Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \lvert \left(\frac{3}{2}\right)^n - 2^m \rvert < \frac{1}{4}\ $?

Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \lvert \left(\frac{3}{2}\right)^n - 2^m \rvert < \frac{1}{4}\ $ ?

I have tried for the first few integers $\ n,m\ $ up until $\ m\approx30\ $ with no $\ n,m\ $ satisfying the inequality. However, I can't think of techniques for trying to prove it False. So I'm stuck.

Edit: To be honest, I'm not even sure, for example, how to try to find $\ p,q\in\mathbb{N}_{\geq 2}\ $ such that $\ \lvert 5^p - 7^q \rvert < 10,\ $ which might be an easier type of problem (or harder? I'm not sure...).

Edit:

$$\left(\frac{3}{2}\right)^n - 2^m = \left(\left(\frac{3}{2}\right)^{n/m}\right)^m - 2^m = \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\left( \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-1} + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-2} \cdot 2 + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-3} \cdot 2^2 + \ldots + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{2} \cdot 2^{m-3} + \left(\left(\frac{3}{2}\right)^{n/m} \right) \cdot 2^{m-2} + \left(\left(\frac{3}{2}\right)^{n/m} \right) \cdot 2^{m-1} \right). $$

Since $\left(\frac{3}{2}\right)^{n/m}\ $ is close to $\ 2,\ $ we therefore have:

$$\left(\frac{3}{2}\right)^n - 2^m \approx \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\left( 2^{m-1} +2^{m-2} \cdot 2 + 2^{m-3} \cdot 2^2 + \ldots +2^2 \cdot 2^{m-3} +2 \cdot 2^{m-2} + 2 \cdot 2^{m-1} \right) = \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\cdot m \cdot 2^{m-1}.$$

I'm not sure if this helps, but maybe it relates to mjqxxxx's answer. Maybe this is what he/she means by "where "very close" means exponentially close as a function of that rational's denominator".

Edit: This is an open problem in number theory, so perhaps this means the question here is also an open problem?

I've very recently found a very nice formula giving a lower bound for your difference-term. Copying (& adapted) from my answer in another thread:

(...) Looking at old entries in my literature-database, I found an interesting limiting formula for a lower bound of $2^{n+m}-3^n$. Some short tinkering with it seem to show, that you can prove your conjecture for all $n+m > 27$ with it.
The formula is (W.J.Ellison cited in Stroeker/Tijdeman,'71): $$ \mid 2^x - 3^y \mid \gt \exp(x (\log 2- \frac1{10})) \qquad \text{for all } x,y \in \mathbb N \quad \text{and } x\gt27 \quad \;^{[1]}\tag 1$$ This can be applied to your equation. By (1) we can write $$ \mid 3^n - 2^{n+m} \mid \gt \mu ^{n+m} \qquad \text{where } \mu =1.80967483607... \tag 2$$ (...)

Now to compare this with your term $\frac14$ we look at logarithms.
For the following steps we assume first, that $m+n$ is such that $2^{m+n} \gt 3^n$ (case 1). (If it is $\lt$ then let us call this case 2 )
We'll write in the following $ \gamma=\log_2(3) \approx 1.585$

• case 1: By logarithms we have $ (m+n ) \log 2 \gt n \log 3 $ and thus we must have $$m \gt n (\gamma-1) \approx n \cdot 0.585 \tag{case 1}$$ Now the logarithm of the rhs in (3) is $ (m+n)(\log2-0.1)-n\log2$ and this can be reduced to $$ (m+n)(\log2-0.1)-n\log2 = m (\log 2-0.1) - 0.1n \approx 0.593m - 0.1 n \tag 4$$ which - with expanded $m$ - is: $$ 0.593 (0.585 n) - 0.1n \approx (0.347 - 0.1)n = 0.247 n $$ So the rhs in eq (3) is always greater than $0.247 n$ and of course this is for all $n$ larger than the $\log$ of your testvalue: $\log \frac14 \lt 0$.
  Of course, since Ellison gave his low bound only for $(m+n) \gt 27$ and thus $n \gt 17$, all the comparision for the remaining cases $n=2..16$ must (and can) be done manually and give the same result: that there is no solution for your inequality.

• case 2: We have that $m$ must be decreased by at least $1$ $$m \lt n (\gamma-1) -1 \approx n \cdot 0.585 -1\tag{case 2}$$ We don't repeat the complete analysis here, just note, that the reduction of $m$ by $1$ gives $$ 0.593 (0.585 n-1) - 0.1n \approx (0.347 - 0.1)n -0.592 = 0.247 n - 0.593 $$ We get the same result, that for all $n \gt 1$ this is larger than $\log \frac14$ and by checking the cases $n=2..16$ we find as well no solution for your inequality.

Result: there are no cases $n \gt 1$ (resp $(n+m)\gt 3$) where your inequality holds, and your difference term in your first equation is for all $n \gt 1$ larger than the rhs.

$\;^{[1]}$The citation of formula (1) is from

R.J.STROEKER & R.TIJDEMAN 
Diophantine equations (with appendix by P.L.Cijsouw, A.Korlaar & R.Tijdeman) 
in: MATHEMATICAL CENTRE TRACTS 154, COMPUTATIONAL METHODS IN NUMBER THEORY; PART I;
MATHEMATISCH CENTRUM, AMSTERDAM  1982

and they attribute this result to W.J.Ellison in 1970/1971

[25] ELLISON,W.J., Recipes for solving diophantine problems by Baker's method,
Sèm.Th.Nombr.,1970-1971,Exp.No.11, Lab.Thèorie Nombres,
C.N.R.S.,Talence,1971.

If this were going to happen, then $\frac{\log 2}{\log {3/2}}$ would need to be very close to a rational number (where "very close" means exponentially close as a function of that rational's denominator). In turn, that would mean that its continued fraction would need to have a large term early on. Looking at that continued fraction, it doesn't: it starts with $[1;1,2,2,3,1,5,2,\ldots]$. This is powerful evidence that there is no such $(m,n)$ pair, but falls short of a proof... I suspect all you can actually prove is that there aren't infinitely many such pairs.

As an example of what a more positive result would look like, suppose you wanted $|(5/2)^m - 2^n|$ to be small instead. The continued fraction for $\log (5/2) / \log 2$ is $[1;3,9,\ldots]$; the truncation before the large term is $[1;3]=4/3$; and indeed $|(5/2)^3 - 2^4|=3/8$ is pretty small.

You can also use this (relies on Baker/Rhin): $2^n<2^l-3^n<3^n-2^n$, where the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{2\}$ and $l={\lceil n \log_23\rceil}$ is the smallest exponent of $2$ making $2^l-3^n$ positive.

As stated by blamethelag, we can write your inequality like this:

$$ \lvert 3^n - 2^{m+n} \rvert < 2^{n-2}$$

What is next is similar to what Gottfried exposed in his answer (use of transcendence theory). There are 2 cases:

• case 1: $2^{m+n}>3^n$ and since $l$ is the smallest possible exponent "$m+n$" for this case we have $$2^{m+n}-3^n\geq 2^l-3^n>0$$ and using the inequality from first line $$2^{m+n}-3^n> 2^n>2^{n-2}$$ Note: that taking $2^{n-2}$ as reference removes the exception list mentioned in the introduction.

• case 2: $3^n>2^{m+n}$ and since $l-1$ is the largest possible exponent "$m+n$" for this case we have $$3^n-2^{m+n}\geq 3^n-2^{l-1}>0$$ and using the inequality from first line $$2^l-3^n<3^n-2^n$$ $$2\cdot 2^{l-1}-2\cdot 3^n<-2^n$$ $$3^n-2^{l-1}>2^{n-1}$$ you end up with $$3^n-2^{m+n}>2^{n-1}>2^{n-2}$$ except for $n=2$ from the exception list where we can have equality

which leads to $$ \lvert 3^n - 2^{m+n} \rvert \ge 2^{n-2}$$ or $\ \lvert \left(\frac{3}{2}\right)^n - 2^m \rvert \ge \frac{1}{4}\ $