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

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

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Perpendicularity in Two Tangent Circles
steven_zhang123   1
N 14 minutes ago by aaravdodhia
Source: 2025 Hope League Test 2 P3
Circle \(O_1\) and circle \(O_2\) are externally tangent at point \(T\). From a point \(X\) on circle \(O_2\), a tangent is drawn intersecting circle \(O_1\) at points \(A\) and \(B\). The line \(XT\) is extended to intersect circle \(O_1\) at point \(S\). A point \(C\) is taken on the arc \(TS\) of circle \(O_1\). The line \(SC\) is extended to intersect the angle bisector of \(\angle BAC\) at point \(I\). The circle passing through points \(A, T, X\) and the circle passing through points \(C, T, I\) intersect at another point \(E\). Prove that \(EO_2 \perp XI\).
Proposed by Luo Haoyu
1 reply
steven_zhang123
Yesterday at 8:24 AM
aaravdodhia
14 minutes ago
Peru IMO TST 2022
diegoca1   0
27 minutes ago
Source: Peru IMO TST 2022 D1 P4
Let $\Omega$ be the circumcircle of triangle $ABC$, with $\angle BAC > 90^\circ $ and $ AB > AC $. The tangents to $\Omega$ at points $B$ and $C$ intersect at $D$. The tangent to $\Omega$ at point $A$ intersects line $BC$ at $E$. The line through $D$ parallel to $AE$ intersects line $BC$ at $F$. The circumference $\Gamma$ with diameter $EF$ intersects line $AB$ at points $P$ and $Q$, and line $AC$ at points $X$ and $Y$.
Prove that one of the angles $\angle AEB$, $\angle PEQ$, $\angle XEY$ is equal to the sum of the other two.
0 replies
diegoca1
27 minutes ago
0 replies
number theory
Hoapham235   5
N 27 minutes ago by Jjesus
Let $x >  y$ be positive integer such that \[ \text{LCM}(x+2, y+2)+\text{LCM}(x, y)=2\text{LCM}(x+1, y+1).\]Prove that $x$ is divisible by $y$.
5 replies
Hoapham235
Wednesday at 4:51 AM
Jjesus
27 minutes ago
centroid lies outside of triangle (not clickbait)
Scilyse   2
N 35 minutes ago by EthanWYX2009
Source: 数之谜 January (CHN TST Mock) Problem 5
Let $P$ be a convex polygon with centroid $G$, and let $\mathcal P$ be the set of vertices of $P$. Let $\mathcal X$ be the set of triangles with vertices all in $\mathcal P$. We sort the elements $\triangle ABC$ of $\mathcal X$ into the following three types:
[list]
[*] (Type 1) $G$ lies in the strict interior of $\triangle ABC$; let $\mathcal A$ be the set of triangles of this type.
[*] (Type 2) $G$ lies in the strict exterior of $\triangle ABC$; let $\mathcal B$ be the set of triangles of this type.
[*] (Type 3) $G$ lies on the boundary of $\triangle ABC$.
[/list]
For any triangle $T$, denote by $S_T$ the area of $T$. Prove that \[\sum_{T \in \mathcal A} S_T \geq \sum_{T \in \mathcal B} S_T.\]
2 replies
Scilyse
Jan 26, 2025
EthanWYX2009
35 minutes ago
Peru IMO TST 2022
diegoca1   0
41 minutes ago
Source: Peru IMO TST 2022 D1 P3
Consider an $n$-gon (a polygon with $n$ sides) with $n \geq 3$. Distinct non-negative integers are assigned to each side and diagonal of the n-gon in such a way that, for any triangle whose vertices are vertices of the n-gon, the numbers assigned to its three sides form an arithmetic progression. Determine the maximum value of $n$ for which this is possible
0 replies
diegoca1
41 minutes ago
0 replies
Peru IMO TST 2022
diegoca1   0
an hour ago
Source: Peru IMO TST 2022 D1P2
Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ such that:
\[ f(xy + f(x))+f(y) = xf(y)+f(x+y) \]for all real numbers $x,y$
0 replies
diegoca1
an hour ago
0 replies
Peru IMO TST 2022
diegoca1   0
an hour ago
Source: Peru IMO TST 2022 D1 P1
Find all positive integers $n$ such that for every integer $k$ with $1 \leq k \leq \sqrt{n}$ is satisfied that :
$\lfloor \frac{n}{k} \rfloor - k$ is odd
0 replies
diegoca1
an hour ago
0 replies
Greatest algebra ever
EpicBird08   21
N an hour ago by smileapple
Source: ISL 2024/A2
Let $n$ be a positive integer. Find the minimum possible value of
\[
S = 2^0 x_0^2 + 2^1 x_1^2 + \dots + 2^n x_n^2,
\]where $x_0, x_1, \dots, x_n$ are nonnegative integers such that $x_0 + x_1 + \dots + x_n = n$.
21 replies
EpicBird08
Jul 16, 2025
smileapple
an hour ago
Creative Inequality
EthanWYX2009   1
N an hour ago by EthanWYX2009
Source: 2025 April 谜之竞赛-4
Given positive integers \( n, k \), let \( S = \{1, 2, \cdots, n\} \). For each subset \( I \) of \( S \), assign a non-negative real number \( X_I \), and define $Y_I = \sum_{J \subseteq I} X_J.$

It is given that \( Y_S = 1 \), and for any subsets \( I, J \subseteq S \), whenever \( I \subseteq J \), $X_I \leq X_J.$ Determine the minimum possible value of
\[\sum_{I \subseteq S} (-1)^{n-|I|} Y_I^{k},\]where \( |I| \) denotes the number of elements in the finite set \( I \).

Created by Cheng Jiang
1 reply
EthanWYX2009
Yesterday at 2:19 PM
EthanWYX2009
an hour ago
Help my diagram has too many points
MarkBcc168   31
N an hour ago by HamstPan38825
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
31 replies
MarkBcc168
Jul 17, 2024
HamstPan38825
an hour ago
Number theory
MathsII-enjoy   1
N 2 hours ago by aaravdodhia
Find all positive intergers $m$ that satisfies: $$\sigma(m^2+8)+\phi(m^2+8)=2m^2+24$$
1 reply
MathsII-enjoy
Jul 19, 2025
aaravdodhia
2 hours ago
I miss Turbo
sarjinius   39
N 2 hours ago by CatinoBarbaraCombinatoric
Source: 2025 IMO P6
Consider a $2025\times2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile.

Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.

Proposed by Zhao Yu Ma and David Lin Kewei, Singapore
39 replies
sarjinius
Jul 16, 2025
CatinoBarbaraCombinatoric
2 hours ago
Guess the leader's binary string!
cjquines0   81
N 2 hours ago by ray66
Source: 2016 IMO Shortlist C1
The leader of an IMO team chooses positive integers $n$ and $k$ with $n > k$, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an $n$-digit binary string, and the deputy leader writes down all $n$-digit binary strings which differ from the leader’s in exactly $k$ positions. (For example, if $n = 3$ and $k = 1$, and if the leader chooses $101$, the deputy leader would write down $001, 111$ and $100$.) The contestant is allowed to look at the strings written by the deputy leader and guess the leader’s string. What is the minimum number of guesses (in terms of $n$ and $k$) needed to guarantee the correct answer?
81 replies
cjquines0
Jul 19, 2017
ray66
2 hours ago
IMO Shortlist 2010 - Problem G1
Amir Hossein   141
N 2 hours ago by TwentyIQ
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
141 replies
Amir Hossein
Jul 17, 2011
TwentyIQ
2 hours ago
Can this sequence be bounded?
darij grinberg   71
N Jun 17, 2025 by Jndd
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
71 replies
darij grinberg
Jan 19, 2005
Jndd
Jun 17, 2025
Can this sequence be bounded?
G H J
G H BBookmark kLocked kLocked NReply
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
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OronSH
1841 posts
#60 • 2 Y
Y by megarnie, cubres
No it cannot be. Consider two consecutive terms $a,b$ with $a<b$ which clearly must exist. (take either $a_0,a_1$ or $a_1,a_2$) then either the next term is $b+a$ giving us another pair $b,b+a$ or it is $b-a$ and then the next term will be $2b-a$ giving us pair $b-a,2b-a.$ In this process the greater term of each pair increases by either $a$ or $b-a.$ Thus the increase is $\ge \min(a,b-a)=x$ which are the difference and the smaller term in our pair. However now notice that $\min(b,(b+a)-b)=\min(a,b)\ge\min(a,b-a)=x$ and $\min(b-a,(2b-a)-(b-a))=\min(b-a,b)\ge\min(a,b-a)=x$ so as long as we repeat this both the smaller term and the difference in each of these pairs will be $\ge x.$ Thus at each step the largest term increases by at least $x$ so it must be unbounded.
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Assassino9931
1527 posts
#61 • 1 Y
Y by cubres
blackbluecar wrote:
The sequence is unbounded...

Notice that $a_2$ and $a_3$ are certainly positive so lets say wlog that $a_0$ and $a_1$ are positive. Notice that $a_k=u \cdot a_0+v \cdot a_1$ for some positive integers $u$ and $v$. Thus, if $b_1,b_2,...$ is an increasing subsequence of $a_1,a_2,...$ then it must be unbounded.

Why must it be unbounded? If e.g. $a_0 = 1$ and $a_1 = 2$, so $a_k = u_k + 2v_k$ and we might have $u_{k+1} = u_k + 2A$, $v_{k+1} = v_k - A$. This would imply $a_{k+1} = a_k$ which is not really possible, as almost all solutions show, but the situation here is evidence that the unboundedness is not too obvious perhaps? What if the increments can be some very small linear combinations of $a_0$ and $a_1$?
This post has been edited 1 time. Last edited by Assassino9931, Mar 12, 2024, 11:20 PM
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asdf334
7584 posts
#62 • 3 Y
Y by MihaiT, Assassino9931, cubres
Assume the sequence is bounded. Notice that no value is equal to zero and so all values are positive (furthermore no adjacent values are equal).

For any value $i$ let $f(i)\in \{i+1,i+2\}$ be a value such that $a_{f(i)}>a_i$. This value exists as it is impossible to have $a_{i+1},a_{i+2}<a_i$.

Then we can create a sequence of values $s_i$ such that
\[a_{s_1}<a_{s_2}<\dots<a_{s_n}<\dots\]and also choose the smallest constant $C$ which is at least each of these values. Write $b_i=a_{s_i}$ for convenience.

Now we can analyze the relationship between three consecutive $s_n$, $s_{n+1}$, and $s_{n+2}$ for large $n$.
  • Notice that $s_{n+2}=s_n+2$ and $s_{n+1}=s_n+1$ cannot happen; this would imply $b_{n+2}=b_{n+1}+b_n>C$.
  • Notice that $s_{n+2}=s_n+4$ and $s_{n+1}=s_n+2$ cannot happen; this would imply
    \[(b_{n+1}-b_n)+(b_{n+2}-b_{n+1})=b_{n+1}.\]
  • Notice that all remaining cases necessarily have $b_n$, $b_{n+1}$, and $b_{n+2}$ in an arithmetic sequence.
Hence we violate the boundedness condition. Done.
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Ywgh1
141 posts
#63 • 1 Y
Y by cubres
First of all, it's clear that $a_n$ is nonnegative for all $n$
and that $a_{n+2}= a_{n+1}-a_{n}$ or $a_{n+2}= a_{n+1}+a_{n}$

Claim : All consecutive terms are different.

Assume FTSOC that $a_n=a_{n+1}$ are the first two consecutive terms.
Then this means $a_{n-1}=0$ implying that $a_{n-2}=a_{n-3}$. A contradiction.

Now let $N$ be a positive integer such that $a_N$ is the maximum term in the sequence. This means that $a_{N+1} \neq a_{N}+a_{N-1}$ as $a_N > a_{N+1}, a_{N-1}$.
So $a_{N+1}= a_{N}-a_{N-1}$, but this forces $a_{N+2}$ to be more than $a_N$, contradiction. Implying that the sequence is unbounded.
This post has been edited 3 times. Last edited by Ywgh1, Aug 9, 2024, 12:18 PM
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Mathandski
776 posts
#65 • 1 Y
Y by cubres
Subjective Rating (MOHs) $       $
Attachments:
This post has been edited 1 time. Last edited by Mathandski, Sep 6, 2024, 11:24 PM
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blueprimes
387 posts
#66 • 1 Y
Y by cubres
This problem I would say lies in a gray area between 5-10 MOHS as it is very easy to fakesolve (i.e, proving a subsequence is increasing but not showing divergence, assuming a maximum term.) Anyways, here is a solution that hasn't been posted yet I believe.
The answer is no, the sequence cannot be bounded. FTSOC, assume it can. Clearly $a_n = a_{n - 1} \pm a_{n - 2}$ for all $n \ge 2$ so define:
  • n is additive if $a_n = a_{n - 1} + a_{n - 2}$
  • n is subtractive if $a_n = a_{n - 1} - a_{n - 2}$
It is easy to show that all terms must be positive. Moreover, we must have an infinite number of subtractive integers as otherwise, the sequence eventually assumes a Fibonacci recurrence which diverges.

Lemma 1: For every additive $n$, the smallest $g > n$ that is subtractive satisfies $a_g \ge a_{n - 2}$.
Proof. Since $a_{n - 2}, a_{n - 1}, \dots, a_{g - 3}$ has a Fibonacci recurrence it is increasing, then
\[a_g = a_{g - 1} - a_{g - 2} = (a_{g - 2} + a_{g - 3}) - a_{g - 2}\ = a_{g - 3} \ge a_{n - 2}. \]
Claim 1: For every subtractive $p$, there exists a larger subtractive $q$ where $2a_p < a_q$.
Proof. Let $s$ be the minimal subtractive integer after $p$, we consider cases on what $s$ is. For brevity let $a_{p - 2} = x, a_{p - 1} = y$, it is easy to check the case $s = p + 1$ fails.

Case 1: $s = p + 2$.
The sequence necessarily has
\[a_{p - 2}, a_{p - 1}, a_p, a_{p + 1}, a_{p + 2}, a_{p + 3} = x, y, y - x, 2y - x, y, 3y - x \]and applying Lemma 1 and letting $n = p + 3$ and $q = g$ yields $a_q \ge 2y - x > 2(y - x) = 2a_p$ as wanted.

Case 2: $s = p + 3$.
We use a similar approach, the sequence has
\[a_{p - 2}, a_{p - 1}, a_p, a_{p + 1}, a_{p + 2}, a_{p + 3}, a_{p + 4} = x, y, y - x, 2y - x, 3y - 2x, y - x, 4y - 3x \]and applying Lemma 1 again with $n = p + 4, q = g$ yields $a_q \ge 3y - 2x > 2(y - x) = 2a_p$ which works.

Case 3: $s \ge p + 4$.
So
\[a_{p - 2}, a_{p - 1}, a_p, a_{p + 1}, a_{p + 2}, a_{p + 3}, a_{p + 4} = x, y, y - x, 2y - x, 3y - 2x, 5y - 3x, \dots \]now Lemma 1 where $n = p + 4, q = g$ gives $a_q \ge 2y - x > 2(y - x) = 2a_p$.

The claim is proven, now recurring it we can extract a subsequence of the original sequence where every proceeding term is more than twice the previous which must diverge. We are done.
This post has been edited 5 times. Last edited by blueprimes, Sep 7, 2024, 5:36 PM
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ezpotd
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#67 • 1 Y
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Assume that the sequence is bounded, this then forces a maximum. We then show some other number in the sequence is bigger than the maximum.

Claim: No two consecutive numbers in the sequence are equal.
Proof: Assume there exists a smallest $i$ with $a_{i} = a_{i + 1}$. This then forces $a_{i - 1} = 0, a_{i - 2} = a_i, a_{i -3} = a_{i - 2}$, so if $i > 3$ we get a contradiction. Otherwise, if $i = 3$ this forces $a_2 = 0$, contradiction, if $i = 2$ this forces $a_1 = 0$, contradiction, if $i = 1$ this violates the problem conditions.

Now take the maximum $m = a_j$, then take $a_{j + 1}$. If it is greater than $m$ we are done, if it is equal to done it violates the claim, contradiction, if it is less than $m$ it forces $a_{j + 2} = m + a_{j + 1}$, done ($a_{j + 2} > m$ since $a_{j + 1} \neq 0$ since $a_{j + 2} = a_{j+ 3}$ by the claim), done.
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dgrozev
2482 posts
#68 • 1 Y
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ezpotd wrote:
Assume that the sequence is bounded, this then forces a maximum. We then show some other number in the sequence is bigger than the maximum.
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Now take the maximum $m = a_j$, then take $a_{j + 1}$. If it is greater than $m$ we are done, if it is equal to done it violates the claim, contradiction, if it is less than $m$ it forces $a_{j + 2} = m + a_{j + 1}$, done ($a_{j + 2} > m$ since $a_{j + 1} \neq 0$ since $a_{j + 2} = a_{j+ 3}$ by the claim), done.

There is an issue. The sequence may not have a maximum term (it's infinite). So, it's not so trivial. But, it can be modified as in #46, and you can take $\limsup a_n$ and get a contradiction.
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ezpotd
1358 posts
#69 • 1 Y
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oops. quick and easy fix:
Assume the sequence is bounded, then the sequence does have a limit supremum $A$. Call a number good if it is at least $0.99A$. Some good number must exist, otherwise the limit supremum would be less than $0.99A$. We force a contradiction

Consider a good number $a_i$ that is followed by a smaller number $a_{i + 1}$. Then we must have $a_{i + 2} = a_{i + 1} + a_{i}$. If $a_{i + 3} = a_{i + 2} - a_{i + 1} = a_{i}$, we have two consecutive good numbers that are decreasing, this forces $a_{i + 4} = a_{i + 3} + a_{i + 2} > A$. If $a_{i + 3} = a_{i + 2} + a_{i + 1}$, either this is greater than $A$ and we are done or we have $a_{i + 3} > a_{i + 2}$. If $a_{i + 4} = a_{i + 3} + a_{i + 2}$ the sum is greater than $A$ and we are instantly done, if $a_{i + 4} = a_{i + 3}  -a_{i + 2} = a_{i + 1}$, we have then proved that $a_{i + 3} = a_{i} + 2a_{i + 1}$, so $a_{i + 3n} = a_{i} + 2n $, obviously contradiction since it would imply $a$ unbounded.

If no good number is ever followed by a smaller number, at some point all the remaining numbers in the sequence are good, implying that the difference between them is at most $0.01A$, which obviously fails.

i just realized how horrible this writeup is but im too lazy.
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shendrew7
818 posts
#70 • 1 Y
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$\boxed{\text{No.}}$ Our solution follows from three observations:
  • All terms must positive and nonzero, with no two consecutive terms equal.
  • Every term is a linear combination of $a_1$, $a_2$, so terms must differ by a multiple of $a_1-a_2$.
  • We always have $a_n < a_{n+1}$ or $a_n < a_{n+2}$, so the max increases by $\ge |a_1-a_2|$ every two terms, which finishes. $\blacksquare$
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Ilikeminecraft
734 posts
#71 • 1 Y
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Note that we have either: $$a_{n + 1} - a_n = a_{n + 2}$$or $$a_{n + 1} + a_n = a_{n + 2},$$and all terms in the sequence are nonnegative. For $n \geq 0, $ if $a_{n + 2} = a_{n + 1} - a_n$, label it with S, and otherwise with A.

I claim that all terms need to be positive. Assume not. Then, there has to be an $a_{n} = a, a_{n + 1} = a, a_{n + 2} = 0.$ If $a_{n + 1}$ is an S, then $a_{n - 1} = 0.$ If it is an A, then $a_{n}$ is still 0. By continuing in this fashion, we will require $a_0$ and $a_1$ to either be equal or one to be 0. This is not allowed.

I claim that we can't have two SS in a row. Assume the previous two are $a, b.$ Then, the first S would be $b - a,$ and the second S would be $b - a - b = -a,$ which contradicts the fact that all terms are positive.

Thus, every time our sequence has an S, both sides must be wrapped in As(if the third term is an S, we just shift the sequence index down by 1).

Now, consider when there is a chain of ASASA\dots . We assume previous two terms are $a, b.$ Then, the sequence would be like $a, b, a + b, a, 2a + b, a + b, 3a + 2b, \ldots.$ Clearly, this sequence is unbounded. Furthermore, by inserting any A anywhere, we are only making the sequence grow faster. Thus, we are done.
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Lemmas
18 posts
#72 • 1 Y
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Easy. we can take that $a_1>a_0$ Take $d=min(a_1 , a_0 , a_1-a_0)$. You can easily prove that $a_i \ge d$ and prove that after some time $a_i$ raises by $d$.
This post has been edited 2 times. Last edited by Lemmas, Apr 28, 2025, 7:20 AM
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Maximilian113
592 posts
#73 • 1 Y
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Clearly each $a_i$ is nonnegative. But if some $a_i=0$ it follows that $a_{i-1}=a_{i-2} \implies a_{i-3}=0,$ and moving downwards yields either one of $a_0, a_1 = 0$ or $a_0=a_1,$ contradiction. It then also follows that no adjacent numbers are equal, and all numbers are positive.

Now for the sake of a contradiction suppose that the sequence is bounded, and the maximum is $N.$ If $a_0=N,$ observe that $a_2=a_1 \pm N$ which is a contradiction in either case. If $a_1=N, a_0=x < N,$ note that $a_2=N-x \implies a_3=a_2 \pm a_1=N-x \pm N \implies a_3=2N-x > N,$ contradiction.

Now if $a_k = N$ with $k \geq 2$ let $a_{k-1}=y, a_{k-2}=x$ for positive $x, y.$ Then $$N=a_{k-1} \pm a_{k-2} \implies N=x+y \implies a_{k+1}=a_k \pm a_{k-1} \implies a_{k+1}=N-y=x.$$Thus $a_{k+2}=N-x \pm N,$ so $a_{k+2} = 2N-x > N,$ contradiction.

Thus the answer is NO.

Oops, this is a fakesolve since the maximum might not be attanable.. perhaps changing $N$ to the running maximum at some point fixes it though, as we could show that this running maximum strictly increases (and such a running maximum exists and is attainable)
This post has been edited 3 times. Last edited by Maximilian113, Apr 28, 2025, 9:49 PM
Reason: bruh
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ezpotd
1358 posts
#74 • 1 Y
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The answer is no.

Claim: The sequence has a smallest element.
Proof: Assume there is no smallest element, say for all $c$ there exists $a_i <c$, then take $c$ small enough such that the first $i$ with $a_i < c$ satisfies $i > 3$, which must exist since the first $3$ terms have a minimum. Now take the first such $i $. However then $a_{i - 1} - a_{i} = a_{i - 2}$, so $a_{ i - 3} = a_i$, so $i$ could not have possibly been the first such index to satisfy $a_i < c$, contradiction.

Let $x$ be the smallest element of the sequence. Call a good element of the sequence any element such that it is greater than its predecessor. We show that given a good element of the sequence of size $b$, we can find a good element with size at least $b + x$ in a finite number of terms. If the element after $b$ is $b - s \ge x$ for some $s$, the next element must be $2b - s \ge b + x$ and this element is good. If the element after $b$ is $b + s$, we are automatically done as $s \ge x$ since $s$ must be in the sequence.

Note that we must have at least one good element, as either $a_2$ or $a_3$ is good, so applying the above claim infinitely gives us that the sequence is unbounded.
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Jndd
1437 posts
#75 • 1 Y
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Consider the case where $a_n>a_{n+1}$ for some $n$. Then, we are forced to have $a_{n+2}=a_{n+1}+a_n$, so $a_{n+2}>a_n>a_{n+1}$. After this, we can either have $a_{n+3}=a_{n+2}-a_{n+1}$, or $a_{n+3}=a_{n+2}+a_{n+1}$.

In the former case, we would have $a_{n+2}>a_{n+3}=a_n>a_{n+1}$, and so we are back in the situation that $a_k>a_{k+1}$ for some $k$ (this time $k=n+2$), and $a_{n+2}$ is larger than $a_n$ by $a_{n+1}$. Additionally, $a_{n+3}>a_{n+1}$.

In the case that $a_{n+3}=a_{n+2}+a_{n+1}$, we have $a_{n+3}>a_{n+2}>a_n>a_{n+1}$. We can continue having $a_k>a_{k-1}$ for each subsequent $k$, but notice if we repeat this for all following terms, the sequence cannot possibly converge, as we add $a_{k}$ to $a_{k+1}$ to get $a_{k+2}$ for each $k$, and each $a_{k+1}$ is greater than $a_k$. Thus, at some point, we must have $a_{m+2}=a_{m+1}-a_m$ for some $m$. Here, we would have $a_{m+1}>a_{m+2}=(a_m+a_{m-1})-a_m=a_{m-1}$, which brings us back to our original situation of having $a_k>a_{k+1}$ for some $k$. Similarly to the previous case, notice that $a_{m+1}$ is larger than $a_n$ by at least $a_{n+1}$. Additionally, $a_{m+2}>a_{n+1}$

Now, suppose we started out with $a_0<a_1$. Then, notice that we are forced to have $a_{k+2}=a_{k+1}-a_k$ for some $k$ instead of having $a_{k+2}=a_{k+1}+a_k$ for all $k$, because in that case, the sequence cannot converge (as explained in the previous paragraph).

We therefore must have $a_k>a_{k+1}$ for infinitely many values of $k\in\mathbb{N}$. Notice that since we have shown that if $m,n$ with $m>n$ are both numbers that satisfy $a_k>a_{k+1}$, then $a_m$ is larger than $a_n$ by at least $a_{n+1}$, which is always at least the value of $a_{t+1}$ where $t$ is the least value in the sequence of integers that satisfies $a_k>a_{k+1}$. Thus, the sequence is unbounded.
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