When does 2^m - 2^n divide 3^m - 3^n?

I think the only such pairs [m,n] are [2, 1], [3, 1], [5, 1], [4, 2], [6, 2], [8, 2], [14, 2], [5, 3], [7, 3], [9, 3], [15, 3], [8, 4], [16, 4]. At any rate these all work and my computer tells me there are no more with 1001>m>n. If I think of a proof that there are no more I will return -- but don't hold your breath.

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EDIT:

I just saw Doctor D little's comment below. Unfortunately it is not correct, because the denominator can split its primes between the two numerators, for example when m=5 and n=1, 3^m - 3^n = 240, 2^m - 2^n = 30, so 30 divides 240 but 3^n/(2^(m-n)-1) = 3/15 = 1/5 not an integer, i.e. (2^(m-n)-1) does not divide 3^n.

There may still be some usefulness in Doctor D little's comment, but anyone trying to use it needs to be aware that as it stands it is not correct.

(3^m - 3^n) / (2^m - 2^n)

= (3^n / 2^n) * [3^(m-n) - 1] / [2^(m-n) - 1]

In order for this to be an integer, and noting the fact that 3^n and 2^n have no common prime factors, then

2^(m-n) - 1 must divide 3^n and

2^n must divide 3^(m-n) - 1

Not a complete answer, but maybe someone else can do something with this.

*EDIT*

Yes I realized that error. I'm just not sure as yet how to change it into a form that can actually solve the problem without trial and error.

properly you will never study something via having others do ur hw yet i can coach you....the 1st one is 28n^6...it is what you do once you have an exponent exterior a parenthesis you upload it to notwithstanding is interior (x^2)^3= x^5...make experience? and in case you had this (2x^2)^3 (sqaure 2 and then upload 2 and 3!) so it equals 8x^5 with branch its an identical ingredient different than you subtract!

(3^m - 3^n) / (2^m - 2^n)

= (3/2)^n * [3^(m-n) - 1] / [2^(m-n) - 1]

we know

a^p = a mod p , for prime p.

a^(p-1) = 1 mod p

a^(p-1) - 1 = 0 mod p

so that settles any prime factor within [3^(m-n) - 1] and [2^(m-n) - 1], because they share them.

that means, we need m - n = p - 1

m - n + 1 = prime p.

i'll came back when i can add some more.

because [3^(m-n) - 1] still need factors of 2 and [2^(m-n) - 1] factors of 3. but knashha said he'll be here, so all is well.

i checked and p is only up to 13. why is that so?

Interesting Question:

I tried to generalize solution set for (a^m - a^n) / (b^m - b^n),

(where a>b & m>n and a & b are prime / co-prime numbers)

Here goes few of the observations:

1) If a & b prime numbers then m-n+1 can only belong to set {2, 3, 4, 5, 6, 7, 9, 11, 13, 19, 27, 37, 53.....}

I have not come across any case where m-n+1 does not belong to above solution set but sequence may continue, who knows! This is also applicable if a & b are co-primes

2) Solution exists if only if "b" is one of the factor of (a+1) or

(a-1) I do not know why this happens but looks interesting!

3) Number of possible solutions depends upon how many times "b" appears as a factor of (a+1) or (a-1).

Thus elements in a solution set will be will be highest if "a+1" or

"a-1" is equal to b^x for some x

(Try taking a & b as 9 & 2 (Co-prime case), 17& 2 (Prime case) etc )

@ miryana kina:

Thus m-n+1 = p is not always true. :) but thanks for giving idea!

Since the original question is already answered I have posted only other relevant part as this may be useful to generalize the answer.

= (2^m - 2^n)/(3^m - 3^n)

= 2/3

Answer: 2/3

Checking back to 2^m - 2n:

= 2/3(3^m - 3^n)

= 2^m - 2^n