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A problem in group theory (indices of imprimitive groups) gives rise to the following conjectures in number theory. Suppose a positive integer $n$ has binary and ternary expansions $n=\sum_{k\geqslant0}b_k2^k=\sum_{k\geqslant0}t_k3^k$. For example, $6=2^2+2=2\cdot 3$; $81=2^6+2^4+1=3^4$.

Conjecture 1. If $\sum_{k\geqslant0}b_k=\sum_{k\geqslant0}t_k=2$, then $n\in\{6,10,12,18,36\}$.

Conjecture 2. If $\sum_{k\geqslant0}(b_k+t_k)=4$, then $n\in\{6,10,12,18,36,81\}$.

Terry Tao discusses the separation of powers of 2 and powers of 3; see his blog:

http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/

Conjectures 1 and 2 are related to separation problems. For example, the four $n$ with $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,2)$ or $(2,1)$ are related to the solutions to $|3^p-2^q|=1$, namely $(p,q)=(1,1),(2,3),(0,1),(1,2)$. More trivially, $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,1)$ is related to the solution to $|3^p-2^q|=0$, namely $(p,q)=(0,0)$. I wonder whether one needs the values for the `effective constants' in Tao's blog, or whether elementary arguments suffice to prove these conjectures.

Answers to the question " $3^n - 2^m = \pm 41$ is not possible. How to prove it? " may help.

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You can prove your conjecture by combining local methods, linear forms in logarithms, and computations (either brute force or continued fractions).

Suppose you have $\sum t_k=\sum b_k = 2$, i.e. $2^x+2^y=3^u+3^v$, $x>y$, $u\geq v$. Then $2^{2(x-y)}-1$ is divisible by $3^v$. Since 2 is a primitive root modulo 9, it is primitive modulo $3^k$, hence $2\cdot 3^{v-1}|2(x-y)$. In particular, $3^{v-1}\leq x$. In the same way we find that $2^{y-2}\leq u$.

Baker's method gives lower bounds for linear forms in logarithms. In this case the most appropriate result google gave me is due to Bugeaud ( http://www.birs.ca/workshops/2012/12ss131/files/bugeaud_LFL.pdf , Theorem C):

Let $a_1, a_2, b_1, b_2$ be rational integers, $a_1, a_2$ multiplicative independent. Let $A_1, A_2$ be real numbers satisfying $A_i\geq\max(3, a_i)$. Put $B=\frac{b_1}{\log A_2}+\frac{b_2}{\log A_1}$. Then for $\Lambda := |b_1\log a_1 + b_2\log a_2|$ we have the bound $$ \log \Lambda \geq -30.9\log A_1\log A_2\left(\max\left(21, 0.66+\log B\right)\right)^2. $$ In our case we have $a_1=2$, $a_2=3$, $b_1=x$, $b_2=u$. Then Baker gives us a lower bound for $\Lambda$, on the other hand we have $$ \Lambda = \log\frac{2^x}{3^u} = \log\frac{2^x+2^y}{3^u+3^v} + \log\frac{2^x}{2^x+2^y} + \log\frac{3^u+3^v}{3^u}, $$ and the first of the three terms on the right is 0, while the other two are very small and differ in sign, hence $\Lambda<\max(\frac{6u}{2^x}, \frac{3x}{3^u})$. I don't optimize $A_1$ and $A_2$, and just put $A_1=A_2=3$ and get $$ \max\left(\frac{6u}{2^x}, \frac{3x}{3^u}\right) > \exp\left(-37.3(0.57+\log(x+u))^2\right). $$ For $x\geq 10000$ this is impossible.

You could now check all quadruples $(x,y,u,v)$ with $x<10000$ by a computer, which is easy, since $u$ is uniquely determined by $x$, and $y$ and $v$ are small, or check that $2^x-3^u$ is large for all $x,u$ in this range by computing the continued fraction expansion of $\frac{\log 3}{\log 2}$.

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A paper of Stewart [J. reine angew. Math., 1980] proves that your $\sum_{k \geq 0} (b_k+t_k)$ grows like $\log\log n/\log\log\log n$. This result is effective and rather much more general, but not explicit.

To actually resolve your conjectures, one can mostly argue locally. A paper where this is done somewhat systematically is one of Brenner and Foster [Pacific J. Math., 1982]. I think a couple of the exponential equations you require solved might need something more, like lower bounds for linear forms in $p$-adic and complex logarithms. These are applied to a problem very similar to yours in a pair of papers of Tijdeman and Wang [Pacific J. Math., 1988] and Wang [Indagationes Math., 1989]. THe applications of these techniques to your problem date back somewhat earlier to work of Ellison and of Stroeker and Tijdeman, both in conference proceedings (sorry, can't recall the references).

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Both conjectures and generalizations of them follow from the n-term conjecture.

You want sums of small multiples of powers of $2$ and $3$ to be equal and the the number of terms to be small. In the n-term conjecture the radical is $6$, which implies effective bound on the largest power.

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  • $\begingroup$ The $abc$-conjecture, and the more general $abcd$-conjecture which is needed to prove Conjectures 1 and 2 above, may be much harder to prove than my two conjectures. If they are equally hard, it would surprise me. That said, I am no expert on the $n$-term conjecture. $\endgroup$ – Glasby Sep 25 '14 at 12:06

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