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Let $F:=\{F_n\}_{n=1}^\infty $ be the Fibonacci sequence (defined by $ F_n=F_{n-1}+F_{n-2}$ with $F_1=F_2=1$).

  • Is it true that there exists an infinite set of integers $B$ such that $(F-F)\cap (B-B)=\{ 0\}$?

(note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$)

user133281
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M.H.Hooshmand
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1 Answers1

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Claim: Given a set $T$ of integers in $\{1,2,\ldots,n\}$, we can find a subset $B$ of size at least $n/2|T|$ such that $(B-B) \cap T$ is empty.

Now let $T=(F-F) \cap \{1,2,\ldots,n\}$. Note that $|T|=O(\log^2n)$ because the $n$th Fibonacci number is approximately $1.61^n$. Thus you can find a $B$ such that $(B-B) \cap (F-F)$ is $\{0\}$ and in fact has at least $\Omega(\dfrac{n}{\log ^2n})$ elements in the first $n$ natural numbers.

Proof of Claim: Initialize $B=\{1\}$. While $|B|<n/2|T|$, add $b$ to $B$ such that $(b-B) \cup (B-b)$ avoids $T$ except for zero. This is possible if $H=(B-T) \cup (T-B)$ does not exhaust $\{1,2,\ldots,n\}$. Since $|H| \leq 2|B||T|$, it is possible if $|B| <\dfrac{n}{2|T|}$.

Aravind
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