2022 USAJMO Problems/Problem 1

Revision as of 22:43, 29 December 2023 by Kscv (talk | contribs) (Solution 1)

Problem

For which positive integers $m$ does there exist an infinite arithmetic sequence of integers $a_1,a_2,\cdots$ and an infinite geometric sequence of integers $g_1,g_2,\cdots$ satisfying the following properties?

$\bullet$ $a_n-g_n$ is divisible by $m$ for all integers $n>1$;

$\bullet$ $a_2-a_1$ is not divisible by $m$.

Solution 1

We claim that $m$ satisfies the given conditions if and only if $m$ is a perfect square.

To begin, we let the common difference of $\{a_n\}$ be $d$ and the common ratio of $\{g_n\}$ be $r$. Then, rewriting the conditions modulo $m$ gives: \[a_2-a_1=d\not\equiv 0\pmod{m}\text{         (1)}\] \[a_n\equiv g_n\pmod{m}\text{             (2)}\]

Condition $(1)$ holds if no consecutive terms in $a_i$ are equivalent modulo $m$, which is the same thing as never having consecutive, equal, terms, in $a_i\pmod{m}$. By Condition $(2)$, this is also the same as never having equal, consecutive, terms in $g_i\pmod{m}$:

\[(1)\iff g_l\not\equiv g_{l-1}\pmod{m}\text{ for any integer }l>1\] \[\iff g_{l-1}(r-1)\not\equiv 0\pmod{m}.\text{        (3)}\]


Also, Condition $(2)$ holds if \[g_{l+1}-g_l\equiv g_l-g_{l-1}\pmod{m}\] \[g_{l-1}(r-1)^2\equiv0\pmod{m}\text{        (4)}.\]

Restating, $(1),(2) \textrm{if} (3),(4)$, and the conditions $g_{l-1}(r-1)\not\equiv 0\pmod{m}$ and $g_{l-1}(r-1)^2\equiv0\pmod{m}$ hold if and only if $m$ is a perfect square.

[will finish that step here]

See Also

2022 USAJMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6
All USAJMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png