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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/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.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
Inspired by Bet667
sqing   3
N 23 minutes ago by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
3 replies
sqing
an hour ago
sqing
23 minutes ago
Geometry marathon
HoRI_DA_GRe8   846
N 26 minutes ago by ItzsleepyXD
Ok so there's been no geo marathon here for more than 2 years,so lets start one,rules remain same.
1st problem.
Let $PQRS$ be a cyclic quadrilateral with $\angle PSR=90°$ and let $H$ and $K$ be the feet of altitudes from $Q$ to the lines $PR$ and $PS$,.Prove $HK$ bisects $QS$.
P.s._eeezy ,try without ss line.
846 replies
+1 w
HoRI_DA_GRe8
Sep 5, 2021
ItzsleepyXD
26 minutes ago
Find all functions $f$ is strictly increasing : \(\mathbb{R^+}\) \(\rightarrow\)
guramuta   0
31 minutes ago
Find all functions $f$ is strictly increasing : \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such that:
i) $f(2x)$ \(\geq\) $2f(x)$
ii) $f(f(x)f(y)+x) = f(xf(y)) + f(x) $
0 replies
guramuta
31 minutes ago
0 replies
Partitioning coprime integers to arithmetic sequences
sevket12   3
N 35 minutes ago by quacksaysduck
Source: 2025 Turkey EGMO TST P3
For a positive integer $n$, let $S_n$ be the set of positive integers that do not exceed $n$ and are coprime to $n$. Define $f(n)$ as the smallest positive integer that allows $S_n$ to be partitioned into $f(n)$ disjoint subsets, each forming an arithmetic progression.

Prove that there exist infinitely many pairs $(a, b)$ satisfying $a, b > 2025$, $a \mid b$, and $f(a) \nmid f(b)$.
3 replies
sevket12
Feb 8, 2025
quacksaysduck
35 minutes ago
purple comet discussion
ConfidentKoala4   66
N Today at 3:43 AM by wuwang2002
when can we discuss purple comet
66 replies
ConfidentKoala4
May 2, 2025
wuwang2002
Today at 3:43 AM
9 ARML Location
deduck   36
N Today at 2:31 AM by idk12345678
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
36 replies
deduck
Tuesday at 4:19 PM
idk12345678
Today at 2:31 AM
how prestigious is hsmc
ConfidentKoala4   3
N Today at 1:20 AM by ConfidentKoala4
been wonderin this for a while

how prestigious is it? ik its not as good as mathily (they rejected me :mad: ) but Idk how good it actually is
3 replies
ConfidentKoala4
Today at 12:46 AM
ConfidentKoala4
Today at 1:20 AM
9 Does Mental Health Actually Matter?
heheman   9
N Today at 12:47 AM by maxamc
Looking at the goals I once had, it was all just so silly and stupid

I didn't even reach my "Low" goal for AIME... so pathetic

Missed JMO by a huge margin, after missing by only 12.5 points last year

(BTW i didn't slack off one bit)

I guess the most important thing is just to keep my head up and keep going. I can't let failures stop me. Honestly I don't care about setting goals anymore. They only give me a lot of internal pressure to do well. I think the most important thing is to focus on what I do everyday, consistently, and pay attention to the beautiful things in life (like math).

I'm going to try getting more involved in real life. After coming back from COVID, I had trouble to make as many friends with non-math people. But I was reconnecting with some of my friends that I had prepandemic and I realized how precious those friendships really were.

Now the last thing to do is grind my last bit of nonexistent ego to dust and focus on the present, stop looking back

(Note: This doesn't mean I'm going to quit, I just mean I'm going to do math on my own and try to not feel any pressure to do well. Cause i feel like that pressure really beat me a lot.)

I love this community and am happy for everyone who qualified olympiad but at this point competition math just reminds me only of my failures. (Even if it's my own fault.) So I'm probably going to take a break for a while. Thanks everyone for being nice to me and stuff. Sorry if this sounds cringe (it will in a week)

9 replies
heheman
Mar 8, 2024
maxamc
Today at 12:47 AM
HCSSiM results
SurvivingInEnglish   60
N Today at 12:32 AM by Vivaandax
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
60 replies
SurvivingInEnglish
Apr 5, 2024
Vivaandax
Today at 12:32 AM
Rational sequences
tenniskidperson3   57
N Yesterday at 10:25 PM by OronSH
Source: 2009 USAMO problem 6
Let $s_1, s_2, s_3, \dots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \dots.$ Suppose that $t_1, t_2, t_3, \dots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$. Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.
57 replies
tenniskidperson3
Apr 30, 2009
OronSH
Yesterday at 10:25 PM
1:1 Physics Tutors
DinoDragon186   3
N Yesterday at 7:26 PM by talhee
I am looking for 1:1 physics tutor.
I am a beginner in physics and am in 9th grade.
I want to make it to IPhO in the coming years.
3 replies
DinoDragon186
Dec 10, 2024
talhee
Yesterday at 7:26 PM
Looking for Physics or USAPhO Tutor
physicsplease   4
N Yesterday at 7:20 PM by talhee
Hii I am looking for a USAPhO tutor for next year's season. I think I have tried literally everything possible to improve but I feel like I just hit a massive roadblock right now.

It would be ideal if I can find someone who have a lot of experience with physics olympiads. My goal is medal/gold in usapho next year, and I am very determined & willing to put in a lot of hours, especially more so in the summer. Please recommend anyone or dm in aops, thank you.

Have qualified usapho before (last year), took both physics c and sufficient higher math.
4 replies
physicsplease
Apr 11, 2025
talhee
Yesterday at 7:20 PM
MathILy 2025 Decisions Thread
mysterynotfound   40
N Yesterday at 4:11 PM by bjump
Discuss your decisions here!
also share any relevant details about your decisions if you want
40 replies
mysterynotfound
Apr 21, 2025
bjump
Yesterday at 4:11 PM
Mathcounts state
happymoose666   39
N Yesterday at 1:54 PM by Inaaya
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
39 replies
happymoose666
Mar 24, 2025
Inaaya
Yesterday at 1:54 PM
a, b subset
MithsApprentice   20
N Apr 25, 2025 by Ilikeminecraft
Source: USAMO 1996
Determine (with proof) whether there is a subset $X$ of the integers with the following property: for any integer $n$ there is exactly one solution of $a + 2b = n$ with $a,b \in X$.
20 replies
MithsApprentice
Oct 22, 2005
Ilikeminecraft
Apr 25, 2025
a, b subset
G H J
G H BBookmark kLocked kLocked NReply
Source: USAMO 1996
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MithsApprentice
2390 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Determine (with proof) whether there is a subset $X$ of the integers with the following property: for any integer $n$ there is exactly one solution of $a + 2b = n$ with $a,b \in X$.
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Agrippina
126 posts
#2 • 2 Y
Y by Adventure10, Mango247
I posted (or meant to post) essentially the same problem a few weeks ago, here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354033.

I will try to put up a solution soon.
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t0rajir0u
12167 posts
#3 • 10 Y
Y by quangminhltv99, JasperL, Wizard_32, Adventure10, Mango247, aidan0626, kiyoras_2001, and 3 other users
Let $ f(x) = \sum_{a \in X} x^a$; the given condition is equivalent to $ f(x) f(x^2) = \frac {1}{1 - x}$, which immediately gives it away: recall that unique binary expansion implies the identity

$ \frac {1}{1 - x} = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)...$

so take $ f(x) = (1 + x)(1 + x^4)(1 + x^{16}) ...$. In other words, $ X$ consists of the set of positive integers whose binary expansions in base $ 4$ contain only $ 0$s and $ 1$s. (The bijective proof is straightforward: consider the base-$ 4$ expansion of $ n$ and isolate its digits of $ 2$ and $ 3$, etc.) Agrippina's problem is similar: the identity is $ f(x) f(x^2) f(x^4) = \frac {1}{1 - x}$ and we can take $ f(x) = (1 + x)(1 + x^8)(1 + x^{64})...$.
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dgreenb801
1896 posts
#4 • 2 Y
Y by Adventure10, Mango247
Ingenious! How did you think of that?
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t0rajir0u
12167 posts
#5 • 2 Y
Y by Adventure10, Mango247
Well, first of all this problem's been posted before (although without the source) and this solution given by several MOPpers, so it wasn't hard to remember. It was also given as a very nice list of examples in this thread. Generally, it is very natural to analyze solutions to equations like $ a + b = n, a \in A, b \in B$ by studying the generating functions of $ A$ and $ B$ because $ AB$ gives all the possible sums at once. For example, the following can be solved by similar means.

Putnam 2003 A6: For a set $ S$ of non-negative integers let $ r_S(n)$ denote the number of ordered pairs $ (s_1, s_2) \in S^2, s_1 \neq s_2$ such that $ s_1 + s_2 = n$. Is it possible to partition the non-negative integers into disjoint sets $ A$ and $ B$ such that $ r_A(n) = r_B(n)$ for all $ n$?

The important step is to realize that if $ S(x) = \sum_{s \in S} x^s$, then the set of all sums of distinct elements of $ S$ is given by $ S(x)^2 - S(x^2)$ (why?). The rest is computation, and once you've figured out what the answer should be it is not hard to give a direct proof.

I'll note that even if you didn't think of binary expansion, repeated application of the problem condition allows you to perform the following calculation:

$ f(x) = \frac {1}{(1 - x) f(x^2)} = \frac { f(x^4) (1 - x^2)}{1 - x} = \frac {(1 - x^2)}{(1 - x)(1 - x^4) f(x^4)} = \frac {(1 - x^2)(1 - x^{8}) f(x^{16})}{(1 - x)(1 - x^4)} = ...$

Even if this is not rigorous, it tells you what the answer should look like, but even more it strongly suggests that this is the only answer, not just an answer that works.
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t0rajir0u
12167 posts
#6 • 2 Y
Y by Adventure10, Mango247
My apologies; I misread "integer" as "non-negative integer," and the solution I gave doesn't work as is. I'll keep thinking.
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aznlord1337
130 posts
#7 • 3 Y
Y by Delray, vsathiam, Adventure10
Use induction: Suppose we have integers $ x_1...x_n$ such that every $ x_i + 2x_j$ is distinct. Suppose this set misses the value $ n$. Then add to the set $ k, n-2k$, so now we have $ n$ included as a sum. It is clear that if you take $ k$ arbitrarily large it wont overlap with any previous sums. So such a set exists.
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tenniskidperson3
2376 posts
#8 • 5 Y
Y by Delray, Adventure10, Mango247, and 2 other users
Since nobody has come up with a (complete) solution, let me post mine (inspired by Kalva):

First, let us verify that a "base -4" can work. That is, every number can be expressed uniquely as $a_1-4a_2+16a_3-64a_4+\ldots$ where $a_i\in\{0, 1, 2, 3\}$. Uniqueness is evident: if we have $a_1-4a_2+\ldots=b_1-4b_2+\ldots$ then suppose $k$ is the first natural number such that $a_k\neq b_k$. Then

$(-4)^{k-1}a_k+(-4)^ka=(-4)^{k-1}b_k+(-4)^kb$

for the rest of the integer $a$ and $b$. Then that means that, dividing by $(-4)^{k-1}$, we have $a_k-b_k=4(a-b)$, so $a_k-b_k$ is divisible by 4, which is impossible because $a_k$ and $b_k$ are not equal and between 0 and 3. So the uniqueness is proved.

Now take the $4^k$ numbers $a_1-4a_2+16a_3-\ldots+(-4)^{k-1}a_k$ where $a_i\in\{0, 1, 2, 3\}$. The minimal number possible is $3(-4-64-\ldots)$ and the maximal number is $3(1+16+256+\ldots)$, which cover a range of

$3(1+16+256+\ldots)-3(-4-64-\ldots)+1=3(1+4+16+\ldots+4^{k-1})+1=4^k$

numbers. Since there are $4^k$ numbers in the possible range and all numbers are expressed at most once, all numbers must be expressed exactly once. Thus base -4 exists and we can work with it like any normal base.

So now as before, we place all numbers with only 0's and 1's in their base -4 expansions in the set. Then for any integer $n$, take its base -4 expansion $n_1-4n_2+16n_3-\ldots$. If $n_i=0$, let $a_i=b_i=0$; if $n_i=1$, let $a_i=1$ and $b_i=0$; if $n_i=2$, let $a_i=0$ and $b_i=1$; and if $n_i=3$, let $a_i=b_i=1$, so that in any case, $a_i+2b_i=n_i$. Then let $a=a_1-4a_2+16a_3-\ldots\in X$ and $b=b_1-4b_2+16b_3-\ldots\in X$ also. Then clearly $a+2b=n$ by construction.

We must show uniqueness. For any $a, b\in X$, we have

$a-b=(a_1-b_1)-4(a_2-b_2)+16(a_3-b_3)-\ldots$.

The first nonzero $a_i-b_i$ is either $1$ or $-1$. Thus $a-b=\pm(-4)^{k-1}+(-4)^kx$ where $x$ is an integer. Hence $|a-b|$ is divisible by $4^{k-1}$ and not $2\cdot4^{k-1}$, so the highest power of two that divides $a-b$ is also a power of 4.

Now if $a+2b=c+2d$ for $a, b, c, d\in X$, then $a-c=2(d-b)$. If $a\neq c$ then let us look at the highest power of 2 that divides this common difference. It must be a power of 4 that divides $a-c$, but also is a power of 4 that divides $d-b$ and so is twice a power of 4 that divides $2(d-b)$. No number is both a power of 4 and twice a power of 4, so this contradiction shows uniqueness and we're done.
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zero.destroyer
813 posts
#9 • 2 Y
Y by Adventure10, Mango247
Using generating functions since they generalize to alot of counting problems, (though this uses neg exponents, but it still is legit)

Let $f(x)=(1+x^{1})(1+x^{-4})(1+x^{16})(1+x^{-64})*...(1+x^{(-4)^{a}})$ (the same base -4 thing)
Then
$f(x)f(x^{2})=\frac{(1+x^{1})(1+x^{2})(1+x^{4})(1+x^{8})....(1+x^{2^{n}})}{x^{r}}$, where $r$ is a pretty huge number,

which means at the limiting case, as $a$ tends to infinity,
$f(x)f(x^{2})=...+1/x^{3}+1/x^{2}+1/x^{1}+1+x^{1}+x^{2}+x^{3}...$ which solves the problem.
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tenniskidperson3
2376 posts
#10 • 2 Y
Y by Adventure10, Mango247
No that's not legit. It's only legit if the generating function converges for some $x$, which this one does not. The generating function approach does not prove the answer, it just points you in the direction. You need another approach to show that it actually works. In the limit, what is $r$? How do you know that the generating function doesn't just become $\ldots+\frac{1}{x^4}+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1$ and stop there?
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zero.destroyer
813 posts
#11 • 2 Y
Y by Adventure10, Mango247
Sorry, I was in a rush, and put only the general idea down. It doesn't HAVE to converge for some X, because of the fact that all it's doing is manipulating the exponents of the terms, as a representation of a sumset problem. I'm not ever actually going to evaluate the function $f(x)$. I know it makes absolutely no sense if you considered $x$ as an actual number, but this wasn't the point here.

And sorry, I can't calculate right now, but essentially it's pretty easy to show through calculation that the most negative exponent and most positive exponents are increasing arbitrarily large as $a$ approaches infinity.
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tenniskidperson3
2376 posts
#12 • 3 Y
Y by Adventure10, Mango247, and 1 other user
My point is that you cannot just rush into these calculations. You need the theory of formal power series or (in this case) Laurent series. The calculations you do, taking products of sums and expanding them, is only justified when the power (Laurent) series converges. That's why everyone was up in arms when Euler calculated $\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$ by setting $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\ldots=x(x^2-\pi^2)(x^2-4\pi^2)\ldots$; he had no rigorous justification for saying two power series were equal just because they had exactly the same roots. And this is a bit like what you're doing; you're saying that the power series $\frac{(1+x)(1+x^2)(1+x^4)\ldots}{x^r}=\ldots+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1+x+x^2+x^3+\ldots$, whatever $r$ means in the limiting case.

Now I'm not saying I don't believe you, because I just gave the same solution set as you. I'm just pointing out that your taking the limit of $f(x)f(x^2)$ as $a$ or $n$ goes to infinity and saying that it equals the product of $\lim_{n\rightarrow\infty}f(x)$ and $\lim_{m\rightarrow\infty}f(x^2)$ is unjustified. It would be justified if it converged for some number $x$, but it doesn't.
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zero.destroyer
813 posts
#13 • 2 Y
Y by Adventure10, Mango247
Sorry if I was unclear with my words, by limit, I didn't actually mean that the numerical value for some particular $x$, of $f(x)*f(x^2)$; by limit, I just meant that the larger integers (in absolute value) could be represented uniquely once we increased the size of our sequence. I'm not ever using $f(x)*f(x^2)$ as any evaluation for a numerical answer, I'm just using $f(x)*f(x^2)$ conveniently because it just simplified the concept of "looking at all sums".
Your example indeed uses these generating functions as actual polynomials, which evaluate numerical answers, which isn't the same as what I'm doing.

I can easily just as well put this into an argument without generating functions, but still using the same concept of "the terms acting like a binary string". It's just that the algebraic manipulations are more representative/clear of what the concept is.
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ZetaX
7579 posts
#14 • 6 Y
Y by v_Enhance, starchan, StarLex1, Adventure10, Mango247, and 1 other user
zero.destroyer's argument is correct. You don't need convergence for formal power series at all, but it is indeed necessary to be careful with Laurent series' that are infinite in both directions; but this is mostly due to the fact that these do not form a ring, not even a module over the power series' (but over polynomials or finite Laurent sums).

The argument here never compares terms in a non-formal nature, unlike Euler, who, as said above, plugged in real numbers to to speak of "roots". It is enough to show that additional factors in the product do not contribute to those summands of exponent $s$, where $|s|$ is bounded by some growing bound dependent on the number of factors. This is the case here.

If one wants to do it very formally, an algebraic version would be to state that the module of Laurent series is the projective limit over $n$ of Laurents sums whose exponent's modulus is bounded by $n$. This is just a less comprehensible way to state what I said in the previous paragraph, though.



Also, note this "fact": $\sum_{n \in \mathbb Z} x^ n = \sum_{n=1}^ \infty x^{-n} + \sum_{n=0}^ \infty x^n = \frac{x^{-1}}{1-x^ {-1}} + \frac{1}{1-x} = \frac{1}{x-1} + \frac{1}{1-x} = 0$. The error here is that to apply geometric series, you would need it to be a module over power series' (i.e. multiplying any Laurent series with a power series would need to make sense; try to multiply the above one by $\sum_{n=0}^ \infty x^n $ to see the problem). But the problem is not that we lack a common are of convergence for those sums.
Actually, the fractions are the meromorphic continuations of the sums (which in turn are the Laurent expansions around $\infty$ and $0$) and as an identity of meromorphic functions, this is completely correct!
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Delray
348 posts
#15 • 2 Y
Y by Adventure10, Mango247
Not clear or not if $a$ and $b$ are distinct.
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HamstPan38825
8860 posts
#16 • 2 Y
Y by StarLex1, Mango247
Let $f(X) = \sum_{a \in X} X^a$. The necessary and sufficient condition for $X$ to satisfy the condition is for $$f(X)f(X^2) = \frac 1{1-X}$$for all $x \neq 1$. One can notice that we have $$f(X^4) = \frac 1{(1-X^2)f(X^2)} = \frac 1{(1-X^2) \cdot \frac 1{(1-X)f(X)}} = \frac{f(X)(1-X)}{1-X^2} = \frac{f(X)}{1+X},$$so the infinite product $$\frac{f(X)}{(1+X)(1+X^4)(1+X^{16}) \cdots} = f(X^{2^n}) \to f(0) = 1,$$so the function $$f(X) = (1+X)(1+X^4)(1+X^{16}) \cdots$$can be checked to work.

In more concrete terms, we may pick $X$ to be the set of numbers that can be represented as the sum of some distinct nonnegative powers of 4.
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RedFireTruck
4223 posts
#17
Y by
We want to find $X$ such that $(\sum_{i\in X} x^i)(\sum_{i\in X} x^{2i})=\dots+x^{-2}+x^{-1}+1+x^1+x^2+\dots$. $(1+x)(1+x^2)=1+x+x^2+x^3$. $(1+x+x^2+x^3)(1+x^{-4}+x^{-8}+x^{-12})=x^{-12}+\dots+x^3$. $(x^{-12}+\dots+x^3)(1+x^{16}+x^{32}+x^{48})=x^{-12}+\dots+x^{51}$. We could keep going like this forever, extending in both directions. Therefore, $X$ is $\{1, -4, 16, -64, \dots\}$ works.
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pinkpig
3761 posts
#18
Y by
solution
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shendrew7
795 posts
#19
Y by
We interpret this using generating functions. Defining $A(x) = \sum_{k \in \mathcal{X}} x^k$, our condition requires
\[A(x) A(x^2) = \ldots + x^{-2} + x^{-1} + x^0 + x^1 + x^2 + \ldots.\]
to cover all integers exactly once. From here, we note that the functions
\begin{align*}
A(x) &= \prod \left(1+x^{(-4)^i}\right) = (1+x^1)(1+x^{-4})(1+x^{16})(1+x^{-64}) \ldots \\
A(x^2) &= \prod \left(1+x^{2 \cdot (-4)^i}\right) = (1+x^2)(1+x^{-8})(1+x^{32})(1+x^{-128}) \ldots
\end{align*}
indeed have the desired product, so our construction for $\mathcal{X}$ is simply
\[\boxed{\{\mathcal{X}\} = \text{Integers with only 0 and 1 as digits in base -4}}. \quad \blacksquare\]
This post has been edited 1 time. Last edited by shendrew7, Jan 26, 2024, 2:09 PM
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Maximilian113
575 posts
#20
Y by
Posting for storage, this is basically the same as multiple solutions above

Let $f(x)=\sum_{k \in X} x^k,$ then we require $$f(x)f(x^2)=\cdots+x^{-3}+x^{-2}+x+1+x+x^2+x^3+\cdots.$$However observe that $f(x)=(1+x)(1+x^{-4})(1+x^{16})(1+x^{-64})\cdots$ works since every number has a unique representation in base $-2.$
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Ilikeminecraft
619 posts
#21
Y by
Pick $X = \{4^k\mid k\in\mathbb Z_{\geq0}\}.$ To prove this, just use the generating function.
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