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The musical scale of $5$ fifths has good temperament because $3^5$ is quite close to a power of $2$, namely $2^8-3^5=13$, and I suspect that in the subsequent powers of $3$ none is so close to a power of $2$. Analogously, our usual chromatic scale with $12$ notes has good temperament because $3^{12}$ is very close to a power of $2$, namely $3^{12}-2^{19}=7153$.

Does the Diophantine equation $|3^n-2^m|\le 13$ admit a solution with $n>5$?

Does the Diophantine equation $|3^n-2^m|\le 7153$ admit a solution with $n>12$?

A direct inspection of the first $35$ scales of fifths with good temperament shows that any such solution must be $n>10^{18}$.

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    $\begingroup$ related: math.stackexchange.com/q/2481119/87355 $\endgroup$
    – Carlo Beenakker
    Commented Feb 26 at 19:40
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    $\begingroup$ the difference grows exponentially with increasing $n$, see the estimate $\left| 3^n - 2^ m\right| = \mathcal{O}\left( \frac{3^n}{n} \right)$ for $n\gg 1$. $\endgroup$
    – Carlo Beenakker
    Commented Feb 26 at 20:23
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    $\begingroup$ @CarloBeenakker The estimate you quote is an upper bound for infinitely many $n$'s. Exponential lower bounds are available for all $n$, see e.g. my post below. $\endgroup$
    – GH from MO
    Commented Feb 26 at 20:26
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    $\begingroup$ Also related: mathoverflow.net/questions/116840/… $\endgroup$
    – Pietro Majer
    Commented Feb 26 at 21:15
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    $\begingroup$ @TimothyChow. Emilio Gagliardo used to have an office in the Math Department of the University of Pavia, with a soundproof door, where he would practice compositions on such a scale. $\endgroup$
    – Yaakov Baruch
    Commented Feb 26 at 21:30

1 Answer 1

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Ellison (1971), see MR0417051, proved that $$|3^n-2^m|>1.8^m,\qquad m>27.$$ From this bound and the SAGE program below, we see that the only solution of $|3^n-2^m|\leq 13$ with $n\geq 5$ is $(m,n)=(8,5)$, and the only solution of $|3^n-2^m|\leq 7153$ with $n\geq 12$ is $(m,n)=(19,12)$.

for m in range(28):
    for n in range(5,18):
        if abs(3^n-2^m)<14:
            print(m,n,3^n-2^m)
for m in range(28):
    for n in range(12,18):
        if abs(3^n-2^m)<7154:
            print(m,n,3^n-2^m)

P.S. It is worth pointing out that the solutions $(8,5)$ and $(19,12)$ above correspond to the continued fraction approximants $8/5$ and $19/12$ of $\log 3/\log 2$.

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    $\begingroup$ Note that the ideal base in Ellison's result is $2e^{-1/10}$ which is slightly over 1.8. There is a followup paper by Tijdeman MR0325549 which gets an asymptotically tighter, but not effective bound. (I think that one can be made effective though.) $\endgroup$
    – JoshuaZ
    Commented Feb 26 at 21:18
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    $\begingroup$ @JoshuaZ Sure, I used $1.8$ for stylistic reasons. There is a better effective bound by Rhin (1987), see MR1017910. His theorem gives that $\|m\log 2+n\log 3\|\geq\max(m,n)^{-13.3}$ as long as $\max(m,n)\geq 2$. $\endgroup$
    – GH from MO
    Commented Feb 26 at 21:31
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    $\begingroup$ Ah, didn't know about the Rhin result. Good to know. $\endgroup$
    – JoshuaZ
    Commented Feb 26 at 21:39
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    $\begingroup$ @JoshuaZ I was lazy to work out what effective lower bound follows for $|3^n-2^m|$ from the result of Rhin (1987). $\endgroup$
    – GH from MO
    Commented Feb 26 at 21:41
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    $\begingroup$ MR0417051: Ellison - On a theorem of S. Sivasankaranarayana Pillai. Do you know if there is an exposition of the proof available anywhere widely accessible? $\endgroup$
    – LSpice
    Commented Feb 26 at 23:44

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