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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.

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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]
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0 replies
1 viewing
jlacosta
Mar 2, 2025
0 replies
k i Suggestion Form
jwelsh   0
May 6, 2021
Hello!

Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form:
Suggestion Form

To keep all new suggestions together, any new suggestion threads posted will be deleted.

Please remember that if you find a bug outside of FTW! (after refreshing to make sure it’s not a glitch), make sure you’re following the How to write a bug report instructions and using the proper format to report the bug.

Please check the FTW! thread for bugs and post any new ones in the For the Win! and Other Games Support Forum.
0 replies
jwelsh
May 6, 2021
0 replies
k i Read me first / How to write a bug report
slester   3
N May 4, 2019 by LauraZed
Greetings, AoPS users!

If you're reading this post, that means you've come across some kind of bug, error, or misbehavior, which nobody likes! To help us developers solve the problem as quickly as possible, we need enough information to understand what happened. Following these guidelines will help us squash those bugs more effectively.

Before submitting a bug report, please confirm the issue exists in other browsers or other computers if you have access to them.

For a list of many common questions and issues, please see our user created FAQ, Community FAQ, or For the Win! FAQ.

What is a bug?
A bug is a misbehavior that is reproducible. If a refresh makes it go away 100% of the time, then it isn't a bug, but rather a glitch. That's when your browser has some strange file cached, or for some reason doesn't render the page like it should. Please don't report glitches, since we generally cannot fix them. A glitch that happens more than a few times, though, could be an intermittent bug.

If something is wrong in the wiki, you can change it! The AoPS Wiki is user-editable, and it may be defaced from time to time. You can revert these changes yourself, but if you notice a particular user defacing the wiki, please let an admin know.

The subject
The subject line should explain as clearly as possible what went wrong.

Bad: Forum doesn't work
Good: Switching between threads quickly shows blank page.

The report
Use this format to report bugs. Be as specific as possible. If you don't know the answer exactly, give us as much information as you know. Attaching a screenshot is helpful if you can take one.

Summary of the problem:
Page URL:
Steps to reproduce:
1.
2.
3.
...
Expected behavior:
Frequency:
Operating system(s):
Browser(s), including version:
Additional information:


If your computer or tablet is school issued, please indicate this under Additional information.

Example
Summary of the problem: When I click back and forth between two threads in the site support section, the content of the threads no longer show up. (See attached screenshot.)
Page URL: http://artofproblemsolving.com/community/c10_site_support
Steps to reproduce:
1. Go to the Site Support forum.
2. Click on any thread.
3. Click quickly on a different thread.
Expected behavior: To see the second thread.
Frequency: Every time
Operating system: Mac OS X
Browser: Chrome and Firefox
Additional information: Only happens in the Site Support forum. My tablet is school issued, but I have the problem at both school and home.

How to take a screenshot
Mac OS X: If you type ⌘+Shift+4, you'll get a "crosshairs" that lets you take a custom screenshot size. Just click and drag to select the area you want to take a picture of. If you type ⌘+Shift+4+space, you can take a screenshot of a specific window. All screenshots will show up on your desktop.

Windows: Hit the Windows logo key+PrtScn, and a screenshot of your entire screen. Alternatively, you can hit Alt+PrtScn to take a screenshot of the currently selected window. All screenshots are saved to the Pictures → Screenshots folder.

Advanced
If you're a bit more comfortable with how browsers work, you can also show us what happens in the JavaScript console.

In Chrome, type CTRL+Shift+J (Windows, Linux) or ⌘+Option+J (Mac).
In Firefox, type CTRL+Shift+K (Windows, Linux) or ⌘+Option+K (Mac).
In Internet Explorer, it's the F12 key.
In Safari, first enable the Develop menu: Preferences → Advanced, click "Show Develop menu in menu bar." Then either go to Develop → Show Error console or type Option+⌘+C.

It'll look something like this:
IMAGE
3 replies
slester
Apr 9, 2015
LauraZed
May 4, 2019
k i Community Safety
dcouchman   0
Jan 18, 2018
If you find content on the AoPS Community that makes you concerned for a user's health or safety, please alert AoPS Administrators using the report button (Z) or by emailing sheriff@aops.com . You should provide a description of the content and a link in your message. If it's an emergency, call 911 or whatever the local emergency services are in your country.

Please also use those steps to alert us if bullying behavior is being directed at you or another user. Content that is "unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable" (AoPS Terms of Service 5.d) or that otherwise bullies people is not tolerated on AoPS, and accounts that post such content may be terminated or suspended.
0 replies
dcouchman
Jan 18, 2018
0 replies
Right angles on incircle
DynamoBlaze   37
N 7 minutes ago by sangsidhya
Source: RMO 2018 P6
Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $I$ be the incentre of triangle $ABC$, and let $D,E,F$ be the points where the incircle touches the sides $BC,CA,AB,$ respectively. Let $BI,CI$ meet the line $EF$ at $Y,X$ respectively. Further assume that both $X$ and $Y$ are outside the triangle $ABC$. Prove that
$\text{(i)}$ $B,C,Y,X$ are concyclic.
$\text{(ii)}$ $I$ is also the incentre of triangle $DYX$.
37 replies
1 viewing
DynamoBlaze
Oct 7, 2018
sangsidhya
7 minutes ago
Where is the equality?
AndreiVila   2
N 44 minutes 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
44 minutes ago
Truth or lie at a table
SinaQane   7
N an hour ago by Oksutok
Source: 239 2019 S4
There are $n>1000$ people at a round table. Some of them are knights who always tell the truth, and the rest are liars who always tell lies. Each of those sitting said the phrase: “among the $20$ people sitting clockwise from where I sit there are as many knights as among the $20$ people seated counterclockwise from where I sit”. For what $n$ could this happen?
7 replies
SinaQane
Jul 31, 2020
Oksutok
an hour ago
2 var inquality
sqing   4
N an hour ago by sqing
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$$ a+b+ab\geq2$$
4 replies
+1 w
sqing
Today at 4:06 AM
sqing
an hour ago
k 2023 IMO
ostriches88   4
N Mar 14, 2025 by jlacosta
As outlined in this (locked) post, the forum/blog creation message is out of date. It has been over a year since that post, and it still has not changed. I don't know if this got lost somewhere in the "passing this along" chain or if it was determined to be irrelevant, but it seems like a relatively simple fix ;)
4 replies
ostriches88
Mar 10, 2025
jlacosta
Mar 14, 2025
Pressing &#039;go down button&#039; always creates a gray box on the last post
Craftybutterfly   13
N Mar 14, 2025 by Craftybutterfly
Summary of the problem: Pressing go down to last post button always creates a gray box overlapping last post
Page URL: any forum
Steps to reproduce:
1. Go to any topic in a forum
2. The gray box at the bottom overlaps part of the first post
Expected behavior: Should not show a gray box
Frequency: 100% of the time
Operating system(s): Linux HP EliteBook 835 G8 Notebook PC
Browser(s), including version: Chrome 133.0.6943.142 (Official Build) (64-bit) (cohort: Stable)
Additional information: It works on any other device, on my iPhone XR, a MacOS, and my iPad. Took the screenshot a month ago. The gray box still appears
13 replies
Craftybutterfly
Mar 12, 2025
Craftybutterfly
Mar 14, 2025
k Somehow I broke the unique contest collection naming...
Equinox8   1
N Mar 14, 2025 by jlacosta
Somehow I broke the unique contest collection naming scheme; for what it's worth, I also ran into an AJAX timeout error while trying to make this particular post collection for the first time, and pressed "create" twice.

Not sure if there's anything to be done here, but are there potentially unforeseen consequences here?

https://artofproblemsolving.com/community/c4258765_2024_puerto_rico_team_selection_test (this is the one I'm working on; it's also not finished yet)
https://artofproblemsolving.com/community/c4258764_2024_puerto_rico_team_selection_test
1 reply
Equinox8
Mar 12, 2025
jlacosta
Mar 14, 2025
k New Forums Duplicates
k1glaucus   3
N Mar 14, 2025 by jlacosta
Summary of the problem: In the New Forums collection, many of the forums are duplicated, although some have slightly different information (likely due to the forum being edited by its admin(s)). Typically only one of the two has threads, and I believe this is the only way to access
Page URL: https://artofproblemsolving.com/community/c74_new_forums
Steps to reproduce:
1. Go to the link
2. You should see duplicates of some forums
Expected behavior: Each forum appears once
Frequency: Every time
Additional information: Typically only one of the two has threads, and from creating forums in the past I believe this is the only way to access the duplicate. Also not all of the forums are duplicated. Another reason to suspect these are duplicates are the forums with similar names have the same admin.
3 replies
k1glaucus
Mar 12, 2025
jlacosta
Mar 14, 2025
k Pi Day!!
SomeonecoolLovesMaths   66
N Mar 14, 2025 by LawofCosine
Happy $\pi$ day everyone!
66 replies
SomeonecoolLovesMaths
Mar 14, 2025
LawofCosine
Mar 14, 2025
k Help with beast academy
tyrantfire4   1
N Mar 14, 2025 by tyrantfire4
So my siblings do BA and it won't work if anyone can fix it that would be great!
All browers won't work
Apple and fire
Open BA account
1 reply
tyrantfire4
Mar 14, 2025
tyrantfire4
Mar 14, 2025
Search function in private messages not working
WisteriaV   2
N Mar 12, 2025 by mathlearner2357
For the past while, the search function in private messages hasn’t been working. Whenever I search for anything, it says, “No topics here!” after trying to load for a while. I’ve tried different devices (laptop and ipad) and browsers (chrome on both devices, safari on ipad, and microsoft edge on laptop), and the results are the same. I’ve also had friends say the same happens for them.
2 replies
WisteriaV
Mar 12, 2025
mathlearner2357
Mar 12, 2025
Viewing next classes
lilorocks11   2
N Mar 11, 2025 by jlacosta
Hello,

I'm wondering how to look at my next classes. I'm not entirely sure if I'm registered for a class. I tried looking around on the My AoPS page, but didn't find anything that had to do with the course I wanted.
2 replies
lilorocks11
Mar 8, 2025
jlacosta
Mar 11, 2025
Error when abandoning quest
dragonborn56   3
N Mar 11, 2025 by pb0975
I attempted to abandon the Alcumus quest Hit The Gym: Accuracy. I clicked on it and selected abandon, but the popup disappeared and the quest was still there. There were no changes to the log, and my daily abandon was used, as when I clicked the quest again it said that all of my abandons were used.
3 replies
dragonborn56
Mar 10, 2025
pb0975
Mar 11, 2025
What????
LawofCosine   21
N Mar 10, 2025 by LawofCosine
Umm....what is happening? The $-\infty$ seems to be squashed but the $\infty$ is normal. This is from the AoPS Calculus ebook.
21 replies
LawofCosine
Mar 7, 2025
LawofCosine
Mar 10, 2025
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
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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
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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
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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
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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.
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jaescl
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#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.
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darij grinberg
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#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.
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darij grinberg
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#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.
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jaescl
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#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
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#9
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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 $k^{F_{2n}-F_{2(n-1)}}$.
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