M and n are positive integers such that $2^n - 3^m > 0$. Prove (or disprove) that $2^n - 3^m \geqslant 2^{n-m}-1$.

Solution 1:

Answer: Possible counter-examples must satisfy $m<e^{28}$ and $n<e^{29}$, hence only a finite number of them may exists. This follows by Baker's theorem about lower bounds of linear forms of logarithms.

For prove it, let $(m,n)$ be a counter example, that's $3^m <2^n <3^m+2^{n-m}-1$. We get $$\begin {align} 0 <n\log (2)-m\log (3) &<\log\frac {1-3^{-m}}{1-2^{-m}}\\ &<\frac {1-3^{-m}}{1-2^{-m}}-1\\ &<2^{-m} \end {align}$$

By Baker's theorem there exists an effective constant $C$ (not depending on $n,m$) such that $$n\log (2)-m\log(3) >n^{-C} $$ for all $n,m$ such that $n\log (2)-m\log(3) >0$.

This leads to $n^{-C}<2^{-m } $ which implies $\frac m{\log(n)}<\frac C{\log(2)}$ hence $\frac m {\log (n)} $ is bounded by an effective constant. On the other hand $n\log (2)-m\log (3)<2^{-m}<1$ gives $\log (n)<\log (m)+1$ hence \begin{align*} \frac m {\log (m)} &\leq\frac{3m}{1+\log(m)}\\ &<\frac{3m}{\log(n)}\\ &<\frac{3C}{\log(2)} \end{align*}

that's $\frac m {\log (m)} $ is bounded as well by an effective constant.

By effectivness of the constant, is enough to verify the impossibility of $3^m <2^n <3^m+2^{n-m}-1$ for finitely many values of $m,n $.

In particular, the explicit result by Baker and Wüstholz states: \begin{align*} &\log|\Lambda|>-C'h(\alpha_1)h(\alpha_2)\log(\max\{|\beta_1|,|\beta_2|\})\\ &C'=18(n+1)!n^{n+1}(32d)^{n+2}\log(2nd) \end{align*} where $\Lambda=\beta_1\lambda_1+\beta_2\lambda_2$ and in our case \begin{align*} &\beta_1=n& &\lambda_1=\log(\alpha_1)& &\alpha_1=2\\ &\beta_2=-m& &\lambda_2=\log(\alpha_2)& &\alpha_2=3 \end{align*} hence $h(\alpha_i)=\alpha_i$, $d=1$ and $n=2$. Since $C=C'h(\alpha_1)h(\alpha_2)$, with these values, we get: $$C=18\cdot 3!\cdot 2^3\cdot 32^4\cdot\log(4)\cdot 2\cdot 3<8\cdot 10^9$$ from which $$\frac m{\log(m)}<4\cdot 10^{10}$$ which is satisfied for $m<e^{28}$. Thus, possible counter-examples are bounded by $m<e^{28}$. Since $\log(n)<\log(m)+1$ we have $n<e^{29}$, hence only a finite number of counter-examples may exists.

Solution 2:

Answer: There are at most finitely many counterexamples. This is related to the irrationality measure $\mu$ of $\log 3/\log 2$, which is finite (and effectively boundable) according to this article by Yann Bugeaud, Theorem 1.1. What I don't know is whether we can bound the possible counterexamples.

We have by the mean value theorem for $\exp$: $$\begin{align*}2^n-3^m &\geq 3^m (n\log 2-m\log 3)\\ &\geq 3^m n\log 2 \cdot n^{-\mu-\varepsilon}\end{align*}$$ for (say) $\varepsilon=0.1$, except for possibly a finite number of pairs $(m,n)$ coming from exceptionally good approximations of $\log3/\log 2$.

To finish, the idea is that either the inequality is trivial, or $3^m$ is sufficiently close to $2^n$ to show that this is larger than $2^{n-m}$.

Note that (less important)

• $m\leq n-1$ except for small values
• We may ignore the $+1$ in the RHS because the LHS is odd

and more importantly:

• If $2^{n-1} \geq 3^m$ it is trivial, becaue $2^n-3^m > 2^{n-1} \geq 2^{n-m}$

so that we may assume $3^m\geq 2^{n-1}$.

Now $$6^m \geq 6^{(n-1)\log2/\log 3} > 3^{n-1}$$ (use a calculator for the last inequality) so that $$3^m >2^n/2^m \cdot 1.5^n \cdot\tfrac13=2^{n-m}\cdot 1.5^n\cdot\tfrac13$$ where we keep the $1.5^n$ to take care of factors $n^{-\ldots}$.

Combining everything, $$\begin{align*} 2^n-3^m &\geq 3^mn\log 2 \cdot n^{-\mu-\varepsilon}\\ &>2^{n-m}1.5^n\cdot\tfrac13\cdot n\log 2 \cdot n^{-\mu-\varepsilon}\\ &>2^{n-m}\end{align*}$$ for $n$ sufficiently large (because $\mu$ is finite!), except for possibly a finite number of pairs $(m,n)$.