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Consider the following (easy) lemma.

Lemma. There is a subset $Q$ of the positive integers and a fixed constant $N > 0$ such that
1)$Q$ has positive asymptotic density and

2)for each rational numbers $\alpha,\beta$ it results $\alpha \beta^n \in Q$ for at most $N$ positive integers $n$.

Proof (sketch). Take $Q$ as the set of squarefree positive integers and $N=2$.

My question is: For a fixed rational number $r$, can we replace $\alpha \beta^n$ with $\alpha \beta^n + r$ in the lemma above?

Thank you in advance for any help.

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Let us write $r=\frac{s}{t}$ with $s$ and $t$ coprime. The set $S$ of positive integers which are square free and congruent to $-s\pmod t$ has positive density (essentially by the same argument which is used to show that the square free integers have positive density). Let $Q=\{\frac{m+s}{t}:m \in S\}$, which also has positive density. Now assume that $\alpha \beta^n+r\in Q$. Then $t\alpha\beta^n\in S$. However, $S$ contains only square free integers, so $t\alpha \beta^n\in S$ can hold for at most two values of $n$ (unless $\beta=\pm 1$, which should be excluded from the question), so $N=2$ works.

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Carl Pomerance sketched a solution for me, which I didn't write down in detail, but I think it goes like this:

If $r$ is an integer, we can take $S$ to be the squarefree integers, shifted by $r$, and $N=2$.

If $r$ is not an integer, then if $\beta$ is not an integer there is at most one value of $n$ such that $\alpha\beta^n+r$ is an integer, and we can take $S$ to be the set of all integers, with $N=1$.

Finally, if $r$ is not an integer, and $\beta$ is an integer, then we can write $\alpha\beta^n+r$ as $(st^n+p)/q$ for some integers $p$, $q$, $s$, $t$, with $\gcd(p,q)=1$. Let $U$ be the set of numbers that are squarefree and congruent to $-p$ modulo $q$; this is a set of positive asymptotic density. Let $S$ be the set of numbers of the form $(u+p)/q$, with $u$ in $U$. Then $S$ has positive asymptotic density, and if $(st^n+p)/q$ is in $S$, then $st^n$ is squarefree, so we can use this set $S$, with $N=2$.

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  • $\begingroup$ ---If $r$ is not an integer, then if $\beta$ is not an integer there is at most one value of $n$ such that $\alpha\beta^n+r$ is an integer--- Really? Cannot I always solve a system $\alpha\beta^n+r=a,\alpha\beta^{n+1}+r=b$ of two linear equations with two unknowns $\alpha,r$? $\endgroup$ – fedja May 22 '14 at 11:37
  • $\begingroup$ Hmmm. Clearly, I have to rethink this. $\endgroup$ – Gerry Myerson May 22 '14 at 12:55

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