2022 USAJMO Problems/Problem 1

We claim that $m$ satisfies the given conditions if and only if $m$ is squareful.

To begin, we let the common difference be $d$ and the common ratio 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 iff 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 $g_i\pmod{m}$ :

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

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

Whee! Restating, $(1),(2)\iff (3),(4)$ , and the conditions $g_{l-1}(r-l)\not\equiv 0\pmod{m}$ and $g_{l-1}(r-1)^2\equiv0\pmod{m}$ hold if and only if $m$ is not squareful.

[will finish that step here]

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2022 USAJMO Problems/Problem 1