Let $p\neq 3$ be a prime number. Show that there is a non-constant arithmetic sequence of positive integers $x_1,x_2,\ldots ,x_p$ such that the product of the terms of the sequence is a cube.
Problem
Source: Baltic Way 2011
Tags: arithmetic sequence, number theory proposed, number theory
Swistak
06.11.2011 16:19
(Of course $p \neq 3$) Let's start from sequence $1, 2, ..., p$. If prime $q$ divides product of the terms of the sequence in $k$-th power and $k=3l+1$ or $k=3l+2$, multiply every term by $q$ or $q^2$
yunxiu
06.11.2011 16:37
If $p = 3k + 2$, let ${x_i} = ip!$, then $\prod\limits_{i = 1}^p {{x_i}} = {(p!)^p}p! = {(p!)^{3k + 3}}$. If $p = 3k + 1$, let ${x_i} = i{(p!)^2}$, then $\prod\limits_{i = 1}^p {{x_i}} = {(p!)^{2p}}p! = {(p!)^{6k + 3}}$.
mahanmath
06.11.2011 16:57
Exactly same idea : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1435455&sid=6c903424c9bfca0bd78c4f94fb6d97a3#p1435455
E.A.K
20.04.2018 14:34
why the answer $x_i=i^3$ is not correct?
blawho12
20.04.2018 14:54
E.A.K wrote: why the answer $x_i=i^3$ is not correct? That is not an arithmetic sequence.