Finitely many palindromes in two consecutive number bases, for fixed and distinct numbers of digits

Solution 1:

The almost-counterexample I gave in the comments has factor 2 in the denominators, and this is not without a reason. In fact, this factor prevents existence of an infinite series of solutions of a fixed length. Here is a proof.

First notice that in an infinite series of solutions, values of $b$ cannot be bounded. This immediately proves the case $|d_1 - d_2|>1$ as one palindrome in this case is asymptotically at least factor $b$ times larger than the other. Hence, it remains to consider the case $|d_1-d_2|=1$.

Let $d=2l+1$ be the length of one palindrome and $d-1=2l$ be the length of the other. If $b$ is the base of the first palindrome, then the second must be in base $b+1$ (not $b-1$ as this palindrome divisible by the base plus 1). Then we need to solve $$\sum_{i=0}^{l-1} a_i (b^i + b^{2l-i}) + a_l b^l = \sum_{i=0}^{l-1} c_i ((b+1)^i + (b+1)^{2l-1-i})$$ in integers $a_0\in[1,b-1]$, $c_0\in[1,b]$, $a_i\in [0,b-1]$ and $c_i\in[0,b]$ for $i\in\{1,2,\dots,l\}$.

Linearizing this equation as explained in my MO answer to a related question and expressing $a_0$, $a_1$, and $c_0$, we get $$\begin{cases} a_0 = -k_d,\\ a_1 = -\frac{d}2 k_0 b + k_1 b - k_0 - \frac{d}2 k_d,\\ c_0 = a_1 - k_d b + k_{d-1}, \end{cases} $$ where we have $k_0,k_1,k_{d-1},k_d$ are some integers whose lower and upper bounds depend on $d$ but not on $b$.

(Argument below is simplified.)

To keep $a_1\in[0,b-1]$ and $c_0\in[1,b]$ for large $b$, the coefficients of $b$ in $a_1$ and $c_0$ must be between $0$ and $1$. Together with $a_0\geq 1$ (i.e. $k_d\leq -1$) this implies that $k_d=-1$ and the coefficient of $b$ in $a_1$ and $c_0$ equal $1$ and $0$, respectively. Then, however, $a_1$ is a half-integer, which is impossible. Thus, an infinite series of solutions does not exist. QED