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

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Be sure to mark your calendars for the following events:
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
postaffteff
JetFire008   9
N 3 minutes ago by drago.7437
Source: Internet
Let $P$ be the Fermat point of a $\triangle ABC$. Prove that the Euler line of the triangles $PAB$, $PBC$, $PCA$ are concurrent and the point of concurrence is $G$, the centroid of $\triangle ABC$.
9 replies
JetFire008
Mar 15, 2025
drago.7437
3 minutes ago
A scary fish and a fiend
nukelauncher   96
N 10 minutes ago by Mathandski
Source: USA November TST for IMO 2021 and TST for EGMO 2021, Problem 2, by Zack Chroman and Daniel Liu
Let $ABC$ be a scalene triangle with incenter $I$. The incircle of $ABC$ touches $\overline{BC},\overline{CA},\overline{AB}$ at points $D,E,F$, respectively. Let $P$ be the foot of the altitude from $D$ to $\overline{EF}$, and let $M$ be the midpoint of $\overline{BC}$. The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and $Q$, respectively. Show that the incenter of triangle $GQM$ coincides with $D$.

Zack Chroman and Daniel Liu
96 replies
+1 w
nukelauncher
Nov 16, 2020
Mathandski
10 minutes ago
Iterated FE on positive integers
MarkBcc168   61
N 15 minutes ago by pi271828
Source: ELMO 2020 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that $$f^{f^{f(x)}(y)}(z)=x+y+z+1$$for all $x,y,z \in \mathbb{N}$.

Proposed by William Wang.
61 replies
MarkBcc168
Jul 28, 2020
pi271828
15 minutes ago
IMO 2023 P2
799786   89
N an hour ago by kaede_Arcadia
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
89 replies
799786
Jul 8, 2023
kaede_Arcadia
an hour ago
combinatorıc
o.k.oo   0
an hour ago
A total of 3300 handshakes were made at a party attended by 600 people. It was observed
that the total number of handshakes among any 300 people at the party is at least N. Find
the largest possible value for N.
0 replies
o.k.oo
an hour ago
0 replies
Problem 2, Olympic Revenge 2013
hvaz   66
N an hour ago by MonkeyLuffy
Source: XII Olympic Revenge - 2013
Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.
66 replies
hvaz
Jan 26, 2013
MonkeyLuffy
an hour ago
Functional equation wrapped in f's
62861   35
N 2 hours ago by ihatemath123
Source: RMM 2019 Problem 5
Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]for all real numbers $x$ and $y$.
35 replies
62861
Feb 24, 2019
ihatemath123
2 hours ago
JBMO Shortlist 2023 C1
Orestis_Lignos   6
N 2 hours ago by zhenghua
Source: JBMO Shortlist 2023, C1
Given is a square board with dimensions $2023 \times 2023$, in which each unit cell is colored blue or red. There are exactly $1012$ rows in which the majority of cells are blue, and exactly $1012$ columns in which the majority of cells are red.

What is the maximal possible side length of the largest monochromatic square?
6 replies
Orestis_Lignos
Jun 28, 2024
zhenghua
2 hours ago
Number Theory
MuradSafarli   6
N 2 hours ago by krish6_9
Find all natural numbers \( a, b, c \) such that

\[
2^a \cdot 3^b + 1 = 5^c.
\]
6 replies
1 viewing
MuradSafarli
6 hours ago
krish6_9
2 hours ago
Equilateral triangle geo
MathSaiyan   1
N 2 hours ago by ricarlos
Source: PErA 2025/3
Let \( ABC \) be an equilateral triangle with circumcenter \( O \). Let \( X \) and \( Y \) be two points on segments \( AB \) and \( AC \), respectively, such that \( \angle XOY = 60^\circ \). If \( T \) is the reflection of \( O \) with respect to line \( XY \), prove that lines \( BT \) and \( OY \) are parallel.
1 reply
MathSaiyan
Yesterday at 1:47 PM
ricarlos
2 hours ago
IMO 2009, Problem 5
orl   86
N 3 hours ago by Ilikeminecraft
Source: IMO 2009, Problem 5
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b + f(a) - 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

Proposed by Bruno Le Floch, France
86 replies
orl
Jul 16, 2009
Ilikeminecraft
3 hours ago
Diagonals BD,CE concurrent with diameter AO in cyclic ABCDE
WakeUp   10
N 3 hours ago by zhenghua
Source: Romanian TST 2002
Let $ABCDE$ be a cyclic pentagon inscribed in a circle of centre $O$ which has angles $\angle B=120^{\circ},\angle C=120^{\circ},$ $\angle D=130^{\circ},\angle E=100^{\circ}$. Show that the diagonals $BD$ and $CE$ meet at a point belonging to the diameter $AO$.

Dinu Șerbănescu
10 replies
WakeUp
Feb 5, 2011
zhenghua
3 hours ago
Parallel lines in two-circle configuration
Tintarn   3
N 3 hours ago by zhenghua
Source: Francophone 2024, Senior P3
Let $ABC$ be an acute triangle, $\omega$ its circumcircle and $O$ its circumcenter. The altitude from $A$ intersects $\omega$ in a point $D \ne A$ and the segment $AC$ intersects the circumcircle of $OCD$ in a point $E \ne C$. Finally, let $M$ be the midpoint of $BE$. Show that $DE$ is parallel to $OM$.
3 replies
Tintarn
Apr 4, 2024
zhenghua
3 hours ago
IMO Shortlist 2013, Algebra #5
lyukhson   33
N 3 hours ago by HamstPan38825
Source: IMO Shortlist 2013, Algebra #5
Let $\mathbb{Z}_{\ge 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\ge 0} \rightarrow \mathbb{Z}_{\ge 0} $ satisfying the relation
\[ f(f(f(n))) = f(n+1 ) +1 \]
for all $ n\in \mathbb{Z}_{\ge 0}$.
33 replies
lyukhson
Jul 9, 2014
HamstPan38825
3 hours ago
Can this sequence be bounded?
darij grinberg   66
N Yesterday at 2:34 AM by shendrew7
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
66 replies
darij grinberg
Jan 19, 2005
shendrew7
Yesterday at 2:34 AM
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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megarnie
5533 posts
#55 • 1 Y
Y by ImSh95
The answer is no.

Suppose $(a_i)$ was bounded.

First we get that $a_i\ge 0$ for all $i$.

Claim: $a_n \ge 0$ for all $n$.
Proof: Suppose some $n$ satisfied $a_n = 0$. Note $n\ge 2$. Then $a_{n+1} = a_{n-1}, a_{n-1} = a_{n-2}$, which gives that $a_{n-3} = 0$. Repeating this process we find that $a_0 = 0, a_1= 0$, or $a_2 = 0$, which implies $a_2 = 0$. But then $a_0 = a_1$, contradiction. $\square$


Claim: $a_{n+3}\ge a_n$ for all $n$.
Proof: Note that $a_{n+2}\in \{a_{n+1} - a_n, a_{n+1} + a_n\}$ and $a_{n+3}\in \{a_{n+2} - a_{n+1}, a_{n+2} + a_{n+1}\}$ by the problem condition.
If $a_{n+2} = a_{n+1} + a_n$, then $a_{n+3} \in \{a_n, 2a_{n+1} + a_n\}$, which implies $a_{n+3} \ge a_n$.

If $a_{n+2} = a_{n+1} - a_n$, then $a_{n+3}\in \{-a_n, 2a_{n+1}  - a_n\}$. Since $a_{n+3}\ge 0$, $a_{n+3} = 2a_{n+1} - a_n$. Since $a_{n+1}\ge a_n$, $a_{n+3}\ge 2a_n - a_n = a_n$, as desired. $\square$

Now, the three sequences $(a_{3i}), (a_{3i+1}), (a_{3i+2})$ are each increasing. Since $(a_i)$ is bounded they are all eventually constant, so $a_n = a_{n+3}$ for all $n\ge C$ for some constant $C$.

Let $a_C = x, a_{C+1} = y$. Note that $a_{C+2} \in \{y-x, y+x\}$ and $y = |a_{C+3} - a_{C+2}| = |x - a_{C+2}|$.

If $a_{C+2} = y-x$, then $|2x-y| = y$. Since $x>0$, we have $x=y$. This is absurd because it implies $a_{C-1} = 0$, contradiction.
Thus, $a_{C+2} = y+x$. Now we have $y+ x = a_{C+2}  = |a_{C+3} - a_{C+4}| = |x-y|$, which implies either $x=0$ or $y=0$, contradiction.
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awesomeming327.
1664 posts
#56 • 1 Y
Y by ImSh95
Clearly, $a_n$ is nonnegative for all $n$ and $a_{n+2}=a_{n+1}-a{n}$ or $a_{n+1}+a_n$. Let $n$ be the smallest index such that $a_n=a_{n+1}$. Then, $a_{n-1}=0.$ However, that implies $a_{n-2}=a_{n-3}$, contradiction. Therefore, consecutive terms are different.

Suppose that the sequence is bounded, then let $m$ be the maximum value and $M$ be the index such that $a_M=m$. Note that $a_{M-1}$, $a_{M+1} < m$. $a_{M+1}$ can't be $a_M+a_{M-1}$ so it has to be $a_M-a_{M-1}$. Now, $a_{M+1} < a_M$ so $a_{M+2}$ cannot be $a_{M+1}-a_M$ so it must be $a_{M+1}+a_M=2a_M-a_{M-1}>a_M$, contradiction. Therefore, the sequence is not bounded.
This post has been edited 1 time. Last edited by awesomeming327., Feb 17, 2023, 9:38 PM
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BobsonJoe
811 posts
#57 • 1 Y
Y by ImSh95
Note that $a_{n+2} = a_{n+1} + a_n$ or $a_{n+2} = a_{n+1} - a_n$. If the former case is always true, then we are trivially done. Otherwise, there exists an $n$ with $a_n > a_{n+1}$. Shift the sequence so that $a_0 > a_1$.

Lemma 1: For all $n$, $a_n \ge a_1$.
Proof: We use induction, base cases $n=0, 1$. We can check that $a_2 = a_0 + a_1 > a_1$. For $n > 2$, if $a_{n} = a_{n-1} + a_{n-2}$, we are done. Otherwise,
\[a_{n} = a_{n-1} - a_{n-2} = a_{n-3} + a_{n-2} - a_{n-2} = a_{n-3} \ge a_1\]
Lemma 2: For all $n$, either $a_{n+1} \ge a_n + a_1$ or $a_{n+2} \ge a_n + a_1$.
Proof: If $a_{n+1} = a_n + a_{n-1} \ge a_n + a_1$ are we done. Otherwise
\[a_{n+1} = a_n - a_{n-1} < a_n \implies a_{n+2} = a_{n+1} + a_n \ge a_n + a_1\]
Hence, we can construct an infinite subsequence, each term at least $a_1$ greater than the previous, so the sequence must be unbounded.
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popop614
264 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
1865 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
1720 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
1197 posts
#61
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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
7579 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
136 posts
#63
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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
711 posts
#65
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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
303 posts
#66
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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
1246 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
2459 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.
---
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
1246 posts
#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
790 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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