Problem 1721

By inducting on $n$, we show that there exist relatively consistent $n$ distinct positive integers.

$= 1+2$, and hence $2$ and $3$ is a consistent pair.

Now let $x_1 < x_2 < \dots < x_n$ be relatively consistent $n$ distinct positive integers. Then for any given $i < j$ there exist distinct positive divisors of $x_i$, say $d_1, d_2, \dots, d_k$, such that $d_1 + d_2 + \dots + d_k = x_j$. Note that for any positive integer $a$, we have that $a d_1, a d_2, \dots, a d_k$ are distinct divisors of $a x_i$ and their sum is equal to $a x_j$. Therefore, $a x_1 < a x_2 < \dots < a x_n$ are relatively consistent for every positive integer $a$.

Let $b$ be an integer greater than $x_1$, and consider $b x_1$, $(b+1)x_1$, $(b+1)x_2, \dots, (b+1)x_n$. By the observation above, $(b+1)x_1$, $(b+1)x_2, \dots, (b+1)x_n$ are relatively consistent. Note that by the induction hypothesis, for any given $i > 1$ there exist positive divisors of $x_1$, say $d_1, \dots, d_m$, such that $d_1 + d_2 + \dots + d_m = x_i$. Then, because of the choice of $b$, we see that $d_1, d_2, \dots, d_m, b d_1, b d_2, \dots, b d_m$ are distinct divisors of $b x_1$ and their sum is $(b+1)x_i$. Therefore, $b x_i$ and $(b+1)x_i$ are consistent for every $i > 1$. Furthermore, $b x_1$ and $(b+1)x_1$ are consistent as well, since $x_1 + b x_1 = (b+1)x_1$.

Let $a > 1$ be an integer. It is easy to verify that
$$a^{2017},\ a^{2017} + a^{2016},\ a^{2017} + a^{2016} + a^{2015},\ \dots,\ a^{2017} + a^{2016} + \dots + a + 1$$
are pairwise relatively consistent.