# Total digit sum in distinct bases grow unboundedlly

For positive integers $$n$$ and $$d\ge 2$$, let $$S_d(n)$$ be the digit sum of $$n$$ in base $$d$$.

When $$a$$ and $$b$$ are not powers of each other, is it always true that $$\liminf_{n\to\infty} S_a(n)+S_b(n)=\infty?$$

The limit superior is infinite, e.g. let $$x_n=(ab)^{b^n}(a^n-1)+b^n-1$$ then $$S_a(x_n)+S_b(x_n)\ge (a+b-2)n$$, and we know that if there is a counterexample, it will be sparse. This is the case I'm having difficulty eliminating: we need to prove that for any fixed $$k$$ there are at most finite $$n$$ such that $$S_a(n)+S_b(n)=k$$. This is true for $$k=2$$ when $$k\ge n$$ by $$S_b(a^n)>1$$, but I can't see a similar way to prove it for e.g. $$k=3$$.

I'm originally trying to prove the weaker claim that $$\liminf_{n\to\infty}\sum_{p\text{ prime}}\log(p)\left\lfloor\frac{S_p(n)}{p-1}\right\rfloor=\infty$$ which arises from the ratio $$(n!)^{-1}\prod_pp^{\left\lfloor\frac{n}{p-1}\right\rfloor}$$. The sum has an upper bound $$O(\sqrt{n\log n})$$, by considering $$\left\lfloor\frac{S_p(n)}{p-1}\right\rfloor$$ having at most $$\sqrt{n/\log n}$$ nonzero values for $$p\ge\sqrt{n\log n}$$, which fits experimental evidence.

• It would follow from this plausible statement (which may be hard to prove on its own): for any fixed integers $m,k\geq1$, $$a^{x_1} + \dots +a^{x_m} = b^{y_1}+\dots+b^{y_k}$$ has a finite number of solutions in nonnegative integers $x_i, y_j$. Commented May 29 at 4:37
• Sounds like a job for the Schmidt subspace theorem (or the theory of the S-unit equation). Commented May 29 at 5:41
• There is $\Omega(\log\log n/\log\log\log n)$ bound for $\log a/\log b$ irrational by C.L. Stewart (see mathoverflow.net/q/30357)
– te4
Commented May 30 at 5:00
• The condition "$a$ and $b$ are not powers of each other" is not quite right; try $a=4$, $b=8$, and $n = 2^{6k}+1$ . . . Commented May 30 at 5:18
• @NoamD.Elkies you're right, it should be rational powers of each other Commented May 30 at 9:44

As observed in comments, it suffices to show that for fixed $$m,k$$, there are only finitely many solutions in non-negative integers to the equation $$a^{x_1} + \dots + a^{x_m} = b^{y_1} + \dots + b^{y_k}. \tag{1}$$ To prove this we can assume inductively that the claim is already proven for smaller values of $$m+k$$. In particular we can assume that no non-trivial subsum of the $$a^{x_1},\dots,a^{x_m}$$ agrees with no non-trivial subsum of the $$b^{y_1},\dots,b^{y_k}$$, since this would split the above equation (1) into two equations of the form (1) with smaller values of $$m+k$$, and this would only generate finitely many solutions to (1) by the induction hypothesis.
Let $$S$$ denote the set of primes dividing at least one of $$a$$ or $$b$$, and define an integer $$S$$-unit to be a (rational) integer with all prime factors lying in $$S$$. Then (1) is asserting that the $$m+k$$ integer $$S$$-units $$a^{x_1}, \dots, a^{x_m}, -b^{y_1},\dots,-b^{y_k}$$ sum to zero. The claim now follows from the following result (a consequence of the Schmidt(-Schlickewei) subspace theorem):
Theorem Let $$S$$ be a finite set of primes, and $$n \geq 1$$. Then up to projective equivalence $$(z_1,\dots,z_n) \mapsto (\lambda z_1,\dots,\lambda z_n)$$, there are only finitely many solutions to the equation $$z_1+\dots+z_n=0$$ where $$z_1,\dots,z_n$$ are integer $$S$$-units with the property that no non-trivial subsum of these integer $$S$$-units sum to zero.
See for instance Corollary 1.1 of these notes of Evertse for a proof. (The fact that $$a,b$$ are not rational powers of each other ensures that each projective class of solutions to $$z_1+\dots+z_{k+m}=0$$ can generate at most one solution to (1).)
Remark 1. One can illustrate the use of the subspace theorem by focusing on the simple example $$3^x = 2^y + 1$$. Here we use a special case of the Schmidt--Schlickewei subspace theorem: For any $$\delta>0$$, and up to projective equivalence, the number of integer solutions $$z_1, z_2$$ to the inequality $$\frac{\|z_1\|_\infty \|z_2\|_\infty}{\|z\|_\infty} \frac{\|z_1\|_2 \|z_2\|_2}{\|z\|_2} \frac{\|z_1+z_2\|_3 \|z_2\|_3}{\|z\|_3} \leq H(z)^{-2-\delta}$$ is finite, where $$\|z_i\|_p$$ is the $$p$$-adic valuation of $$z_i$$ (equal to $$p^{-\nu}$$ if $$z_i$$ is divisible by exactly $$\nu$$ powers of $$p$$), $$\|z\|_\infty = |z|$$ is the Archimedean valuation, $$\|z\|_v := \max(\|z_1\|_v, \|z_2\|_v)$$, and $$H(z)$$ is the absolute height of $$z$$ (which, if $$z_1,z_2$$ are coprime, is just $$\max(|z_1|, |z_2|)$$). But if one plugs in $$z_1 = 2^y$$ and $$z_2=1$$ here for some solution to $$3^x = 2^y + 1$$, one can calculate that the left-hand side decays like $$H(z)^{-3}$$, so there are only finitely many solutions to $$3^x=2^y+1$$.
Remark 2. Due to the use of the subspace theorem, this result is ineffective. It would be interesting to get a more effective bound on the solutions here. I experimented with trying to use Baker's theorem, but could not use this theorem to cover all cases (the gaps between consecutive $$x_i$$ or $$y_j$$ seem to play a role in whether this theorem is helpful).