China National Team Selection Test

Suppose on the contrary that there exist 40 distinct positive integers in arithmetic progression such that each term can be written as $2^k+3^l$, and denote this sequence by $a,a+d,a+2d,\dots ,a+39d$, where $a,d$ are positive integers. Let $$m=\lfloor {\log }_2(a+39d)\rfloor ,n=\lfloor {\log }_3(a+39d)\rfloor .$$ In what follows, we first show that at most one of $a+26d,a+27d,\dots ,a+39d$ cannot be written as $2^m+3^l$ or $2^k+3^n$ ($k,l$ are nonnegative integers). Suppose that $a+hd$ cannot be written as $2^m+3^l$ or $2^k+3^n$, for some $26\le h\le 39$. Then, by assumption, $a+hd=2^b+3^c$ for some nonnegative integers $b,c$. By the definition of $m$ and $n$, it is clear that $b\le m,c\le n$. Since $a+hd$ cannot be written as $2^m+3^l$ or $2^k+3^n$, we have $b\le m-1,c\le n-1$. If $b\le m-2$, then $$\begin{array}{rl}a+hd& \le 2^{m-2}+3^{n-1}=\frac{1}{4}\times 2^m+\frac{1}{3}\times 3^n\\ & \le \frac{7}{12}\times (a+39d)<a+26d,\end{array}$$ a contradiction. If $c\le n-2$, then $$\begin{array}{rl}a+hd& \le 2^{m-1}+3^{n-2}=\frac{1}{2}\times 2^m+\frac{1}{9}\times 3^n\\ & \le \frac{11}{18}\times (a+39d)<a+26d,\end{array}$$ also a contradiction. It follows that $b=m-1,c=n-1$, which implies that at most one of $a+26d,a+27d,\dots ,a+39d$ cannot be written as $2^m+3^l$ or $2^k+3^n$. In these 14 numbers, at least 13 numbers can be written as $2^m+3^l$ or $2^k+3^n$. By the pigeonhole principle, at least 7 numbers can be written in the same form. We shall discuss two cases. Case 1: There are 7 numbers in the form of $2^m+3^l$, denoted by $$2^m+3^{l_1},2^m+3^{l_2},\dots ,2^m+3^{l_7},$$ where $l_1<l_2<\cdots <l_7$. Thus, $3^{l_1},3^{l_2},\dots ,3^{l_7}$ are the 7 terms of an arithmetic progression with 14 terms and the common difference $d$. However, $$13d\ge 3^{l_7}-3^{l_1}\ge (3^5-\frac{1}{3})\times 3^{l_2}>13(3^{l_2}-3^{l_1})\ge 13d,$$ a contradiction. Case 2: There are 7 numbers in the form of $2^k+3^n$, denoted by $$2^{k_1}+3^n,2^{k_2}+3^n,\dots ,2^{k_7}+3^n,$$ where $k_1<k_2<\cdots <k_7$. Thus $2^{k_1},2^{k_2},\dots ,2^{k_7}$ are the 7 terms of an arithmetic progression with 14 terms and the common difference $d$. However, $$13d\ge 2^{k_7}-2^{k_1}\ge (2^5-\frac{1}{2})\times 2^{k_2}>13(2^{k_2}-2^{k_1})\ge 13d,$$ a contradiction. It follows from the above arguments that our assumption at the very beginning is false, which completes the proof.