Some help with a number theory problem

[Martin Rattigan]

Since ${b < a^{k+1}}$ and ${deg(Q_k) < k+1}$, the rightmost expression in (*) is bounded above by ${1}$
when ${n}$ is sufficiently large. Thus ${r_{k,n} = 0}$ for large ${n}$, since it is an integer.

For all large ${n}$, this yields
${b^n p_k + Q_k(a^n) = 0}$ and thus ${p_k(b/a^k )^n + Q_k(a^n) /(a^n )^k = 0}$.


This forces ${p_k = 0}$, since otherwise the left side is unbounded as ${n\rightarrow +\infty}$.
We now conclude that ${Q_k(a^n) = 0}$ for all ${n}$ and thus ${Q_k}$ is the zero polynomial
and we are done.

 

[Martin Rattigan]

I have noticed a couple of misprints in the previous two posts. Unfortunately I can no longer edit them. No one has pointed out the misprints so hopefully they haven't caused too much confusion, but I have corrected them and combined the posts below.

Define a sequence ${(Q_k)}$ of polynomials with ${deg(Q_k) \leq k}$ by ${Q_0 = -1}$, and ${Q_{k+1}(x) = a^{k+1}(x-1)Q_k(ax)-b(a^{k+1}x-1)Q_k(x)}$ for ${k \geq 0}$.

Observe that ${Q_{k+1}(0) = (b - a^{k+1})Q_k(0)}$. Iterating this and employing ${Q_0 = -1}$ leads to ${Q_k(0) = -(b - a^k )(b - a^{k-1} )...(b - a)}$.

Assume that ${b\neq a^j}$ for every non-negative integer ${j}$, so that ${Q_k(0)\neq 0}$. We will obtain a contradiction by identifying a ${k}$ such that ${Q_k}$ is identically ${0}$.

Let ${r_{0,n} = (b^n - 1)/(a^n - 1)}$ for ${n>0}$. By assumption ${r_{0,n}}$ is an integer.

For ${k\geq 0}$ define ${r_{k+1,n}}$ recursively by ${r_{k+1,n}=a^{k+1}r_{k,n+1}-br_{k,n}}$.

Let ${p_0=1}$ and ${{p_{k+1} = b(1 - a^{k+1} )p_k}}$.

By induction on ${k}$ it follows for ${n \geq 1}$ and ${k \geq 0}$ that

${r_{k,n}=[b^np_k + Q_k(a^n) ]/[(a^{n+k} - 1)(a^{n+k-1} - 1)...(a^n - 1)]}$

Now fix ${k}$ so that $a^k<b<a^{k+1}$. Since

$(a^n-1)(a^{n+1}-1)\dots(a^{n+k}-1)=a^{n(k+1)}(1-a^{-n})(a-a^{-n})\dots(a^k-a^{-n})\geq a^{n(k+1)}/2$

we see that

$|r_{k,n}|\leq |b^np_k+Q_k(a^n)|/[a^{n(k+1)}/2]\leq 2(|p_k|(b/a^{k+1})^n+|Q_k(a^n)|/(a^n )^{k+1})\;\;(*)$

Since ${b < a^{k+1}}$ and ${deg(Q_k) < k+1}$, the rightmost expression in (*) is bounded above by ${1}$ when ${n}$ is sufficiently large. Thus ${r_{k,n} = 0}$ for large ${n}$, since it is an integer.

For all large ${n}$, this yields ${b^n p_k + Q_k(a^n) = 0}$ and thus ${p_k(b/a^k )^n + Q_k(a^n) /(a^n )^k = 0}$.

This forces ${p_k = 0}$, since otherwise the left side is unbounded as ${n\rightarrow +\infty}$. We now conclude that ${Q_k(a^n) = 0}$ for all ${n}$ and thus ${Q_k}$ is the zero polynomial and we are done.