Plan ahead for the next school year. Schedule your class today!

G
Topic
First Poster
Last Poster
k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Wednesday, Jul 16 - Oct 29
Sunday, Aug 17 - Dec 14
Tuesday, Aug 26 - Dec 16
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10

Prealgebra 2 Self-Paced

Prealgebra 2
Friday, Jul 25 - Nov 21
Sunday, Aug 17 - Dec 14
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Tuesday, Jul 15 - Oct 28
Sunday, Aug 17 - Dec 14
Wednesday, Aug 27 - Dec 17
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Wednesday, Jul 2 - Sep 17
Sunday, Jul 27 - Oct 19
Monday, Aug 11 - Nov 3
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8

Introduction to Number Theory
Tuesday, Jul 15 - Sep 30
Wednesday, Aug 13 - Oct 29
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Friday, Jul 18 - Nov 14
Thursday, Aug 7 - Nov 20
Monday, Aug 18 - Dec 15
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30

Introduction to Geometry
Monday, Jul 14 - Jan 19
Wednesday, Aug 13 - Feb 11
Tuesday, Aug 26 - Feb 24
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
Friday, Aug 8 - Feb 20
Tuesday, Aug 26 - Feb 24
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4

Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24

Intermediate Number Theory
Wednesday, Sep 24 - Dec 17

Precalculus
Wednesday, Aug 6 - Jan 21
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31

Advanced: Grades 9-12

Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)

MATHCOUNTS/AMC 8 Advanced
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Tuesday, Aug 26 - Nov 11
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)

AMC 10 Problem Series
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 10 - Nov 2
Thursday, Aug 14 - Oct 30
Tuesday, Aug 19 - Nov 4
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 10 Final Fives
Friday, Aug 15 - Sep 12
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30

AMC 12 Problem Series
Wednesday, Aug 6 - Oct 22
Sunday, Aug 10 - Nov 2
Monday, Aug 18 - Nov 10
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28

AIME Problem Series A
Thursday, Oct 23 - Jan 29

AIME Problem Series B
Tuesday, Sep 2 - Nov 18

F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT


Programming

Introduction to Programming with Python
Thursday, Aug 14 - Oct 30
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3

Intermediate Programming with Python
Friday, Oct 3 - Jan 16

USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3

Physics

Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11

Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Jul 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
The Bank of Oslo
mathisreaI   62
N 26 minutes ago by teen.x.ken
Source: IMO 2022 Problem 1
The Bank of Oslo issues two types of coin: aluminum (denoted A) and bronze (denoted B). Marianne has $n$ aluminum coins and $n$ bronze coins arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k \leq 2n$, Gilberty repeatedly performs the following operation: he identifies the longest chain containing the $k^{th}$ coin from the left and moves all coins in that chain to the left end of the row. For example, if $n=4$ and $k=4$, the process starting from the ordering $AABBBABA$ would be $AABBBABA \to BBBAAABA \to AAABBBBA \to BBBBAAAA \to ...$

Find all pairs $(n,k)$ with $1 \leq k \leq 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.
62 replies
mathisreaI
Jul 13, 2022
teen.x.ken
26 minutes ago
A geometry problem
MathMaxGreat   1
N 28 minutes ago by Razorrizelim
Source: 2025 Summer NSMO
In $\triangle ABC$, $D$ is on the angle bisector of $\angle BAC$, $ED\perp BC$ and $E$ is on $BC$.
Prove: the radius of the circumcircles of the pedal triangles for $\triangle ABD$ and $\triangle ACD$ is equal.
1 reply
MathMaxGreat
2 hours ago
Razorrizelim
28 minutes ago
ISI UGB 2025 P8
SomeonecoolLovesMaths   11
N 29 minutes ago by Cats_on_a_computer
Source: ISI UGB 2025 P8
Let $n \geq 2$ and let $a_1 \leq a_2 \leq \cdots \leq a_n$ be positive integers such that $\sum_{i=1}^{n} a_i = \prod_{i=1}^{n} a_i$. Prove that $\sum_{i=1}^{n} a_i \leq 2n$ and determine when equality holds.
11 replies
SomeonecoolLovesMaths
May 11, 2025
Cats_on_a_computer
29 minutes ago
Make a square of rectangles
a_507_bc   5
N 33 minutes ago by quantam13
Source: ARO Regional stage 2024 9.1=10.1=11.1
There are $2024$ rectangles $1 \times n$ for $n=1, 2, \ldots, 2024$. Is it possible to make a square using some of them, such that the side length of the square is greater than $1$?
5 replies
a_507_bc
Feb 2, 2024
quantam13
33 minutes ago
Inspired by JBMO 2025
sqing   2
N 35 minutes ago by sqing
Source: Own
Let $  a, b >0, \frac{a+1}{b} + \frac{b+1}{a} + a+b =7 .$ Prove that
$$\frac{(a^2 + b)^2}{b + 1} + \frac{(b^2 + a)^2}{a + 1} + \frac{(ab + 1)^2}{a + b} \geq \frac{121ab(a + b +1)^2}{50(a + b +ab)}$$Equality holds when $(a,b)=(\frac12,1)$ or $(a,b)=(2,2) . $
2 replies
sqing
6 hours ago
sqing
35 minutes ago
Differences of fractions
SinaQane   2
N 40 minutes ago by quantam13
Source: 239 2019 S1
The following fractions are written on the board $\frac{1}{n}, \frac{2}{n-1}, \frac{3}{n-2}, \ldots , \frac{n}{1}$ where $n$ is a natural number. Vasya calculated the differences of the neighboring fractions in this row and found among them $10000$ fractions of type $\frac{1}{k}$ (with natural $k$). Prove that he can find even $5000$ more of such these differences.
2 replies
SinaQane
Jul 31, 2020
quantam13
40 minutes ago
A square (10 × 10). Find $\min S$
lcbnihhuang   0
an hour ago
A square (10 × 10) is formed by 100 unit squares, which are numbered from 1 to 100 in order from left to right, top to bottom. This square is divided into 50 rectangles, each having an area of 2 square units. Let S be the sum of the products of the two numbers written on each rectangle. Find the minimum possible value of S.
0 replies
lcbnihhuang
an hour ago
0 replies
Inspired by JBMO 2025
sqing   2
N an hour ago by sqing
Source: Own
Let $  a, b >0, \frac{2a+1}{b} + \frac{2b+1}{a} + 4a+4b =14 .$ Prove that
$$\frac{8(2a^2 + b)^2}{2b + 1} + \frac{8(2b^2 + a)^2}{2a + 1} + \frac{(4ab + 1)^2}{a + b} \geq \frac{242ab(2a + 2b +1)^2}{25(a + b +2ab)}$$Equality holds when $(a,b)=(\frac14,\frac12)$ or $(a,b)=(1,1) . $
2 replies
sqing
5 hours ago
sqing
an hour ago
JBMO TST Bosnia and Herzegovina 2025 P3
Rotten_   18
N an hour ago by sqing
Let a, b, c be real numbers such that
\[
a + b + c = 0 \quad \text{and} \quad abc = -16.
\]Find the minimum value of the expression
\[
W = \frac{a^2 + b^2} {c} + \frac{b^2 + c^2} {a} + \frac{c^2 + a^2} {b}.
\]
18 replies
Rotten_
Monday at 9:24 AM
sqing
an hour ago
angles in triangle
AndrewTom   35
N an hour ago by smileapple
Source: BrMO 2012/13 Round 2
The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$.
35 replies
AndrewTom
Feb 1, 2013
smileapple
an hour ago
Computer too strong
Eyed   63
N 2 hours ago by OronSH
Source: 2020 ISL G6
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other
63 replies
Eyed
Jul 20, 2021
OronSH
2 hours ago
sqrt a^6+2abc inequality
Golden_Verse   4
N 2 hours ago by perfect_square
Source: Own (Probably)
$a, ~b, ~c \in \mathbb{R^+}, \quad ab+bc+ca=3$
$$(!)\quad \sqrt{a^6+2abc}+\sqrt{b^6+2abc}+\sqrt{c^6+2abc} \geq \sqrt{3}(a+b+c) $$If it exists I'd appreciate the link.
4 replies
Golden_Verse
Jul 4, 2025
perfect_square
2 hours ago
Number theory
T.N.T   14
N 2 hours ago by Baimukh
Source: Silk way 2017
Prove that for each prime $ P =9k+1$ ,exist natural n such that $P|n^3-3n+1$.
14 replies
1 viewing
T.N.T
Mar 27, 2017
Baimukh
2 hours ago
Inequality
JARP091   4
N 2 hours ago by Victoria_Discalceata1
Source: Own
$x, y, z \ge 0$. Prove that:
\[
\sum \frac{x^2}{51(x+y+z)^2 + 81x^2} \ge \frac{1}{180}
\]
4 replies
JARP091
Yesterday at 7:23 AM
Victoria_Discalceata1
2 hours ago
IMO 2018 Problem 5
orthocentre   85
N Jun 27, 2025 by EVKV
Source: IMO 2018
Let $a_1$, $a_2$, $\ldots$ be an infinite sequence of positive integers. Suppose that there is an integer $N > 1$ such that, for each $n \geq N$, the number
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$$is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \geq M$.

Proposed by Bayarmagnai Gombodorj, Mongolia
85 replies
orthocentre
Jul 10, 2018
EVKV
Jun 27, 2025
IMO 2018 Problem 5
G H J
Source: IMO 2018
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bjump
1070 posts
#83
Y by
Note that for integral $C \ge 0$
$$\frac{a_1}{a_2} + \frac{a_2}{a_3}+ \cdots  +\frac{a_{N+C-1}}{a_{N+C}} + \frac{a_{N+C}}{a_1}  \in \mathbb Z$$$$\frac{a_1}{a_2} + \frac{a_2}{a_3}+ \cdots  +\frac{a_{N+C}}{a_{N+C+1}} + \frac{a_{N+C+1}}{a_1}  \in \mathbb Z$$Subtracting the first expression from the second expression gives:
$$\frac{a_{N+C+1}}{a_1} + \frac{a_{N+C}}{a_{N+C+1}} -\frac{a_{N+C}}{a_1} \in \mathbb Z$$Suppose for some prime $p$, $\nu_p (a_1) = 0$ this implies $\nu_p(a_{N+C}) \ge  \nu_p(a_{N+C+1})$ implying that the sequence is eventually constant. Now if $\nu_p (a_1) \ge 1$ then if $\nu_p(a_1) >\nu_p (a_{N+C+1}) > \nu_p (a_{N+C})$ We have
$$\nu_p (a_{N+C}) - \nu_p (a_{N+C+1}) = \nu_p(a_{N+C+1} - a_{N+C}) -\nu_p (a_1)$$$$\nu_p (a_{N+C}) - \nu_p (a_{N+C+1}) = \nu_p(a_{N+C}) -\nu_p (a_1)$$$$\nu_p (a_{N+C+1}) = \nu_p (a_1)$$If $\nu_p (a_{N+C})= \nu_p (a_1)$ suppose for the sake of contradiction that $\nu_p (a_{N+C+1}) \neq \nu_p (a_1)$, we have:
$$\nu_p (a_{N+C}) - \nu_p ( a_{N+C+1}) = \nu_p(a_{N+C+1}) - \nu_p (a_1)$$$$\nu_p (a_{N+C})+\nu_p (a_1)  = 2 \nu_p(a_{N+C+1}) $$$$\nu_p(a_1)  = \nu_p(a_{N+C+1})$$A contradiction.
If $\nu_p(a_1) > \nu_p (a_{N+C}) > \nu_p (a_{N+C+1})$ we have the middle fraction is an integer and it is impossible for $\nu_p(a_{N+C+1} - a_{N+C}) =\nu_p( a_{N+C+1}) \ge \nu_p(a_1) $ to be true.

Now suppose $\nu_p (a_{N+C}) > \nu_p (a_1)$ we have that
$$\frac{a_{N+C+1}}{a_1}+\frac{a_{N+C}}{a_{N+C+1}} \in \mathbb Z$$If $\nu_p (a_1) < \nu_p( a_{N+C+1})$ we have $\nu_p(a_{N+C+1} \le \nu_p (a_{N+C})$. Otherwise $\nu_p (a_{N+C+1})  \le  \nu_p (a_i)$.

Therefore the sequence will be stuck at a constant with $\nu_p$ less than $\nu_p (a_1)$, $\nu_p(a_1)$ if the sequnce changes at all. It is impossible for the sequence to stay strictly above $\nu_p (a_1)$ due to our last argument. Thus $(a_n)$ is eventually constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
183 posts
#84 • 1 Y
Y by alexanderhamilton124
Easy?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihategeo_1969
269 posts
#85
Y by
No way I still haven't done this.

See that \[\frac{a_{n+1}}{a_1}+\frac{a_n}{a_{n+1}}=b_n \iff a_{n+1}a_1b_n-a_na_1=a_{n+1}(a_{n+1}-a_n) \text{ for large }n \left(\clubsuit \right)\]Say a prime $p \mid a_1$ if it exists (or else $a_{n+1} \mid a_n \implies a_{n+1} \leq a_n$ so by discrete IVT we are done).

See that there finitely many such primes $p$ and say $\nu_p(a_1)=C>0$.

Claim: Either the sequence $(\nu_p(a_n))_{n \geq 1}$ is eventually just $C$ or eventually $\nu_p(a_{n+1}) \leq \nu_p(a_n)$.
Proof: Say $\nu_p(a_{n+1})>\nu_p(a_n)$. See that applying $\nu_p$ in $\clubsuit$ we get \begin{align*}
& C+\nu_p(a_{n+1}b_n-a_n)=\nu_p(a_{n+1})+\nu_p(a_{n+1}-a_n) \iff C+\nu_p(a_n)=\nu_p(a_{n+1})+\nu_p(a_n) \iff \nu_p(a_{n+1})=C
\end{align*}Now see that if $\nu_p(a_{n+2}) \geq \nu_p(a_{n+1})$ then $\nu_p(a_{n+2})=C$. So assume the contrary. Again applying $\nu_p$ in $\clubsuit$ but rearranged and shifting $n \mapsto n+1$; we get \[\nu_p(a_{n+2})+C+\nu_p(b_{n+1})=\nu_p(a_{n+2}^2-a_{n+2}a_{n+1}+a_{n+1}a_1)=2\nu_p(a_{n+2}) \implies \nu_p(a_{n+2}) \geq C\]Which is a contradiction. $\square$

This claim obviously finishes.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathwiz_1207
105 posts
#86
Y by
We will prove that the sequence $\{v_p(a_n)\}$ is constant after a finite number of terms. Note that the condition is equivalent to
\[\frac{a_{n + 1}}{a_n} + \frac{a_n}{a_{n+1}} - \frac{a_n}{a_1} \in \mathbb{Z} \leftrightarrow \frac{(a_{n+1} - a_n)(a_{n + 1} - a_1)}{a_{n + 1}a_1} \in \mathbb{Z}\]Let $n$ be such that $n \geq N$ in what follows.


If $v_p(a_{n+1}) < v_p(a_n)$, we must have $v_p(a_{n+1}) \geq v_p(a_1)$. If $v_p(a_{n + 1}) < v_p(a_1)$, then
\[v_p(a_{n+1} - a_n) + v_p(a_{n + 1} - a_1) - v_p(a_{n+1}) - v_p(a_1) = v_p(a_{n+1}) + v_p(a_{n+1}) - v_p(a_{n+1}) - v_p(a_1) < 0\]a contradiction.


If $v_p(a_{n+1}) > v_p(a_n)$, we must have $v_p(a_{n + 1}) = v_p(a_1)$. If $v_p(a_{n + 1} < v_p(a_1)$, we have
\[v_p(a_{n+1} - a_n) + v_p(a_{n + 1} - a_1) - v_p(a_{n+1}) - v_p(a_1) = v_p(a_n) + v_p(a_{n+1}) - v_p(a_{n + 1}) - v_p(a_1) < 0\]a contradiction. Similarly, if $v_p(a_{n+1}) > v_p(a_1)$, we have
\[v_p(a_{n+1} - a_n) + v_p(a_{n + 1} - a_1) - v_p(a_{n+1}) - v_p(a_1) = v_p(a_n) + v_p(a_1) - v_p(a_{n+1}) - v_p(a_1) < 0\]a contradiction.


If $v_p(a_{n + 1}) = v_p(a_n)$, we must have $v_p(a_{n+1}) \geq v_p(a_1)$. If $v_p(a_{n+1}) < v_p(a_1)$, we have
\[v_p(a_{n+1}^2 - a_na_{n + 1} + a_1a_n) - v_p(a_{n + 1}) - v_p(a_1) = v_p(a_{n+1}) - v_p(a_1) < 0\]a contradiction.


Now, let $b_n = v_p(a_{n})$. Then, $b_n \geq b_1$ for all $n \geq N + 1$, since $b_1 \leq b_{n + 1} < b_n$, $b_n < b_{n + 1} = b_1$ or $b_1 \leq b_{n + 1} = b_n$. This implies that after a finite number of terms, either $\{b_n\}$ is $b_1$ or it is constant, so we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Feb 17, 2025, 9:43 PM
Reason: formatting
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
VideoCake
18 posts
#87
Y by
Solved together with raflikk! :)

Solution. Denote given expression by \(S_n\), and notice that \(S_{n+1} - S_n\) has to be an integer, so
\[S_{n+1} - S_n = \frac{a_{n+1} - a_{n}}{a_{1}} - \frac{a_{n}}{a_{n+1}}\]meaning that \(a_1a_{n+1} \mid a_{n+1}(a_{n+1} - a_{n}) - a_na_1\). This implies \(a_1 \mid a_{n+1}(a_{n+1} - a_{n})\) and \(a_{n+1} \mid a_na_1\). Suppose that a prime \(p\) divides \(a_1\) and \(a_n\). Then,
\[p \mid a_1 \mid a_{n+1}(a_{n+1} - a_n) \implies p \mid a_{n+1}^2\]which means that \(p \mid a_{n+1}\). Thus, \(p \mid a_i\) for all \(i \geq n\). Now we repeat the following step:

Assume that there exists a positive integer \(C\) such that all terms \(a_i\) with \(i \geq C\) are integers, and assume that \(a_1\) is an integer. Pick a prime \(p\) such that \(p \mid a_1\) and \(p \mid a_i\) (with \(i \geq C\)). Since all \(a_j\) with \(j \geq i \geq C\) are integers, we know that \(p \mid a_j\) for all \(j \geq i\). Now we divide every term in the sequence by \(p\). All ratios are still the same. (We allow some terms in the sequence to be non-integers after this step). Note how all \(a_j\) with \(j \geq i\) are still integers, so we pick our new constant \(C\) to be equal to \(i\), and note how \(a_1\) is still an integer.

Eventually, it is not possible to perform the step by picking a prime \(p\), as \(a_1\) only has a finite amount of divisors. Then, \(\gcd(a_1, a_i) = 1\) for all \(i \geq C\). Lastly, this means that for every integer \(n \geq C\), we have:
\[a_{n+1} \mid a_1a_n \implies a_{n+1} \mid a_n \implies a_{n+1} \leq a_n\]We divided all terms in the sequence with the same primes, so \(a_{n+1} \leq a_n\) also holds in the original sequence, so this sequence has to be eventually constant, we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
587 posts
#88
Y by
Note that for all $k \geq N,$ $$\frac{a_{k}}{a_{k+1}}+\frac{a_{k+1}}{a_1} - \frac{a_{k+1}}{a_1} \in \mathbb Z \implies \frac{a_k-a_{k+1}}{a_k}+\frac{a_{k+1}}{a_1} - \frac{a_k}{a_1} \in \mathbb Z.$$Thus $$S_k=(a_{k+1}-a_k) \left( \frac{1}{a_1} - \frac{1}{a_{k+1}} \right) \in \mathbb Z.$$Now fix a prime $p.$ If $v_p(a_N) < v_p(a_{N+1}),$ we have that $$v_p(a_N) = v_p(a_{N+1}-a_N).$$So if $v_p\left( \frac{1}{a_1}\right) \neq v_p \left( \frac{1}{a_{N+1}} \right),$ it follows that $$v_p \left( \frac{1}{a_1} - \frac{1}{a_{N+1}} \right) \leq v_p \left( \frac{1}{a_{N+1}} \right) < v_p \left(\frac{1}{a_N}\right),$$therefore adding this with above yields $v_p(S_k) < 0,$ contradiction.

Thus $v_p\left( \frac{1}{a_1} \right) = v_p \left( \frac{1}{a_{N+1}} \right) \implies v_p(a_1) = v_p(a_{N+1}).$ Now by similar logic to above, $v_p(a_{N+2}) \leq v_p(a_{N+1})$ as it is impossible for $v_p(a_{N+2}) > v_p(a_{N+1})$ and $v_p (a_1) = v_p (a_{N+2})$ to happen at the same time. But if $v_p(a_{N+2}) < v_p(a_{N+1}),$ we see that $$v_p(S_{N+1}) = v_p(a_{N+2}) + v_p \left( \frac{1}{a_1}-\frac{1}{a_{N+2}} \right) = v_p \left(\frac{a_{N+2}}{a_1}-1 \right) < 0,$$contradiction. Thus $v_p(a_{N+2}) = v_p(a_1),$ and applying this yields that the sequence $v_p(a_n)$ is eventually constant.

Now suppose that for the sake of a contradiction $v_p(a_n)$ is not eventually constant, then by above for all $k \geq N$ we have $v_p(a_k) \geq v_p(a_{k+1}).$ But this is a contradiction, as there are a finite number of possible values $v_p(a_k)$ can take, as $a_k$ are positive integers. Since $p$ is not special it follows that $\{ a_n \}$ is eventually constant. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
clarkculus
254 posts
#89 • 1 Y
Y by centslordm
Let the given expression be $f(n)$. Define $g(n)=f(n+1)-f(n)=\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1}$, which is an integer for $n\ge N$. I claim the sequence $\nu_p(a_{N}),\nu_p(a_{N+1}),\dots$ is eventually weakly decreasing via casework.

1) $\nu_p(a_{N+1})>\nu_p(a_N)$: $g(N)=\frac{a_N}{a_{N+1}}+\frac{a_{N+1}}{a_1}-\frac{a_N}{a_1}\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_N}{a_{N+1}}\right)=\nu_p\left(\frac{a_{N+1}-a_N}{a_1}\right)\implies\nu_p(a_N)-\nu_p(a_{N+1})=\nu_p(a_{N+1}-a_N)-\nu_p(a_{1})\]so $\nu_p(a_{N+1})=\nu_p(a_1)$. Now, if $k\ge1$ and $\nu_p(a_{N+k})=\nu_p(a_1)$, $g(N+k)\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_{N+k}}{a_{N+k+1}}+\frac{a_{N+k+1}}{a_1}-\frac{a_{N+k}}{a_1}\right)\ge0\]Since $\nu_p(a_{N+k}/a_1)=0$, if $\nu_p(a_{N+k+1})>\nu_p(a_1)$, then $\nu_p(a_{N+k}/a_{N+k+1})<0$ with the second term an integer, contradiction, and if $\nu_p(a_{N+k+1})<\nu_p(a_1)$, then $\nu_p(a_{N+k+1}/a_1)<0$ with the first term an integer, contradiction. Therefore, $\nu_p(a_{N+k+1})=\nu_p(a_1)$ as well. An inductive argument then shows that $\nu_p(a_1)=\nu_p(a_{N+k})$ for all $k\ge1$.

2) $\nu_p(a_{N+1})<\nu_p(a_N)$: $g(N)\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_{N+1}-a_N}{a_1}\right)\ge0\implies\nu_p(a_{N+1})\ge\nu_p(a_{1})\]For $k\ge1$ and assuming $\nu_p(a_{N+k-1})\ge\nu_p(a_{N+k})\ge\nu_p(a_1)$, $g(N+k)\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_{N+k}}{a_{N+k+1}}+\frac{a_{N+k+1}}{a_1}-\frac{a_{N+k}}{a_1}\right)\ge0\]so either $\nu_p(a_{N+k}/a_{N+k+1})\ge0$ and $\nu_p(a_{N+k+1}/a_1)\ge0$, which implies $\nu_p(a_{N+k})\ge \nu_p(a_{N+k+1})\ge\nu_p(a_1)$, or $\nu_p(a_{N+k}/a_{N+k+1})=\nu_p(a_{N+k+1}/a_1)$, which implies the same thing. By induction, for all $k\ge1$,
\[\nu_p(a_{N+1})\ge\nu_p(a_{N+2})\ge\dots\ge\nu_p(a_{N+k})\ge\nu_p(a_1)\]
3) $\nu_p(a_{N+1})=\nu_p(a_N)$: Trivial, as the decreasingness of the sequence of $\nu_p(a_n)$ is unchanged.

Partition all prime numbers into two sets $P$ and $Q$ such that $p\in P$ iff $\nu_p(a_1)=0$. For all $p\in P$, $g(N)\in\mathbb{Z}$ implies $\nu_p(a_N/a_{N+1})\ge0$, so $\nu_p(a_N)\ge\nu_p(a_{N+1})$, and a similar induction shows $\nu_p(a_n)$ is weakly decreasing for $n\ge N$. Also, by our claim, for each $q\in Q$, there exists a constant $N_q\ge N$ for which $\nu_q(a_n)$ is weakly decreasing for $n\ge N_q$. Since $Q$ is finite (as $a_1$ has a finite number of prime divisors), the union of $\{N\}$ and the set of all the $N_q$ is finite and thus has a maximal element $M$. Hence, for all primes $p$, $\nu_p(a_n)$ is weakly decreasing for $n\ge M$. This implies the sequence $\nu_p(a_n)$ is eventually constant for all primes $p$, so the sequence $a_n$ is eventually constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
chenghaohu
103 posts
#90
Y by
Let $S_n = \frac{a_1}{a_2} + .... + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$. Consider the difference between $S_{n+1}$ and $S_n$, for $n \ge N$. We will show that the fact that this difference, $\frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}}$ must be an integer for all such $n \ge N$ will force $v_p (a_k) = v_p(a_{k+1})$ for all primes $p$ at sufficiently large $k\ge N$.

Consider the following cases:

Case 1: If $v_p(a_n) > v_p(a_{n+1})$.

If $v_p(a_n) > v_p(a_{n+1})$, then for $\frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}}$ to be an integer, it is necessary that $v_p(a_{n+1}) \ge v_p(a_1).$ Thus considering $\frac{a_{n+2} - a_{n+1}}{a_1} + \frac{a_{n+1}}{a_{n+2}}$, we see that $v_p(a_{n+1}) \ge v_p(a_{n+2}) \ge v_p(a_1)$ resulting from $v_p(a_{n+2}) - v_p(a_1) = v_p(a_{n+1}) - v_p(a_{n+2})$. Continuing this iteration, we see that $v_p(a_{n+k})$ are all equal to some number between $v_p(a_{n+1})$ and $v_p(a_1)$, inclusive at the end, so this case indeed provides a constant value for $a_m$ where $m$ is a sufficiently large positive integer.

Case 2: If $v_p(a_n) < v_p(a_{n+1})$

Since $\frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}}$ is an integer, it is necessary that $v_p(a_{n+1} - a_n) - v_p(a_1) = v_p(a_n) - v_p(a_{n+1})$, forcing that $v_p(a_1) = v_p(n+1)$.
Now we can use an inductive process to show that if $v_p(a_z) = v_p(a_1),$ then $v_p(a_{z+1}) = v_p(a_1)$.

Base case: $z = n+1$. This works because it is necessary that $v_p(a_{n+2}) - v_p(a_1) = v_p(a_{n+1}) - v_p(a_{n+2})$ for $\frac{a_{n+2} - a_{n+1}}{a_1} + \frac{a_{n+1}}{a_{n+2}}$ to be an integer. Inductive step is the same idea as the base case.

Case 3: If $v_p(a_n) = v_p(a_{n+1})$.

$v_p(a_k) = v_p(a_{k+1})$ for all $k\ge n$, then we are good. Else it becomes one of Case $1$ or Case $2$, which resolves successfully. Thus this case also provides a constant value for $a_m$ where $m$ is a sufficiently large positive integer.

Since all $3$ possible cases works, we are done with the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
732 posts
#91
Y by
We assume the contrary and work towards a contradiction. First note that, for all $n \ge N$, both
\[\frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots+\frac{a_n}{a_1} \text{ and } \frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots + \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}\]are positive integers. Hence their difference,
\[\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1} - \frac{a_n}{a_1} = \frac{a_na_1+a_{n+1}^2 - a_na_{n+1}}{a_{n+1}a_1}\]is also a positive integer. Hence,
\[\frac{a_na_1+a_{n+1}^2 - a_na_{n+1}}{a_{n+1}a_1} - 1 = \frac{a_na_1+a_{n+1}^2-a_na_{n+1}-a_{n+1}a_1}{a_{n+1}a_1} = \frac{(a_{n+1}-a_1)(a_{n+1}-a_n)}{a_{n+1}a_1}\]is a non-negative integer. We can now show our key claim.

Claim : There exists a positive integer $M$ such that for all $n \ge M$ we have $\nu_p(a_n) \le \nu_p(a_1)$ or $\nu_p(a_n)$ is constant for all primes $p$.

Proof : Since the sequence $(a_i)$ is not eventually constant by assumption, consider any prime $p$ for which $\nu_p(a_i)$ is also not eventually constant. Consider the sequence $\nu_p(a_i)$. Since this is a sequence of non-negative integers which is not eventually constant it cannot be non-increasing. Consider a term $a_{n+1}$ ($n+1>N$ and $a_{n+1} \not \in \{a_1,a_n\}$) such that $\nu_p(a_{n+1})>\nu_p(a_n)$. Then, since
\[\frac{(a_{n+1}-a_1)(a_{n+1}-a_n)}{a_{n+1}a_1}\]is a non-negative integer,
\begin{align*}
    \nu_p(a_{n+1}-a_1) + \nu_p(a_{n+1}-a_n) &\ge \nu_p(a_1)+\nu_p(a_{n+1})\\
    \nu_p(a_{n+1}-a_1)+\nu_p(a_n) & \ge \nu_p(a_1)+\nu_p(a_{n+1})
    \end{align*}Now if $\nu_p(a_{n+1})>\nu_p(a_1)$ we would have,
\[\nu_p(a_1)+\nu_p(a_n) \ge \nu_p(a_1) + \nu_p(a_{n+1})\]which contradicts the condition that $\nu_p(a_{n_1})>\nu_p(a_n)$. Thus, $\nu_p(a_{n+1}) \le \nu_p(a_1)$ for all such terms $a_{n+1}$.

Even if the sequence $\nu_p(a_i) > \nu_p(a_1)$ for some $i > N$ this implies that this sequence may not increase beyond this point. Hence, it can only increase after reaching a value below $\nu_p(a_1)$ beyond which point $\nu_p(a_i) \le \nu_p(a_1)$ which implies the claim.

However, further note that the sequence $\nu_p(a_i)$ may not eventually be non-decreasing since the sequence is not eventually constant and is bounded above by $\nu_p(a_1)$. Thus, we can consider a term $a_m$ such that $m>M$ but $\nu_p(a_{m+1}) < \nu_p(a_m)$. But now note that,
\begin{align*}
    \nu_p(a_{m+1}-a_1)+\nu_p(a_{m+1}-a_m) & \ge \nu_p(a_1)+\nu_p(a_{m+1})\\
    \nu_p(a_{m+1}) + \nu_p(a_{m+1}) & \ge \nu_p(a_1) + \nu_p(a_{m+1})
\end{align*}which is a clear contradiction since this implies $\nu_p(a_m) > \nu_p(a_{m+1}) \ge \nu_p(a_1)$ violating the previous claim. But this means that the sequence $\nu_p(a_i)$ does not decrease beyond a certain point which is a contradiction. Hence, the only possibility is for $\nu_p(a_i)$ to be eventually constant for all primes $p$ which implies that $(a_i)$ is indeed eventually constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1774 posts
#92
Y by
The condition gives $\left\{\frac{a_i}{a_{i+1}}\right\}=\left\{\frac{a_i-a_{i+1}}{a_1}\right\}$. If $\nu_p(a_n)<\nu_p(a_{n+1})$, we have $\nu_p(a_1)=\nu_p(a_{n+1})$ from $i=n$. Then $i=n+1$ gives $\nu_p(a_{n+2})=\nu_p(a_1)$, and so on to get $\nu_p$ becomes constant. Thus if a prime has $\nu_p$ increase, it happens only once, and it must divide $a_1$, so it happens finitely often. Thus eventually all $\nu_p$s are nonincreasing, and can also only decrease finitely often since finitely many of them are nonzero. Thus they are all eventually constant, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Kempu33334
752 posts
#93
Y by
We will show that for any prime $p$, the sequence $v_p(a)$ is eventually constant. Let us denote \[S_n = \frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots
  + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}. \]Then, we know that \[S_{n+1}-S_n = \dfrac{a_n}{a_{n+1}}+\dfrac{a_{n+1}-a_n}{a_1} \in \mathbb{Z}.\]In other words, \[v_p(S_{n+1}-S_n) = v_p\left(\dfrac{a_n}{a_{n+1}}+\dfrac{a_{n+1}-a_n}{a_1}\right) \ge 0.\]Now we make the following claims:

Claim: If $v_p(a_n) > v_p(a_{n+1})$, then $v_p(a_1) \le v_p(a_{n+1})$.

Proof: For the sake of contradiction, assume that $v_p(a_1) > v_p(a_{n+1})$ is possible. Then, we know that $v_p\left(\dfrac{a_n}{a_{n+1}}+\dfrac{a_{n+1}-a_n}{a_1}\right) < 0$, contradiction. $\square$

Claim: If $v_p(a_n) < v_p(a_{n+1})$, then $v_p(a_1) = v_p(a_{n+1})$.

Proof: We note that $v_p\left(\dfrac{a_n}{a_{n+1}}\right) < 0$, so in order for the overall sum to have a valuation of at least $0$, we need the $v_p$ of the denominators to be the same, in order to combine. $\square$

Now, take some $i$ such that $v_p(a_1) = v_p(a_i)$.
  • If $v_p(a_i) < v_p(a_{i+1})$, then $v_p(a_1) = v_p(a_i) = v_p(a_{i+1})$, contradiction.
  • If $v_p(a_i) > v_p(a_{i+1})$, then $v_p(a_i) \le v_p(a_{i+1}) < v_p(a_i)$, which cannot be true.
Thus, we must have $v_p(a_k) = v_p(a_i)$ for all $k \ge i$, which is the desired conclusion. This implies that if there exists any $i$ such that $v_p(a_1) = v_p(a_i)$, the sequence $a$ is eventually constant. We can in fact take this further --- if there exists any $i$ such that $v_p(a_i) < v_p(a_{i+1})$, then we reach the same conclusion as well.

Now, all there is to show is if the sequence $v_p(a)$ is non-increasing. For some sufficiently large $n$, the sequence will eventually reach zero and stay constant (because if it went negative, it wouldn't be integer), or become constant at a different value, which is the desired result.

Finally, we know that this will only be applied to finitely many primes $p$, and since $a_{n+1}\mid a_n$, the sequence will eventually be constant. $\square$
This post has been edited 1 time. Last edited by Kempu33334, Jun 10, 2025, 1:57 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cubres
133 posts
#94
Y by
$\mathfrak{The \;Fourteenth\; of\; June,\; 2025}$ʕ•ᴥ•ʔ
Storage - grinding IMO problems
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lpieleanu
3093 posts
#95
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mariairam
14 posts
#96
Y by
Let $E(n)= \frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots
  + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$.
$E(n+1)-E(n)=\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1}\in \mathbb{Z}$.
Fix a prime $p$. Then $T=\nu_p(a_na_1+a_{n+1}(a_{n+1}-a_n))\ge \nu_p(a_{n+1})+\nu_p(a_1)$.
Let $S=\min(\nu_p(a_n)+\nu_p(a_1);\nu_p(a_{n+1})+\nu_p(a_{n+1}-a_n))$.
$\boldsymbol{Claim}$: If $n\ge N$, then either $\nu_p(a_n)\ge \nu_p(a_{n+1})$
or $\nu_p(a_{n+1})=\nu_p(a_1)$.
If $S=T$, then $3\nu_p(a_1)+3\nu_p(a_{n+1})<
  3\nu_p(a_{n+1})+2\nu_p(a_n)+\nu_p(a_1)\implies
  \nu_p(a_1)<\nu_p(a_n)$.
Assume FTSOC that $\nu_p(a_{n+1})>\nu_p(a_n)$.
Then $S=\nu_p(a_n)+\nu_p(a_1)\ge \nu_p(a_{n+1})+\nu_p(a_1)$, which is false.
So in this case $\nu_p(a_n)\ge\nu_p(a_{n+1})$.
If, however, $S\neq T$ and $\nu_p(a_{n+1})>\nu_p(a_n)$, then
$\nu_p(a_n)+\nu_p(a_1)=\nu_p(a_{n+1})+\nu_p(a_{n+1}-a_n)
  \implies \nu_p(a_1)=\nu_p(a_{n+1})$.
The claim has been proved.
Now, if there is no $n\ge N$, such that $\nu_p(a_n)=\nu_p(a_1)$, the claim
quite clearly implies that the sequence $(\nu_p(a_n))$ turns constant at some point.
Else, let $\nu_p(a_n)=\nu_p(a_1)$, hence $\nu_p(a_m)\le\nu_p(a_1)\forall m\ge n$.
Assume FTSOC that $(\nu_p(a_m))_{m\ge n}$ is not constant,
implying that there is $m$ such that $\nu_p(a_m)=\nu_p(a_1)=k$
and $\nu_p(a_{m+1})=t<k$.
Let $a_i=p^{\nu_p(a_i)}a_i'$, where $\gcd(a_i',p)\neq 1$.
Since $\frac{a_{n+1}^2-a_na_{n+1}+a_na_1}{a_{n+1}a_1}\in\mathbb{Z}$,
then $p^{k+t}|p^{2t}a_{n+1}'^2-p^{k+t}a_n'a_{n+1}'+p^{2k}a_n'a_1'$,
which is definitely false.
So , for each prime, the sequence $(\nu_p(a_n))$ becomes constant eventually,
so the sequence $(a_n)$ becomes constant in the end.

$\boldsymbol{!Important}$ $\boldsymbol{note!}$: I kind of forgot about the ''caveat''.
As noted before, the way to fix that detail is to notice that
the bounds we've found point to the fact that there is a number that is divisible with all the numbers in the sequence from a certain point onward.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EVKV
76 posts
#97 • 1 Y
Y by Nobitasolvesproblems1979
Here is a quick rundown of my solution
for all $n \geq N+1$ the primes dividing $a_{n}$ divide either $a_{N}$ or $a_{1}$ which is a finite set of primes
Now denote $v_{p}(a_{n})$ as $x_{n}$
Now if $x_{n+1}$ > $x_{n}$
Claim: $x_{i}$ = $x_{1}$ for all $i \geq n+1 $
Proof:
Now for the sake of simplicity
$x_{1}$=a
$x_{n}$= b
$x_{n+1}$=c
$x_{n+2}$ =d
As we know
$\frac{a_{n}a_{1} +(a_{n+1})^{2} -a_{n}a_{n+1}}{a_{1}a_{n+1}}$ is an integer
$v_{p}(a_{n}a_{1})$ = a+b
$v_{p}(a_{n+1}^{2})$ =2c
$v_{p}(a_{n}a_{n+1})$ = b+c
$v_{p}(a_{n+1}a_{1})$ = a+c
Case 1: c>a
a+b < 2c
a+b<b+c
So a+c $ \leq$ a+b contradiction

Case 2: c<a
b+c< 2c
b+c < a+b
So 2c< a+c $ \leq$ b+c contradiction
So a=c

$\frac{a_{n+1}a_{1} +(a_{n+2})^{2} -a_{n+1}a_{n+2}}{a_{1}a_{n+2}}$ is an integer
$v_{p}(a_{n+1}a_{1})$ = 2a
$v_{p}(a_{n+2}^{2})$ =2d
$v_{p}(a_{n+2}a_{n+1})$ = a+d
$v_{p}(a_{n+2}a_{1})$ = a+d
Case 1: d>a
2a < 2d
2a<a+d
So a+d $ \leq$ 2a contradiction

Case 2: d<a
2d< 2a
2d < a+d
So a+d $ \leq$ 2d contradiction
So a=d
Now by induction u will be done

Now If $x_{n}$>$x_{n+1}$
Either it will form a decreasing sequence $x_{n} \geq x_{n+1} \geq \cdots $
Which as it can't decrease infinitely will eventually become constant
Or at some point k $x_{k+1}$>$x_{k}$ which will lead to $x_{k+1}$ =$x_{k+2 }$= $ \cdots$ = a (by earlier claim )
Keep in mind this point k only exists when a $ \neq$ 0 for the prime
So after some time for all primes which can divide $a_{n}$ $x_{n}$=$x_{1}$ or $x_{n}$= some constant
So for all possible prime divisors their powers are fixed so $a_{n}$ is constant
This post has been edited 5 times. Last edited by EVKV, Jun 29, 2025, 1:34 PM
Z K Y
N Quick Reply
G
H
=
a