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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
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.

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0 replies
jlacosta
Thursday at 11:16 PM
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
Diophantine Equation with prime numbers and bonus conditions
p.lazarov06   10
N 9 minutes ago by mathbetter
Source: 2023 Bulgaria JBMO TST Problem 3
Find all natural numbers $a$, $b$, $c$ and prime numbers $p$ and $q$, such that:

$\blacksquare$ $4\nmid c$
$\blacksquare$ $p\not\equiv 11\pmod{16}$
$\blacksquare$ $p^aq^b-1=(p+4)^c$
10 replies
+1 w
p.lazarov06
May 7, 2023
mathbetter
9 minutes ago
Concurrence in Cyclic Quadrilateral
GrantStar   39
N 11 minutes ago by ItsBesi
Source: IMO Shortlist 2023 G3
Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$.

Prove that lines $AD, PM$, and $BC$ are concurrent.
39 replies
GrantStar
Jul 17, 2024
ItsBesi
11 minutes ago
IMO Genre Predictions
ohiorizzler1434   22
N 17 minutes ago by rhydon516
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
22 replies
ohiorizzler1434
Today at 6:51 AM
rhydon516
17 minutes ago
Inequality
MathsII-enjoy   1
N an hour ago by arqady
A interesting problem generalized :-D
1 reply
MathsII-enjoy
2 hours ago
arqady
an hour ago
Inequality
lgx57   2
N an hour ago by mashumaro
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
2 replies
lgx57
an hour ago
mashumaro
an hour ago
Find min
lgx57   1
N an hour ago by arqady
Source: Own
Find min of $\dfrac{a^2}{ab+1}+\dfrac{b^2+2}{a+b}$
1 reply
lgx57
an hour ago
arqady
an hour ago
Product is a perfect square( very easy)
Nuran2010   1
N an hour ago by SomeonecoolLovesMaths
Source: Azerbaijan Junior National Olympiad 2021 P1
At least how many numbers must be deleted from the product $1 \times 2 \times \dots \times 22 \times 23$ in order to make it a perfect square?
1 reply
Nuran2010
3 hours ago
SomeonecoolLovesMaths
an hour ago
smo 2018 open 2nd round q2
dominicleejun   7
N an hour ago by Kyj9981
Let O be a point inside triangle ABC such that $\angle BOC$ is $90^\circ$ and $\angle BAO = \angle BCO$. Prove that $\angle OMN$ is $90$ degrees, where $M$ and $N$ are the midpoints of $\overline{AC}$ and $\overline{BC}$, respectively.
7 replies
dominicleejun
Aug 15, 2019
Kyj9981
an hour ago
inequalities
Cobedangiu   4
N an hour ago by Nguyenhuyen_AG
$a,b,c>0$ and $\sum ab=\dfrac{1}{3}$. Prove that:
$\sum \dfrac{1}{a^2-bc+1}\le 3$
4 replies
Cobedangiu
Today at 4:06 AM
Nguyenhuyen_AG
an hour ago
Two equal angles
jayme   2
N 2 hours ago by jayme
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
2 replies
jayme
Yesterday at 6:52 AM
jayme
2 hours ago
minimum of \sqrt{\frac{a}{b(3a+2)}}+\sqrt{\frac{b}{a(3b+2)}}
parmenides51   11
N 2 hours ago by sqing
Source: JBMO Shortlist 2017 A2
Let $a$ and $b$ be positive real numbers such that $3a^2 + 2b^2 = 3a + 2b$. Find the minimum value of $A =\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}} $
11 replies
parmenides51
Jul 25, 2018
sqing
2 hours ago
Interesting inequalities
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ (a+b)^2 (a+c)^2=16abc. $ Prove that
$$ 2a+b+c\leq \frac{128}{27}$$$$ \frac{9}{2}a+b+c\leq \frac{864}{125}$$$$3a+b+c\leq 24\sqrt{3}-36$$$$5a+b+c\leq \frac{4(8\sqrt{6}-3)}{9}$$
1 reply
sqing
Yesterday at 2:35 PM
sqing
3 hours ago
Geometry Problem
Euler_Gauss   0
3 hours ago
Given that $D$ is the midpoint of $BC$, $DM$ bisects $\angle ADB$ and intersects $AB$ at $M$, $I$ is the incenter of $\triangle {}{}{}ABD$, $AT$ bisects $\angle BAC$ and intersects the circumcircle of \(\triangle {}{}ABC\) at $T$, $MS$ is parallel to $BC$ and intersects $AT$ at $S$. Prove that $\angle MIS + \angle BIT = \pi.$
0 replies
Euler_Gauss
3 hours ago
0 replies
Tricky invariant with 3 numbers on the board
Nuran2010   0
3 hours ago
Source: Azerbaijan Junior National Olympiad 2021
Initially, the numbers $1,1,-1$ written on the board.At every step,Mikail chooses the two numbers $a,b$ and substitutes them with $2a+c$ and $\frac{b-c}{2}$ where $c$ is the unchosen number on the board. Prove that at least $1$ negative number must be remained on the board at any step.
0 replies
Nuran2010
3 hours ago
0 replies
Can this sequence be bounded?
darij grinberg   69
N Apr 28, 2025 by Maximilian113
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
69 replies
darij grinberg
Jan 19, 2005
Maximilian113
Apr 28, 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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popop614
271 posts
#58 • 1 Y
Y by ImSh95
If someone can please make sure that this solution is right, that would be appreciated, thank you! (I'm really iffy on this solution but it might work)

The sequence must be unbounded. Observe that all terms are positive for obvious reasons.

As a start, note that $a_{n+2} = a_{n+1} \pm a_n$ for some choice of sign. Now the sequence either starts increasing, or it starts decreasing; but if it starts decreasing, then it must increase immediately after that since we can't have any negative terms. Henceforth assume $a_0 < a_1$. Moreover assume that $a_1 > a_2$, as otherwise we either find a term where that happens or the sequence is trivially unbounded (always addition).

Call a value $a_i$ a $difference$ if $a_i = a_{i-1} - a_{i-2}$.
$\textbf{Claim.}$ The smallest difference is $a_1 - a_0$.
$\textit{Proof.}$ We have our sequence starting $a_0$, $a_1$, $a_1 - a_0$. We are forced to have $a_3 = 2a_1 - a_0$. Subtracting now gets us a difference of $a_1 > a_1 - a_0$. Adding again and subtracting after that, we obtain a difference of $a_1 - a_0$.

Observe that if we have $a$, $b$, $a + b$ as the sequence, the difference will be $a$; i.e. after a sufficient amount of adding, the difference will be $2$ terms back. But after the second addition we just simply keep looking further at the increasing sequence $a_1 - a_0, 2a_1 - a_0, 3a_1 - 2a_0$ $\cdots$ so the next difference will always be at least $a_1-a_0$.

Now if at any point in the sequence we have $a$, $b$, $b-a$, we are forced to add the difference $b-a$ to $b$; if this happens finitely many times again the sequence is trivially unbounded (infinite adding wahoo) but if it happens infinitely many times, we add a difference of at least $a_1 - a_0$ each time; thus it is unbounded.
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cj13609517288
1900 posts
#59
Y by
The answer is no, the sequence is always unbounded.

Clearly each of these terms are positive, and no two consecutive terms are equal. Consider pairs of consecutive terms(the same pair appearing multiple times in the sequence counts as a different pair). If a pair is decreasing, call it a fruity pair.

If there are finitely many fruity pairs, then the sequence is eventually strictly increasing, so it is unbounded. Therefore, there are infinitely many fruity pairs. So consider what happens when you move from one fruity pair to the next. I claim that either (the second number increases) or (it stays the same and the first one increases).

Suppose that the first fruity pair is $(a,b)$. Then the next term has to be $a+b$, and the term after that can either be $a$(in that case we have completed our larger fruity pair) or $a+2b$. The next term can be $b$(in that case we have completed our larger fruity pair) or $2a+3b$. Eventually this is going to get to $F_{k}a+F_{k+1}b$ and $F_{k+1}a+F_{k+2}b$, and when we finally decide to subtract the next term is $F_{k-1}a+F_{k}b$ which is obviously greater than $b$.

Therefore our claim is true. If the second number keeps staying the same, then the first number is unbounded. If the second number doesn't keep staying the same, then the second number is unbounded. $\blacksquare$
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OronSH
1730 posts
#60 • 1 Y
Y by megarnie
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
1317 posts
#61
Y by
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
7585 posts
#62 • 2 Y
Y by MihaiT, Assassino9931
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
139 posts
#63
Y by
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
756 posts
#65
Y by
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
345 posts
#66
Y by
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
1262 posts
#67
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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
2463 posts
#68
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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
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#69
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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
794 posts
#70
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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
612 posts
#71
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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
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#72
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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
574 posts
#73
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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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