Find the solutions to $\left\lfloor\left(\frac{5}{3} \right)^n\right\rfloor = 3^m$

Find all positive integer solutions to $\left\lfloor\left(\dfrac{5}{3} \right)^n\right\rfloor = 3^m$.

Let $a_n = \left\lfloor\left(\dfrac{5}{3} \right)^n\right\rfloor$. Then $$a_n = 1,2,4,7,12,21,35,59,99,165,275,459,765,1276,2126,3544,5907,9846,16410,\ldots.$$ Since a power of $3$ is odd, we only need to look at the odd terms of $a_n$: let these be $b_n$. Then $b_n = 1,7,21,35,59,99,165,275,459,765,5907,\ldots$. There don't seem to be powers of $3$ with a positive exponent in $b_n$ in the first few terms.

Using the Binomial Theorem, we have $$\left(\dfrac{5}{3} \right)^n = \left(1+\dfrac{2}{3}\right)^n = 1+\dfrac{2}{3} \binom{n}{1}+\left(\dfrac{2}{3}\right)^2 \binom{n}{2}+\cdots+\left(\dfrac{2}{3}\right)^n\binom{n}{n}.$$ How can we continue from here?

• Could you tell us the source of this problem? Thanks! – Matthew Conroy Sep 24 '17 at 19:06
• If that helps, the original equation can be written $$0\le\frac nm-\frac{\log3}{\log\dfrac53}<\frac{\log(1+3^{-m})}{m\log\dfrac53}<\frac{3^{-m}}{m\log\dfrac53},$$ meaning the the rational $\frac nm$ is a very tight approximation of $\frac{\log3}{\log\frac53}$. – Yves Daoust Sep 24 '17 at 19:40
• Nowhere near a solution with $n\le200$. In relative terms one of the closest matches is with $n=28$ when $\lfloor(5/3)^{28}\rfloor=1628414$ that is reasonably close to $3^{13}=1594323$. The difference is $>3^9$. With $n=157$ we get $3^m<(5/3)^n<3^m+3^{m-6}, m=73$. Unless my eyes failed me that was the lowest $n$ with error $<3^{m-6}$. – Jyrki Lahtonen Sep 24 '17 at 21:23
• Simplifying it gives ( hopefully ) $$\left\lfloor n\log_3 \left(\dfrac 53\right) - m\right\rfloor = 0$$. – user8277998 Sep 25 '17 at 8:51
• Well, Wolfram|Alpha gives only two integer solutions n=0, m=0 and n=1, m=0. I wouldn't be too sure that those are the only solutions... – Miksu Sep 25 '17 at 13:26

If we write $$\left(\frac{5}{3}\right)^n = \left[ \left(\frac{5}{3} \right)^n \right] + \left\{ \left(\frac{5}{3}\right)^n \right\},$$ then supposing that $$\left[ \left(\frac{5}{3} \right)^n \right]=3^m$$ implies the inequality $$0 < 5^n - 3^{n+m} < 3^n.$$ Applying lower bounds for linear forms in (two complex) logarithms (say from Laurent-Mignotte-Nesterenko) to $$\Lambda = n \log 5 - (n+m) \log 3$$ leads, after a wee bit of work, to the conclusion that $n \leq 1$.

I'm not sure if there's an elementary way to prove this. I can't see one off the top of my head.

It seems the only solution in non-negative integers is $(n,m)=(1,0)$. I try to explain why I believe this.

We have $$\left(\frac53\right)^{\dfrac{m}{\log_35-1}}=3^m$$ therefore for $0\lt x\lt 1$

$$\left\lfloor\left(\frac53\right)^{\dfrac{m}{\log_35-1}}+x\right\rfloor=3^m$$ In order to have a solution $(n,m)$ it is necessary that the equation $$\left(\frac53\right)^{\dfrac{m}{\log_35-1}}+x=\left(\frac53\right)^n$$ have integer solutions for some $x\in (0,1)$ In other words we need to have $$\left(\frac53\right)^n-\left(\frac53\right)^{m\alpha}\in (0,1)$$ where $\alpha\approx2.150660103087123508854$.

The figure 1 below shows in brown color the region where $$0\lt\left(\frac53\right)^n-\left(\frac53\right)^{m\alpha}\lt1$$ We can see that more the integer $n$ increases, the length $d_n$ of the interval of possibilities for $m$ be also integer are reduced more and more.

For example for $$n=1\text{ one has } 0\le m\le0.45\lt1\text { so } d_1\approx0.45\\n=2\text{ one has } 0.52\lt m\le0.92\lt1\text { so } d_2\approx0.40\\n=3\text{ one has } 1.17\lt m\le1.38\lt2\text { so } d_3\approx0.21\\n=4\text{ one has } 1.72\le m\le1.85\lt2\text { so } d_4\approx0.13\\$$ and so on, for example for $n=10$ calculation gives $4.6445\lt m\lt4.6497\lt5$ giving $d_{10}\approx0.0052$ and it is verified that $d_n\to 0$ quickly.

The conclusion is that if a solution exists, it tends to be a point in the curve of figure 2, i.e. it is a very approximate integer solution of the diophantine equation $$\left(\frac53\right)^n-\left(\frac53\right)^{m\alpha}=1$$

it cannot be an exact solution because in that case we would have $$\left\lfloor\left(\frac53\right)^n\right\rfloor=3^m+1$$ APPENDIX FOR FUN.-When $f(x)=\left\lfloor(\frac53)^x\right\rfloor$ one has $f(2.338)=3\\f(4.675)=3^2\\f(7.013)=3^3\\f(9.3499)=3^4\\f(11.688)=3^5\\f(14.025)=3^6\\f(16.363)=3^7\\f(18.6999)=3^8\\f(21.0371)=3^{10}$.  