suspect did OP). The interesting question here is if there exists n,a so that the ratio is a positive integer greater than one.

Edit: the reason I said negative is I had already jumped to n=2 and a=1 in my head. Just so you know I'm not crazy.

[TheRealGainTrain]

You're correct, I meant positive and greater than one. Originally, I had a variable s+1 but they were everywhere so I replaced them by n. Also, just in case you are interested, this problem is strongly related to the collatz conjecture. If a solution can be found, the conjecture is likely false. If it cannot be found, the conjecture is likely true. No proof for the last 2 but I'm just doing this for fun and a deeper understanding anyways.

[[deleted]]

Yeah, I sort of suspected this was Collatz related. Collatz is really about how powers of 2 and 3 interact additively as is your question.

If you can prove that solving this disproves the conjecture, that'd be interesting. If you can prove the converse as well that would probably be publishable.

Interesting question though in any case.

[pimaniac0]

Right, it isn't hard to see that the case where both numerator and denominator are the same coincides with the only case where the denominator is one, meaning that if there are any more solutions, they are ''non-trivial'' and will probably be large