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...).


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

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?

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Adam Rubinson
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    Fun fact: the approximation $(3/2)^{12} \approx 2^7$ is roughly the reason why we have the 12-tone scale in music. (however it's not within $1/4$.) – Jair Taylor Aug 03 '21 at 17:09
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    See https://en.wikipedia.org/wiki/Baker%27s_theorem – Paramanand Singh Aug 03 '21 at 17:26
  • Fix $n$ and define $f(x) = \frac{3^n-2^{x+n}}{2^{n-2}}$, then determine a bound for $x$ such that $f(x)<1$ and another bound such that $ -1 < f (x)$, take the intersection and see if the set is void or has a positive integer etc. – en3trix Aug 03 '21 at 17:27
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    Baker theorem as above implies the inequality cannot hold for $n$ large enough as the separation between $3^n$ and $2^m$ grows at least like $C3^n/n^q$ for some explicit $C,q$ but of course one needs those to see what large enough is; so while we know that the number of pairs that can satisfy the required inequality is finite, one needs to do the work and explicit Baker 's constants in this case to see the upper bound on such potential pairs – Conrad Aug 03 '21 at 22:00
  • Just checked, $m\leq 10^5$ does not work – QC_QAOA Aug 03 '21 at 22:22
  • https://www.desmos.com/calculator/gblx7lsxue This graph might be helpful for checking those lower numbers @Conrad mentioned. We iteratively fix some $a\in\mathbb{N}$, and look at $(3/2)^{a+x}-2^x$ as a function of $x$. We can find its roots analytically, then look at the closest integer to the root. Interestingly, $a=22$ has a root at $x\approx 31.0072$; nonetheless, since the function is crazily decreasing, its value at $x=31$ is about $4$ million! – Zim Aug 03 '21 at 22:25
  • Baker's theorem is whoosh over my head. Is there another more elementary approach? – Adam Rubinson Aug 03 '21 at 23:19
  • Adam - while to apply Baker's theorems directly might be an overkill (it is as well over my head, btw) , there are results derived from it for the special case of distance (or "separation" as Terence Tao names it) of powers of $2$ and $3$. Easily usable bounds, due to work of G. Rhin and W.J.Ellison, are applied here in MSE occasionally. This is then more "elementary" - but of course the machinery behind is too complicated for the hobbyist... – Gottfried Helms Oct 12 '21 at 19:33
  • Adam - just an addendum to your first edit (at considering other bases than 2 and 3) there might something helpful in my older question https://math.stackexchange.com/q/3803905/1714 and which gives links to possibly richer information. – Gottfried Helms Oct 14 '21 at 09:40

5 Answers5


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 $$ \mid 3^n - 2^{n+m} \mid \gt \mu ^{n+m} \qquad \text{where } \mu =1.80967483607... \tag 2$$ (...)

and thus $$ \underset{\text{for } n+m \gt 27}{ \underbrace{\mid 1.5^n - 2^{m} \mid \gt {\mu^{m+n} \over 2^n}}} \quad \{\overset?\lt \frac14 \} \tag 3$$

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

Diophantine equations (with appendix by P.L.Cijsouw, A.Korlaar & R.Tijdeman) 

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,
Gottfried Helms
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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.

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    For a more artificial, but more dramatic, example -- note that the continued fraction for $\log 378661 / \log 5234$ is $[1;1,1,3.6\times 10^{10},\ldots]$. Since $[1;1,1]=3/2$, this tells you that $5234^3 \approx 378661^2$. And indeed they only differ by $17$. – mjqxxxx Aug 03 '21 at 22:48
  • Ok im done w new answer – Al Brown Aug 03 '21 at 22:55
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    The fact that the number of pairs is finite is a straightforward consequence of Baker theorem on linear forms in logarithms – Conrad Aug 03 '21 at 23:03
  • @Conrad Yes, that’s why I said you can prove that much... but I don’t see a way to guarantee any bound on the size of the finitely many examples. – mjqxxxx Aug 03 '21 at 23:29
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    There is an explicit bound in Baker theorem but of course one would have to actually compute it for this particular case – Conrad Aug 04 '21 at 00:23
  • @AlBrown et.al: Bounds, based on Baker's theorems, are already known and used for the case of powers of $2$ and $3$. You might search for `Mignotte`, `G.Rhin`, and recently `W.J.Ellison` here in MSE with tag `Collatz-conjecture` to find where and how those people formulated their practical bounds which were actually used. Of the `G.Rhin` bound I got aware by the work of Simons/DeWeger some years ago and have mentioned it several cases here and of the `J.W.Ellison` bound by an article of Stroeker/Tijdemans just recently. – Gottfried Helms Oct 12 '21 at 19:22

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}\ $

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This is not an answer but it is too long to be posted as a comment. If you multiply the inequality by $2^n$ it becomes $$|3^n - 2^{m+n}| < 2^{n-2}.$$ Now fix $n \in \mathbb{N}$ and study $f : x \mapsto 3^n-2^{x+n}$, it is a decreasing function that has a unique zero at $x_* = n\log_2(3/2)$. So the minimum of $|f|$ on $\mathbb{N}$ is realized either at $\lceil x_* \rceil$ or at $\lfloor x_* \rfloor$ and you "only" need to give necessary conditions for it to be less than $2^{n-2}$.

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Edited from prior incorrect answer:


At $m=n=2, f=\tfrac{1}{4}$, but it doesn’t seem to go below that, and in fact gets large quickly as we count up in $n$ and use the minimizing value of $m$ for that $n$.

I’m relatively confident, but not certain, that there is no such number. See below.

Minimization Attempt

To first approach this, look for any minima in the function on a continuous domain and, if we find any, check the nearest natural numbers.

Find the critical points of $(\frac{3}{2})^n - 2^m$ and of it’s inverse. Consider any min’s of the former and maxes of the latter for any case $ \tfrac{-1}{4} < f < \tfrac{1}{4} $

$$f_n \equiv \frac{\partial f}{\partial n} = log (3/2) (3/2)^n$$ has no zero point but exists everywhere, so no critical point.

$f_m = -\log (2) 2^m$, same.

(If we had critical points, we’d check the sign of the $D=f_{xx}f_{yy} - f_{xy}^2$ at the m and n of the critical point(s) to see if min or max.)

We have no mins nor maxes in the range of interest.

Looking for Solutions

If the domain were continuous, one solution would be to set $$(\frac{3}{2})^n = 2^m \implies f=0$$

$$n \log(3/2) = m \log2 , m/n = \log(3/2)/\log(2)$$

With our problem, pick natural numbers to approximate that.

For any $n$, the best $m$ is $Round(n \log(3/2)/\log(2))$, where $Round()$ is rounding to the nearest natural number.

So, the best attempt at any $n$ for $f(m,n) is:

$$g(n)= |(\frac{3}{2})^n - 2^{(Round(n \log(3/2)/\log(2))}|$$

This also cannot be minimized or maximized, in this case because $g, g_n$ not continuous. This at least limits the $m,n$ space to search within, but cannot find a solution analytically.


One possible answer would be if this converged to some value less than 1/4. The needed convergence:

$$\lim_{n \to -\infty} |(\frac{3}{2})^n - 2^{(Round(n \log(3/2)/\log(2))}|<\tfrac{1}{4}$$

If anyone can solve that I’d be impressed? Checking some high values, it appears to get big very fast. In fact, at $m=n=2, f=0.25$, and tends strongly to increase at higher $n$ (but not monotonically in $n$) and appears to diverge in the limit.

But I can’t guarantee there isn’t some very large, fortuitous $n$ that surprisingly works. Other conjectures in math, such as the Polya Conjecture, have held to very high numbers ($10^{300}$ in that case), only for a surprise nearly knife-edge counterexample to be found (~1.845 E365). And there may yet be an analytical solution out there for this problem.

Al Brown
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    Even if you take really really high values of m and n and approximate the logarithm really well in this way, a small percentual error could still lead the enormous values of $2^m$ and $(3/2)^m$ to differ by way more than 1/4, right? In short, even if $|(3/2)^{\frac{n_k}{m_k}} - 2| \to 0$ for $\frac{n_k}{m_k}$ approximating that log-value, why does $|(3/2)^{n_k} - 2^{m_k}| \to 0$? – Lukas Miaskiwskyi Aug 03 '21 at 20:54
  • @LukasMiaskiwskyi Ok i edited to address that. Thanks! And i agree – Al Brown Aug 03 '21 at 22:55