TITLE: Product of arithmetic progressions
QUESTION [7 upvotes]: Let $(a_1,a_2\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ be two permutations of arithmetic progressions of natural numbers. For which $n$ is it possible that $(a_1b_1,a_2b_2,\dots,a_nb_n)$ is an arithmetic progression?
The sequence is (trivially) an arithmetic progression when $n=1$ or $2$, and there are examples for $n=3,4,5,6$:
$n=3$: $(11,5,8), (2,3,1)$
$n=4$: $(1,11,6,16), (10,4,13,7)$
$n=5$: $(8,6,4,7,5), (4,9,19,14,24)$
$n=6$: $(7,31,19,13,37,25), (35,11,23,41,17,29)$

REPLY [2 votes]: From Yaakov's answer, take the second difference $(u_i-2u_{i+1}+u_{i+2})$ to remove (1,2,...,n) and the constant term.  Then you have  (n-2)x3 matrix $W=[D^2(u),D^2(v),D^2(uv)]$ that we want to have a null vector.  So calculate $det(W^tW)$, and you need the determinant to be zero for a solution.  I ran the calculations and got no solutions for 8 or 9.  Case 9 took 5 hours on Matlab.