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Difference between revisions of "2022 USAJMO Problems/Problem 1"

(→‎Solution 1: why do we have a false solution in the wiki)
(→‎Solution 1: there is no point in having an incorrect solution in the wiki, let's wipe up this mess)
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==Solution 1==
 
==Solution 1==
  
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(This claim below is false, numbers like 12 work, or anything indicated in the previous solution)
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<math>m</math> satisfies the conditions precisely when <math>m</math> is not squarefree.
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We claim that <math>m</math> satisfies the given conditions if and only if <math>m</math> is a perfect square.
 
  
 
To begin, we let the common difference of <math>\{a_n\}</math> be <math>d</math> and the common ratio of <math>\{g_n\}</math> be <math>r</math>. Then, rewriting the conditions modulo <math>m</math> gives:
 
To begin, we let the common difference of <math>\{a_n\}</math> be <math>d</math> and the common ratio of <math>\{g_n\}</math> be <math>r</math>. Then, rewriting the conditions modulo <math>m</math> gives:
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Restating, <math>(1),(2)\quad \textrm{if} \quad(3),(4)</math>, and the conditions <math>g_{l-1}(r-1)\not\equiv 0\pmod{m}</math> and <math>g_{l-1}(r-1)^2\equiv0\pmod{m}</math> hold if and only if <math>m</math> is a perfect square.
 
Restating, <math>(1),(2)\quad \textrm{if} \quad(3),(4)</math>, and the conditions <math>g_{l-1}(r-1)\not\equiv 0\pmod{m}</math> and <math>g_{l-1}(r-1)^2\equiv0\pmod{m}</math> hold if and only if <math>m</math> is a perfect square.
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[will finish that step here]
 
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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 <math>m = at^2</math> then <math>at + 1 \not\equiv 1 \pmod{at^2}</math> or <math>at</math> is not divisble by <math>at^2</math>. If <math>t = 1</math>, this is false and this is not possible. But if it isn't, if <math>m</math> isn't square free, then it should work.
 
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Note: This counter-conjecture is correct. To prove it, it suffices to show that if m is square-free, then (3) and (4) contradict each other. Indeed, if m is square-free, then each prime dividing m only has a power of 1 in the prime factorization, so given (4) that m|<math>x\cdot (r-1)^2</math>, m has at most one of each prime factor of <math>x \cdot (r-1)^2</math>, but then m has at most one of each prime factor of <math>x \cdot (r-1)</math>, so m divides <math>x \cdot (r-1)</math>, contradicting (3).
 
  
 
==See Also==
 
==See Also==
 
{{USAJMO newbox|year=2022|before=First Question|num-a=2}}
 
{{USAJMO newbox|year=2022|before=First Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:56, 9 March 2025

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

$m$ satisfies the conditions precisely when $m$ is not squarefree.

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.

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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