USAJMO

Let the arithmetic sequence be $\{ a, a+d, a+2d, \dots \}$ and the geometric sequence to be $\{ g, gr, gr^2, \dots \}$ . Rewriting the problem based on our new terminology, we want to find all positive integers $m$ such that there exist integers $a,d,r$ with $m \nmid d$ and $m|a+(n-1)d-gr^{n-1}$ for all integers $n>1$ .
Note that $$m | a+nd-gr^n \; (1),$$ $$m | a+(n+1)d-gr^{n+1} \; (2),$$ $$m | a+(n+2)d-gr^{n+2} \; (3),$$
for all integers $n\ge 1$ . From (1) and (2), we have $m | d-gr^{n+1}+gr^n$ and from (2) and (3), we have $m | d-gr^{n+2}+gr^{n+1}$ . Reinterpreting both equations,
$$m | gr^{n+1}-gr^n-d \; (4),$$ $$m | gr^{n+2}-gr^{n+1}-d \; (5),$$
for all integers $n\ge 1$ . Thus, $m | gr^k - 2gr^{k+1} + gr^{k+2} = gr^k(r-1)^2 \; (6)$ . Note that if $m|g,r$ , then $m|gr^{n+1}-gr^n$ , which, plugged into (4), yields $m|d$ , which is invalid. Also, note that (4) $+$ (5) gives
$$m | gr(r-1)(r+1) - 2d \; (7),$$
so if $r \equiv \pm 1 \pmod m$ or $gr \equiv 0 \pmod m$ , then $m|d$ , which is also invalid. Thus, according to (6), $m|g(r-1)^2$ , with $m \nmid g,r$ . Also from (7) is that $m \nmid g(r-1)$ .
Finally, we can conclude that the only $m$ that will work are numbers in the form of $xy^2$ , other than $1$ , for integers $x,y$ ( $x$ and $y$ can be equal), ie. $4,8,9,12,16,18,20,24,25,\dots$ .
~sml1809