Product of permutations of consecutive numbers yields arithmetic sequence Let $n\geq 3$ be an integer, and $a,b$ be positive integers. Let $c_1,\ldots,c_n$ be a permutation of $a,a+1,\ldots,a+(n-1)$, and $d_1,\ldots,d_n$ be a permutation of $b,b+1,\ldots,b+(n-1)$. Is it possible that $c_1d_1,c_2d_2,\ldots,c_nd_n$ form an arithmetic sequence?
For example, for $n=3$, this is not possible. The permutations we need to check are
$ab,(a+1)(b+1)=ab+a+b+1,(a+2)(b+2)=ab+2(a+b)+4$
$ab,(a+1)(b+2)=ab+2a+b+2,(a+2)(b+1)=ab+a+2b+2$
and so on. However, for general $n$ it is harder to check all permutations.
 A: Assume that there exists a sequence $k=(k_1, ... , k_n)$ such that $k_i=c_i d_i$ (1). 
Let $c$ be $k_{i+1}-k_i$. 
Let $n_j = \vert\lbrace x\in k\vert x\equiv 0 (mod$ $j) \rbrace\vert$. Let $n'_3$ be the number of $i$ such that exactly one of $c_i$ and $d_i$ is divisible by $3$, $n'_9$ the number of $i$ such that $c_i$ and $d_i$ are both divisible by $3$. 
Then, since the $c_i$ and the $d_i$ are a permutation of $n$ consecutive numbers, $ 2\lfloor \frac{n}{3}\rfloor \leq n'_3 + 2n'_9 \leq 2\lceil\frac{n}{3}\rceil$ (2)
Moreover from (1) we have $n_3 = n'_3 + n'_9 $ (3) and $n'_9 \leq n_9$ (4).
Now if $3\vert c$ we have $n_3=0$ which contradicts (2) and (3), or $n_3 = n$, which implies by (3) and (2) $n + n'_9 \leq 2\lceil\frac{n}{3}\rceil $, which is impossible for $n \geq 5$, and it is easy to check the case $n = 4$.
If $3$ doesn't divide $c$, then $\lfloor\frac{n}{3}\rfloor\leq n_3 \leq \lceil\frac{n}{3}\rceil$ and $\lfloor\frac{n}{9}\rfloor\leq n_9 \leq \lceil\frac{n}{9}\rceil$. Hence by (3) $-n'_3-n'_9 \geq - 2\lceil\frac{n}{3}\rceil$, and then using (4) and (2) $\lceil\frac{n}{9}\rceil\geq 2\lfloor\frac{n}{3}\rfloor - \lceil\frac{n}{3}\rceil$, which is impossible for $n\geq 12$, so it suffice to check for $3\leq n \leq12$. But this is easy because there are finite cases, which can be solved because they are linear systems where we have to check if integer solutions exists. 
