We have your learning goals covered with Spring and Summer courses available. Enroll today!

G
Topic
First Poster
Last Poster
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21


Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Sunday, Mar 23 - Aug 3
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Friday, Apr 11 - Jun 27
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21
Tuesday, Jun 10 - Aug 26

Calculus
Sunday, Mar 30 - Oct 5
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Friday, Apr 11 - Jun 27
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

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
Monday, Mar 24 - Jun 16
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Sunday, Mar 30 - Jun 22
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Tuesday, Mar 25 - Sep 2
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
Mar 2, 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
Sharygin 2025 CR P17
Gengar_in_Galar   5
N 4 minutes ago by HVP
Source: Sharygin 2025
Let $O$, $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$; $D$, $E$, $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$, $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$, then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$.
Proposed by: P.Puchkov,E.Utkin
5 replies
Gengar_in_Galar
Mar 10, 2025
HVP
4 minutes ago
Not actually combo
MathSaiyan   0
4 minutes ago
Source: PErA 2025/5
We have an $n \times n$ board, filled with $n$ rectangles aligned to the grid. The $n$ rectangles cover all the board and are never superposed. Find, in terms of $n$, the smallest value the sum of the $n$ diagonals of the rectangles can take.
0 replies
MathSaiyan
4 minutes ago
0 replies
Sharygin 2025 CR P13
Gengar_in_Galar   5
N 6 minutes ago by HVP
Source: Sharygin 2025
Each two opposite sides of a convex $2n$-gon are parallel. (Two sides are opposite if one passes $n-1$ other sides moving from one side to another along the borderline of the $2n$-gon.) The pair of opposite sides is called regular if there exists a common perpendicular to them such that its endpoints lie on the sides and not on their extensions. Which is the minimal possible number of regular pairs?
Proposed by: B.Frenkin
5 replies
Gengar_in_Galar
Mar 10, 2025
HVP
6 minutes ago
Nice, simple geo
MathSaiyan   0
7 minutes ago
Source: PErA 2025/4
Let \( ABC \) be an acute-angled scalene triangle. Let \( B_1 \) and \( B_2 \) be points on the rays \( BC \) and \( BA \), respectively, such that \( BB_1 = BB_2 = AC \). Similarly, let \( C_1 \) and \( C_2 \) be points on the rays \( CB \) and \( CA \), respectively, such that \( CC_1 = CC_2 = AB \). Prove that if \( B_1B_2 \) and \( C_1C_2 \) intersect at \( K \), then \( AK \) is parallel to \( BC \).
0 replies
MathSaiyan
7 minutes ago
0 replies
The last nonzero digit of factorials
Tintarn   1
N 8 minutes ago by AshAuktober
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
1 reply
Tintarn
2 hours ago
AshAuktober
8 minutes ago
Stronger than Iran 96, not completely symmetric
MeoMayBe   2
N 9 minutes ago by CHESSR1DER
Source: Own
Let a, b, c\geq 0. Prove that
(bc+ca+ab)\left[\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}+\frac{1}{(a+b)^{2}}\right]\geq\frac{9}{4}+\frac{2abc(b-c)^{2}}{(a+b)^{2}(a+c)^{2}(b+c)}.
PS. 2 is also the best constant.
PS (2). Sorry again, I cannot type LaTeX since new members cannot share image.
2 replies
MeoMayBe
May 14, 2023
CHESSR1DER
9 minutes ago
Equilateral triangle geo
MathSaiyan   0
10 minutes ago
Source: PErA 2025/3
Let \( ABC \) be an equilateral triangle with circumcenter \( O \). Let \( X \) and \( Y \) be two points on segments \( AB \) and \( AC \), respectively, such that \( \angle XOY = 60^\circ \). If \( T \) is the reflection of \( O \) with respect to line \( XY \), prove that lines \( BT \) and \( OY \) are parallel.
0 replies
MathSaiyan
10 minutes ago
0 replies
Inspired by youthdoo
sqing   1
N 11 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $      \frac{3}{a^2+6}+\frac{1}{b^2+2}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $   \frac{2}{a^2+4}+\frac{3}{b^2+6}+\frac{2}{c^2+4}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $    \frac{3}{a^2+6}+\frac{2}{b^2+4}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 8$$Let $ a,b,c $ be real numbers such that $  \frac{3}{a^2+6}+\frac{4}{b^2+8}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 10$$
1 reply
sqing
25 minutes ago
sqing
11 minutes ago
Clean number theory
MathSaiyan   0
13 minutes ago
Source: PErA 2025/2
Let $m$ be a positive integer. We say that a positive integer $x$ is $m$-good if $a^m$ divides $x$ for some integer $a > 1$. We say a positive integer $x$ is $m$-bad if it is not $m$-good.
(a) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-bad positive integers?
(b) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-good positive integers?
0 replies
MathSaiyan
13 minutes ago
0 replies
Many-solutions combogeo
MathSaiyan   1
N 14 minutes ago by Speedysolver1
Source: PErA 2025/1
Let $S$ be a set of at least three points of the plane in general position. Prove that there exists a non-intersecting polygon whose vertices are exactly the points of $S$.
1 reply
MathSaiyan
17 minutes ago
Speedysolver1
14 minutes ago
2 var inquality
sqing   6
N 18 minutes ago by ionbursuc
Source: Own
Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{3}{2}. $ Show that$$ a+b+ab\geq1$$Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{5}{6}. $ Show that$$ a+b+ab\geq2$$
6 replies
sqing
Today at 4:06 AM
ionbursuc
18 minutes ago
Sharygin 2025 CR P2
Gengar_in_Galar   4
N 26 minutes ago by FKcosX
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
4 replies
Gengar_in_Galar
Mar 10, 2025
FKcosX
26 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 31 minutes ago by John_Mgr
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
32 replies
Asjmaj
Dec 31, 2024
John_Mgr
31 minutes ago
Where is the equality?
AndreiVila   2
N an hour ago by MihaiT
Source: Romanian District Olympiad 2025 9.3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2 replies
AndreiVila
Mar 8, 2025
MihaiT
an hour ago
One secuence satisfying condition
hatchguy   8
N Today at 1:59 AM by jaescl
Prove that there exists only one infinite secuence of positive integers $a_1,a_2,...$ with $a_1=1$, $a_2>1$ and $a_{n+1}^3 + 1 = a_na_{n+2}$ for all positive integers $n$.
8 replies
hatchguy
Sep 4, 2011
jaescl
Today at 1:59 AM
One secuence satisfying condition
G H J
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hatchguy
555 posts
#1 • 2 Y
Y by Adventure10, cubres
Prove that there exists only one infinite secuence of positive integers $a_1,a_2,...$ with $a_1=1$, $a_2>1$ and $a_{n+1}^3 + 1 = a_na_{n+2}$ for all positive integers $n$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZetaSelberg
138 posts
#2 • 3 Y
Y by Adventure10, Mango247, cubres
I have an (little) approach

Let's suppouse that $a_2=2$ then se sequence is $1, 2, 5, 13, 34...$ and the definition of this sequence is

$a_1=1$
$a_n=\displaystyle a_{n-1}+\sum_{i=1}^{n-1}a_i$

I could not prove that it is the sequence (neither that it is the only sequence) but at least it works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#3 • 5 Y
Y by Adventure10, Mango247, cubres, and 2 other users
ZetaSelberg, your sequence is wrong. Is it possible that you have misread 3rd power as 2nd power?

About the problem... First it is clear that $a_2$ can only be $2$; otherwise $a_4\notin\mathbb Z$.

The interesting part is to prove that all members of the sequence are integers if $a_2=2$. The sequence is A003818 in Sloane. Armed with this keyword, we get two topics:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=287631
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=299591

No proof, though. Why not give one?

Theorem 1. Let $\left(b_1,b_2,b_3,...\right)$ be a sequence of positive rational numbers defined recursively by

$b_1=1$, $b_2=2$ and $b_{n+1}=\frac{b_n^3+1}{b_{n-1}}$ for every integer $n\geq 2$.

Then, $b_n$ is a positive integer for every positive integer $n$.


Proof of Theorem 1. We are going to show that every integer $N\geq 3$ satisfies the following assertion:

(1) The numbers $b_1$, $b_2$, ..., $b_{N+1}$ are positive integers, and $b_{N-1}$ is coprime to $b_N$.

In fact, we will prove (1) by induction over $N$:

The induction base ($N=3$) is very easy (just notice that the recurrent definition of $\left(b_1,b_2,b_3,...\right)$ yields $b_3=\frac{b_2^3+1}{b_1}=\frac{2^3+1}{1}=9$ and subsequently $b_4=\frac{b_3^3+1}{b_2}=\frac{9^3+1}{2}=365$).

Now for the induction step. Let $n\geq 3$ be an integer. Assume that (1) holds for $N=n$. Now let us prove that (1) holds for $N=n+1$.

Since (1) holds for $N=n$, we know that the numbers $b_1$, $b_2$, ..., $b_{n+1}$ are positive integers, and that $b_{n-1}$ is coprime to $b_n$.

Now, the recurrent definition of $\left(b_1,b_2,b_3,...\right)$ yields $b_{n+2}=\frac{b_{n+1}^3+1}{b_n}$, $b_{n+1}=\frac{b_n^3+1}{b_{n-1}}$ and $b_n=\frac{b_{n-1}^3+1}{b_{n-2}}$. Since $\frac{b_{n-1}^3+1}{b_{n-2}}=b_n$, we have $b_{n-1}^3+1 = b_nb_{n-2}\equiv 0\mod b_n$.

Since $b_{n-1}$ is coprime to $b_n$, we see that $b_{n-1}^3$ is coprime to $b_n$.

Substituting $b_{n+1}=\frac{b_n^3+1}{b_{n-1}}$ in $b_{n+2}=\frac{b_{n+1}^3+1}{b_n}$, we obtain

(2) $b_{n+2}=\frac{\left(\frac{b_n^3+1}{b_{n-1}}\right)^3+1}{b_n}=\frac{\left(b_n^3+1\right)^3+b_{n-1}^3}{b_nb_{n-1}^3}$.

Now, $b_n\equiv 0\mod b_n$, so that $\left(b_n^3+1\right)^3+b_{n-1}^3\equiv \left(0^3+1\right)^3+b_{n-1}^3=1+b_{n-1}^3=b_{n-1}^3+1\equiv 0\mod b_n$. In other words, $b_n\mid \left(b_n^3+1\right)^3+b_{n-1}^3$. On the other hand, $\frac{b_n^3+1}{b_{n-1}}=b_{n+1}$ is an integer, so that $b_{n-1}\mid b_n^3+1$ and thus $b_{n-1}^3\mid \left(b_n^3+1\right)^3$. Hence, $b_{n-1}^3\mid \left(b_n^3+1\right)^3+b_{n-1}^3$ as well (because clearly $b_{n-1}^3\mid b_{n-1}^3$). So we now know that the number $\left(b_n^3+1\right)^3+b_{n-1}^3$ is divisible by both of $b_{n-1}^3$ and $b_n$. Since $b_{n-1}^3$ is coprime to $b_n$, this yields that the number $\left(b_n^3+1\right)^3+b_{n-1}^3$ is also divisible by the product $b_nb_{n-1}^3$. Hence, $\frac{\left(b_n^3+1\right)^3+b_{n-1}^3}{b_nb_{n-1}^3}$ is an integer. Due to (2), this shows that $b_{n+2}$ is an integer. Thus, $b_{n+2}$ is a positive integer (since clearly $b_{n+2}>0$). Combined with the fact that $b_1$, $b_2$, ..., $b_{n+1}$ are positive integers, this yields that $b_1$, $b_2$, ..., $b_{n+2}$ are positive integers.

Besides, $b_{n+1}=\frac{b_n^3+1}{b_{n-1}}$, so that $b_{n-1}b_{n+1}=b_n^3+1$ and thus $1=b_{n-1}b_{n+1}-b_n^3$. Hence, every common divisor of $b_n$ and $b_{n+1}$ must also divide $1$. In other words, $b_n$ and $b_{n+1}$ are coprime. So we have shown that $b_n$ is coprime to $b_{n+1}$.

We have thus shown that $b_1$, $b_2$, ..., $b_{n+2}$ are positive integers, and that $b_n$ is coprime to $b_{n+1}$. In other words, (1) holds for $N=n+1$. This completes the induction step.

Thus, by induction, (1) is proven for all $N\in\mathbb N$. The claim of Theorem 1 now immediately follows.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZetaSelberg
138 posts
#4 • 2 Y
Y by Adventure10, cubres
darij grinberg wrote:
ZetaSelberg, your sequence is wrong. Is it possible that you have misread 3rd power as 2nd power?


Yeess! Now I had to go an visit an optometrist because it is the third time I have this problem.

Thanks for the solution.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jaescl
11 posts
#5 • 4 Y
Y by Adventure10, Mango247, cubres, and 1 other user
Considering the above sequence, prove or disprove whether the following expressions take integer values for every positive integer $n$:
  1. $\frac{b_{n}+b_{n+4}}{b_{n+2}}$
  2. $\frac{b_{n}(b_{n+3}+1)}{b_{n+1}+1}$

I have no solution yet.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#6 • 2 Y
Y by Adventure10, cubres
Very nice observations.

Theorem 2. Let $\left(  b_{1},b_{2},b_{3},...\right)  $ be the sequence of integers defined in Theorem 1 (in post #3 above).

(a) Every integer $n\geq3$ satisfies $b_{n}\mid b_{n-2}+b_{n+2}$.

(b) Every integer $n\geq2$ satisfies $b_{n}+1\mid b_{n-1}\left( b_{n+2}+1\right)  $.


Proof of Theorem 2. From Theorem 1, we know that $b_{n}$ is a positive integer for every positive integer $n$.

From the auxiliary result (1) that we proved during our proof of Theorem 1, we know that

(3) $b_{N-1}$ is coprime to $b_{N}$ for every integer $N\geq3$.

(a) Let $n\geq3$ be an integer. From the auxiliary result (2) that we proved during our proof of Theorem 1, we know that

$b_{n+2}=\dfrac{\left(  b_{n}^3+1\right)  ^{3}+b_{n-1}^{3}}{b_{n}b_{n-1}^{3}}$.

Thus,

$b_{n+2}b_{n}b_{n-1}^{3}=\left(  b_{n}^3+1\right)  ^{3}+b_{n-1}^{3}$.

Since $\left(  b_{n}^3+1\right)  ^{3}\equiv1\mod b_{n}^{2}$ (this congruence even holds modulo $b_{n}^{3}$), this yields

$b_{n+2}b_{n}b_{n-1}^{3}\equiv1+b_{n-1}^{3}=b_{n-1}^{3}+1\mod  b_{n}^{2}$.

But the recurrent definition of $\left(  b_{1},b_{2},b_{3},...\right)  $ yields $b_{n}=\dfrac{b_{n-1}^{3}+1}{b_{n-2}}$, so that $b_{n-1}^{3} +1=b_{n}b_{n-2}$. Thus,

$b_{n+2}b_{n}b_{n-1}^{3}\equiv b_{n-1}^{3}+1=b_{n}b_{n-2}\mod  b_{n}^{2}$.

Now,

$\left(  b_{n+2}+b_{n-2}\right)  b_{n}b_{n-1}^{3}=\underbrace{b_{n+2}b_{n}b_{n-1}^{3}}_{\equiv b_{n}b_{n-2}\mod b_{n}^{2}} +b_{n-2}b_{n}b_{n-1}^{3}$

$\equiv b_{n}b_{n-2}+b_{n-2}b_{n}b_{n-1}^{3}=b_{n}b_{n-2}\underbrace{\left( b_{n-1}^{3}+1\right)  }_{=b_{n}b_{n-2}}=b_{n}b_{n-2}b_{n}b_{n-2}=b_{n}^{2}b_{n-2}^{2}$

$\equiv 0\mod b_{n}^{2}$.

In other words, $b_{n}^{2}\mid\left(  b_{n+2}+b_{n-2}\right)  b_{n}b_{n-1}^{3}$. Cancelling one $b_{n}$ out of this divisibility, we get $b_{n}\mid\left(  b_{n+2}+b_{n-2}\right)  b_{n-1}^{3}$. Since $b_{n-1}^{3}$ is coprime to $b_{n}$ (because (3) (applied to $N=n$) yields that $b_{n-1}$ is coprime to $b_{n}$), this yields that $b_{n}\mid b_{n+2}+b_{n-2} =  b_{n-2}+b_{n+2}$. This proves Theorem 2 (a).

(b) Let $n\geq2$ be an integer. The recurrent definition of $\left(b_{1},b_{2},b_{3},...\right)  $ yields $b_{n+2}=\dfrac{b_{n+1}^{3}+1}{b_{n}}$, so that $b_{n}b_{n+2}=b_{n+1}^{3}+1$. The recurrent definition of $\left( b_{1},b_{2},b_{3},...\right)  $ also yields $b_{n+1}=\dfrac{b_{n}^{3} +1}{b_{n-1}}$, so that $b_{n+1}b_{n-1}=b_{n}^{3}+1 = \left(b_n+1\right)\left(b_n^2-b_n+1\right)\equiv0\mod  b_{n}+1$.

We have

$b_{n}b_{n-1}\left(  b_{n+2}+1\right)  =b_{n}b_{n-1}b_{n+2}+b_nb_{n-1} =b_{n-1}\underbrace{b_{n}b_{n+2}}_{=b_{n+1}^{3}+1}+\underbrace{b_{n}} _{\equiv-1\mod b_{n}+1}b_{n-1}$

$\equiv b_{n-1}\left(  b_{n+1}^{3}+1\right)  +\left(  -1\right) b_{n-1}=b_{n-1}b_{n+1}^{3}+b_{n-1}-b_{n-1}=b_{n-1}b_{n+1}^{3} $

$=\underbrace{b_{n+1}b_{n-1}}_{\equiv0\mod b_{n}+1}b_{n+1}^{2} \equiv 0\mod b_{n}+1$.

In other words, $b_{n}+1\mid b_{n}b_{n-1}\left(  b_{n+2}+1\right)  $. Since $b_{n}+1$ is coprime to $b_{n}$, this yields $b_{n}+1\mid b_{n-1}\left( b_{n+2}+1\right)  $. This proves Theorem 2 (b).

Some educated guessing was used in this proof... I am wondering how much of it could be automated.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
darij grinberg
6555 posts
#7 • 5 Y
Y by Tintarn, Adventure10, Mango247, jaescl, cubres
I just revisited this thread as I was looking for a simple example on the Laurent phenomenon in cluster algebras, and it dawned upon me that both theorems in my posts can be generalized:

Theorem 3. Let $r$ be a positive integer. Let $\left(b_1,b_2,b_3,...\right)$ be a sequence of positive rational numbers defined recursively by

$b_1=1$, $b_2=2$ and $b_{n+1}=\frac{b_n^r+1}{b_{n-1}}$ for every integer $n\geq 2$.

(a) Then, $b_n$ is a positive integer for every positive integer $n$.

(b) If $r \geq 2$, then every integer $n\geq3$ satisfies $b_{n}\mid b_{n-2}+b_{n+2}$.

(c) If $r$ is odd, then every integer $n\geq2$ satisfies $b_{n}+1\mid b_{n-1}\left( b_{n+2}+1\right)  $.


Proof of Theorem 3. (a) The proof of Theorem 3 (a) can be obtained by replacing every $3$ by $r$ in the proof of Theorem 1 (in post #3 above), and making one further change: The induction base (the case $N=3$) becomes a bit more difficult. Here is how the induction base must be proven now: It is clear that $b_1$ and $b_2$ are positive integers, and $b_3$ is an integer as well (since $b_3 = \frac{b_2^r+1}{b_1}$ and $b_1 = 1 \mid b_2^r+1$). Also, $b_1$, $b_2$, $b_3$ and $b_4$ are clearly positive. To prove that $b_4$ is an integer, we need to argue as follows: We have $b_3 = \frac{b_2^r+1}{b_1} = \frac{2^r+1}{1}$ (since $b_1 = 1$ and $b_2 = 2$), so that $b_3 = \frac{2^r+1}{1} = 2^r+1$ is odd, and thus $b_3^r$ is odd, so that $b_3^r+1$ is even, and thus $2 \mid b_3^r + 1$. But $b_4 = \frac{b_3^r+1}{b_2} = \frac{b_3^r+1}{2}$ (since $b_2 = 2$) is an integer (since $2 \mid b_3^r + 1$). Now, in order to complete the induction base, we need to prove that $b_2$ is coprime to $b_3$. This is clear because $b_2 = 2$ whereas $b_3$ is odd.

(b) Assume that $r \geq 2$. The proof of Theorem 3 (b) can be obtained by replacing every $3$ by $r$ in the proof of Theorem 2 (a) (in post #6 above). The assumption $r \geq 2$ is used in the claim that $ \left( b_{n}^r+1\right) ^r \equiv1\mod b_{n}^{2}$.

(c) Assume that $r$ is odd. The proof of Theorem 3 (c) can be obtained by replacing every $3$ by $r$ in the proof of Theorem 2 (b) (in post #6 above), and replacing the equality $b_{n}^{3}+1 = \left(b_n+1\right)\left(b_n^2-b_n+1\right)$ by $b_{n}^r + 1 = \left(b_n+1\right)\left(b_n^{r-1}-b_n^{r-2}+b_n^{r-3}\pm ... +1\right)$ (this makes sense because $r$ is odd).

The proof of Theorem 3 is thus complete.

This all, of course, is related to cluster algebras: Our recurrence equation describes a cluster algebra of a two-vertex quiver with $r$ arrows from one vertex to the other. See §3.1.2 in Philipp Lampe's Cluster algebras notes. The Laurent phenomenon for cluster algebras can thus be used to prove Theorem 3 (a), although it should be applied with care: we need to extend our sequence $\left(b_1, b_2, b_3, ...\right)$ by an extra initial term $b_0 = 1$ (check that the recurrence is still satisfied) in order to ensure that "Laurent polynomials in the initial values" actually translates into "integers". See also §3.1 of Lee/Schiffler, Positivity for cluster algebras, arxiv:1306.2415v3 for combinatorial expressions for the general term of this sequence.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jaescl
11 posts
#8 • 1 Y
Y by cubres
See also Notes on the combinatorial fundamentals of algebra (from page 126 onward)
Let $B_n$ and $k$ be as defined at https://artofproblemsolving.com/community/c6h1065491. I claim that the following expression is an integer for every integer $n\geq 3$:
$$k^{L_{2n-4}}\left(\frac{B_{n+2}+B_{n-2}}{B_{n}}\right)$$Where $L_{n}$ is the nth Lucas number
This post has been edited 1 time. Last edited by jaescl, Mar 2, 2025, 5:38 PM
Reason: Link format issue
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jaescl
11 posts
#9
Y by
Plugging $r=3$ into result ($107$) of the Grinberg's notes, we get:

$$\left(\frac{B_{n+2}+B_{n-2}}{B_{n}}\right) = B_{n-2}B_{n+2} - \left(\frac{(B_{n}^{3}+1)^3-1}{B_{n}^2}\right)$$
Multiplying both sides of the equality by $k^{L_{2n-4}}$ and using the recurrence formula for $B_{n+2}$ gives:


$$k^{L_{2n-4}}\left(\frac{B_{n+2}+B_{n-2}}{B_{n}}\right) = \frac{k^{L_{2n-4}}B_{n-2}(B_{n+1}^3+1)}{B_{n}} - \frac{k^{L_{2n-4}}}{B_{n}}\left(\frac{(B_{n}^{3}+1)^3-1}{B_{n}}\right)$$
Remember that the main theorem of the linked thread states that $k^{F_{2(n-3)}}B_{n}$ is an integer, where $F_{n}$ is the nth Fibonacci number.

Note that $L_{2n-4}=F_{2(n-1)}-F_{2(n-3)}$ and $k^{F_{2(n-1)}}=\frac{k^{F_{2n}}}{k^{F_{2n}-F_{2(n-1)}}}$

The original statement is equivalent to prove that the following expression is divisible by $k^{F_{2n}-F_{2(n-1)}}$:
$$\frac{k^{F_{2n}}B_{n-2}(B_{n+1}^3+1)}{k^{F_{2(n-3)}}B_{n}} - \frac{k^{F_{2n}}}{k^{F_{2(n-3)}}B_{n}}\left(\frac{(B_{n}^{3}+1)^3-1}{B_{n}}\right)$$A previous step in the proof is to prove that the expressions involved in the subtraction are integers.
The main result follows from the fact that the expressions $k^{F_{2n}}B_{n-2}B_{n+1}^{3}$ and $k^{F_{2n}}\left(\frac{(B_{n}^{3}+1)^3-1}{B_{n}}\right)$ leave the same remainder when divided by $k^{F_{2n}-F_{2(n-1)}}$ for every integer $n\geq 3$, and that $k^{F_{2(n-3)}}B_{n}$ is coprime with $k^{F_{2n}-F_{2(n-1)}}$.
Z K Y
N Quick Reply
G
H
=
a