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Difference between revisions of "2022 USAJMO Problems/Problem 1"
m (→‎Solution 1)
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<cmath>a_n\equiv g_n\pmod{m}\text{            (2)}</cmath>
 
<cmath>a_n\equiv g_n\pmod{m}\text{            (2)}</cmath>
  
−
Condition <math>(1)</math> holds iff no consecutive terms in <math>a_i</math> are equivalent modulo <math>m</math>, which is the same thing as never having consecutive, equal, terms, in <math>a_i\pmod{m}</math>. By Condition <math>(2)</math>, this is also the same as never having equal, consecutive, terms in <math>g_i\pmod{m}</math>:
+
Condition <math>(1)</math> holds if no consecutive terms in <math>a_i</math> are equivalent modulo <math>m</math>, which is the same thing as never having consecutive, equal, terms, in <math>a_i\pmod{m}</math>. By Condition <math>(2)</math>, this is also the same as never having equal, consecutive, terms in <math>g_i\pmod{m}</math>:
  
 
<cmath>(1)\iff g_l\not\equiv g_{l-1}\pmod{m}\text{ for any integer }l>1</cmath>
 
<cmath>(1)\iff g_l\not\equiv g_{l-1}\pmod{m}\text{ for any integer }l>1</cmath>
Line 20: Line 20:
  
  
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Also, Condition <math>(2)</math> holds iff
+
Also, Condition <math>(2)</math> holds if
 
<cmath>g_{l+1}-g_l\equiv g_l-g_{l-1}\pmod{m}</cmath>
 
<cmath>g_{l+1}-g_l\equiv g_l-g_{l-1}\pmod{m}</cmath>
 
<cmath>g_{l-1}(r-1)^2\equiv0\pmod{m}\text{        (4)}.</cmath>
 
<cmath>g_{l-1}(r-1)^2\equiv0\pmod{m}\text{        (4)}.</cmath>
  
−
Whee! Restating, <math>(1),(2)\iff (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 squareful.
+
Whee! Restating, <math>(1),(2)\if (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 squareful.
  
 
[will finish that step here]
 
[will finish that step here]

Revision as of 19:01, 12 November 2023

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 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 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 \[g_{l+1}-g_l\equiv g_l-g_{l-1}\pmod{m}\] \[g_{l-1}(r-1)^2\equiv0\pmod{m}\text{ (4)}.\]

Whee! Restating, $(1),(2)\if (3),(4)$ (Error compiling LaTeX. Unknown error_msg), 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 squareful.

[will finish that step here]

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

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