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.