Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 1)
 
(9 intermediate revisions by 4 users not shown)
Line 6: Line 6:
 
<math>\bullet</math> <math>a_2-a_1</math> is not divisible by <math>m</math>.
 
<math>\bullet</math> <math>a_2-a_1</math> is not divisible by <math>m</math>.
  
−
==Solution 1==
+
==Solution==
 +
 
 +
Let the arithmetic sequence be <math>\{ a, a+d, a+2d, \dots \}</math> and the geometric sequence to be <math>\{ g, gr, gr^2, \dots \}</math>. Rewriting the problem based on our new terminology, we want to find all positive integers <math>m</math> such that there exist integers <math>a,d,r</math> with <math>m \nmid d</math> and <math>m|a+(n-1)d-gr^{n-1}</math> for all integers <math>n>1</math>.
 +
 
 +
Note that
 +
<cmath>m | a+nd-gr^n \; (1),</cmath>
 +
<cmath>m | a+(n+1)d-gr^{n+1} \; (2),</cmath>
 +
<cmath>m | a+(n+2)d-gr^{n+2} \; (3),</cmath>
 +
 
 +
for all integers <math>n\ge 1</math>. From (1) and (2), we have <math>m | d-gr^{n+1}+gr^n</math> and from (2) and (3), we have <math>m | d-gr^{n+2}+gr^{n+1}</math>. Reinterpreting both equations,
 +
 
 +
<cmath>m | gr^{n+1}-gr^n-d \; (4),</cmath>
 +
<cmath>m | gr^{n+2}-gr^{n+1}-d \; (5),</cmath>
 +
 
 +
for all integers <math>n\ge 1</math>. Thus, <math>m | gr^k - 2gr^{k+1} + gr^{k+2} = gr^k(r-1)^2 \; (6)</math>. Note that if <math>m|g,r</math>, then <math>m|gr^{n+1}-gr^n</math>, which, plugged into (4), yields <math>m|d</math>, which is invalid. Also, note that (4)<math>+</math>(5) gives
  
−
We claim that <math>m</math> satisfies the given conditions if and only if <math>m</math> is a perfect square.
+
<cmath>m | gr(r-1)(r+1) - 2d \; (7),</cmath>
  
−
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:
+
so if <math>r \equiv \pm 1 \pmod m</math> or <math>gr \equiv 0 \pmod m</math>, then <math>m|d</math>, which is also invalid. Thus, according to (6), <math>m|g(r-1)^2</math>, with <math>m \nmid g,r</math>. Also from (7) is that <math>m \nmid g(r-1)</math>.
−
<cmath>a_2-a_1=d\not\equiv 0\pmod{m}\text{        (1)}</cmath>
 
−
<cmath>a_n\equiv g_n\pmod{m}\text{            (2)}</cmath>
 
  
−
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>:
+
Finally, we can conclude that the only <math>m</math> that will work are numbers in the form of <math>xy^2</math>, other than <math>1</math>, for integers <math>x,y</math> (<math>x</math> and <math>y</math> can be equal), ie. <math>4,8,9,12,16,18,20,24,25,\dots </math>.
  
−
<cmath>(1)\iff g_l\not\equiv g_{l-1}\pmod{m}\text{ for any integer }l>1</cmath>
+
~sml1809
−
<cmath>\iff g_{l-1}(r-1)\not\equiv 0\pmod{m}.\text{        (3)}</cmath>
 
  
 +
==Solution 1==
  
−
Also, Condition <math>(2)</math> holds if
+
<math>m</math> satisfies the conditions precisely when <math>m</math> is not squarefree.
−
<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>
 
  
−
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.
+
Consider three consecutive terms of the arithmetic sequence modulo <math>m</math>: <math>x - d</math>, <math>x</math>, and <math>x + d</math>. To satisfy the given conditions, the arithmetic and geometric sequences must match modulo <math>m</math>. Thus,
 +
<cmath>(x - d)(x + d) \equiv x^2 \pmod{m}</cmath>
 +
which simplifies to
 +
<cmath>x^2 - d^2 \equiv x^2 \pmod{m} \implies d^2 \equiv 0 \pmod{m}.</cmath>
  
−
[will finish that step here]
+
If <math>m</math> is squarefree, the congruence <math>d^2 \equiv 0 \pmod{m}</math> implies <math>m \mid d</math>, so the conditions are not satisfied. Thus, no solution exists for squarefree <math>m</math>, and it remains for us to construct a solution for all other values of <math>m</math> that are not squarefree.
  
−
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.
+
Suppose <math>m</math> is divisible by some prime square <math>p^2</math>. Consider the arithmetic sequence
 +
<cmath>1,\quad 1 + \frac{m}{p},\quad 1 + 2\frac{m}{p},\quad \dots</cmath>
 +
and the geometric sequence
 +
<cmath>1,\quad 1 + \frac{m}{p},\quad \left(1 + \frac{m}{p}\right)^2,\quad \dots</cmath>
 +
For the geometric sequence, the first two terms in the binomial expansion of each entry match the corresponding terms in the arithmetic sequence and subsequent terms are 0 mod m because <math>\frac{m}{p}</math> raised to a power greater than 1 contains enough <math>p</math>'s in its factorization to ensure that it is a multiple of <math>m</math>. Therefore the construction works and as originally stated, solutions exist precisely when <math>m</math> is not squarefree.
  
 
==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}}

Latest revision as of 04:52, 10 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.

Consider three consecutive terms of the arithmetic sequence modulo $m$: $x - d$, $x$, and $x + d$. To satisfy the given conditions, the arithmetic and geometric sequences must match modulo $m$. Thus, \[(x - d)(x + d) \equiv x^2 \pmod{m}\] which simplifies to \[x^2 - d^2 \equiv x^2 \pmod{m} \implies d^2 \equiv 0 \pmod{m}.\]

If $m$ is squarefree, the congruence $d^2 \equiv 0 \pmod{m}$ implies $m \mid d$, so the conditions are not satisfied. Thus, no solution exists for squarefree $m$, and it remains for us to construct a solution for all other values of $m$ that are not squarefree.

Suppose $m$ is divisible by some prime square $p^2$. Consider the arithmetic sequence \[1,\quad 1 + \frac{m}{p},\quad 1 + 2\frac{m}{p},\quad \dots\] and the geometric sequence \[1,\quad 1 + \frac{m}{p},\quad \left(1 + \frac{m}{p}\right)^2,\quad \dots\] For the geometric sequence, the first two terms in the binomial expansion of each entry match the corresponding terms in the arithmetic sequence and subsequent terms are 0 mod m because $\frac{m}{p}$ raised to a power greater than 1 contains enough $p$'s in its factorization to ensure that it is a multiple of $m$. Therefore the construction works and as originally stated, solutions exist precisely when $m$ is not squarefree.

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_USAJMO_Problems/Problem_1&oldid=244618"