2022 USAJMO Problems/Problem 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

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

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, $(1),(2)\quad \textrm{if} \quad(3),(4)$ , 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]

Note: This shouldn't work since we see that m = 12 is a solution. Let the initials for both series by 1, then let the ratio be 7 and the common difference to be 6. We see multiplying by 7 mod 12 that the geometric sequence is alternating from 1 to 7 to 1 to 7 and so on, which is the same as adding 6. Therefore, this solution is wrong. My counter-conjecture is that all non square-free m (4, 8, 9, 16, 18, 25...) should all work, but I don't have a proof. However, if you edit the one above, you can see non square-free m will work. In order to construct a ratio, we could us (4) and find a square multiple of m, take the square root and add 1 to get the ratio. Let $m = at^2$ then $at + 1 \not\equiv 1 \pmod{at^2}$ or $at$ is not divisble by $at^2$ . If $t = 1$ , this is false and this is not possible. But if it isn't, if $m$ isn't square free, then it should work.

2022 USAJMO (Problems • Resources)
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2022 USAJMO Problems/Problem 1

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Solution 1

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