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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.

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[*]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]
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0 replies
jlacosta
May 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
Inequality
Sadigly   3
N 17 minutes ago by pooh123
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
3 replies
Sadigly
Friday at 7:59 AM
pooh123
17 minutes ago
Calculus
youochange   2
N 21 minutes ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
2 replies
+1 w
youochange
Yesterday at 2:38 PM
youochange
21 minutes ago
A strong inequality problem
hn111009   0
27 minutes ago
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
0 replies
hn111009
27 minutes ago
0 replies
help me please,thanks
tnhan.129   0
32 minutes ago
find f: R+ -> R such that:
f(x)/x + f(y)/y = (1/x + 1/y).f(sqrt(xy))
0 replies
tnhan.129
32 minutes ago
0 replies
Inspired by old results
sqing   0
35 minutes ago
Source: Own
Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2 =3$. Prove that$$\frac{9}{5}>(a-b)(b-c)(2a-1)(2c-1)\geq -16$$
0 replies
sqing
35 minutes ago
0 replies
IMO Shortlist 2011, Algebra 2
orl   43
N an hour ago by ezpotd
Source: IMO Shortlist 2011, Algebra 2
Determine all sequences $(x_1,x_2,\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with \[\sum^{2011}_{j=1} j  x^n_j = a^{n+1} + 1\]

Proposed by Warut Suksompong, Thailand
43 replies
orl
Jul 11, 2012
ezpotd
an hour ago
Sequence inequality
BR1F1SZ   1
N an hour ago by IndoMathXdZ
Source: 2025 Francophone MO Seniors P1
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive integers satisfying the following property: for all positive integers $k < \ell$, for all distinct integers $m_1, m_2, \ldots, m_k$ and for all distinct integers $n_1, n_2, \ldots, n_\ell$,
\[
a_{m_1} + a_{m_2} + \cdots + a_{m_k} \leqslant a_{n_1} + a_{n_2} + \cdots + a_{n_\ell}.
\]Prove that there exist two integers $N$ and $b$ such that $a_n = b$ for all $n \geqslant N$.
1 reply
BR1F1SZ
3 hours ago
IndoMathXdZ
an hour ago
Anything real in this system must be integer
Assassino9931   1
N 2 hours ago by lksb
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
1 reply
Assassino9931
Friday at 9:26 AM
lksb
2 hours ago
Radiant sets
BR1F1SZ   0
3 hours ago
Source: 2025 Francophone MO Juniors P1
A finite set $\mathcal S$ of distinct positive real numbers is called radiant if it satisfies the following property: if $a$ and $b$ are two distinct elements of $\mathcal S$, then $a^2 + b^2$ is also an element of $\mathcal S$.
[list=a]
[*]Does there exist a radiant set with a size greater than or equal to $4$?
[*]Determine all radiant sets of size $2$ or $3$.
[/list]
0 replies
BR1F1SZ
3 hours ago
0 replies
Quadratic system
juckter   35
N Yesterday at 8:17 PM by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
Yesterday at 8:17 PM
Non-homogenous Inequality
Adywastaken   7
N Yesterday at 6:45 PM by ehuseyinyigit
Source: NMTC 2024/7
$a, b, c\in \mathbb{R_{+}}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
7 replies
Adywastaken
Yesterday at 3:42 PM
ehuseyinyigit
Yesterday at 6:45 PM
FE with devisibility
fadhool   2
N Yesterday at 6:44 PM by ATM_
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
2 replies
fadhool
Yesterday at 4:25 PM
ATM_
Yesterday at 6:44 PM
f(m + n) >= f(m) + f(f(n)) - 1
orl   30
N Yesterday at 5:36 PM by ezpotd
Source: IMO Shortlist 2007, A2, AIMO 2008, TST 2, P1, Ukrainian TST 2008 Problem 8
Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

Author: Nikolai Nikolov, Bulgaria
30 replies
orl
Jul 13, 2008
ezpotd
Yesterday at 5:36 PM
Surjective number theoretic functional equation
snap7822   3
N Yesterday at 4:28 PM by internationalnick123456
Source: 2025 Taiwan TST Round 3 Independent Study 2-N
Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
[list=i]
[*] For all $m, n \in \mathbb{N}$, if $m > n$ and $f(m) > f(n)$, then $f(m-n) = f(n)$;
[*] $f$ is surjective.
[/list]
Find the maximum possible value of $f(2025)$.

Proposed by snap7822
3 replies
snap7822
May 1, 2025
internationalnick123456
Yesterday at 4:28 PM
Quadratic system
juckter   35
N Yesterday at 8:17 PM by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
Yesterday at 8:17 PM
Quadratic system
G H J
Source: Mexico National Olympiad 2011 Problem 3
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juckter
323 posts
#1 • 7 Y
Y by itslumi, samrocksnature, jhu08, Adventure10, Mango247, Gato_combinatorio, ehuseyinyigit
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
This post has been edited 1 time. Last edited by juckter, Dec 5, 2016, 2:01 AM
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tensor
433 posts
#2 • 4 Y
Y by aerile, samrocksnature, Adventure10, Mango247
Proof 1:This proof is quite flawed but i think it can be repaired plz give ur suggestions

proof 2
This post has been edited 8 times. Last edited by tensor, Jun 22, 2014, 7:57 PM
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aerile
8 posts
#3 • 5 Y
Y by tensor, samrocksnature, Adventure10, Mango247, BR1F1SZ
tensor wrote:
Suppse in the way of contradiction $\exists k$ such that $a_{k-1}>a_k$
$\implies a_k=a_{k-1}^2+a_{k-1}+1>a_k^2+a_k-1\implies |a_k|<1$
$a_{k-1}>a_k$ does not imply $|a_k|<1$, for example, $a_{k-1}=-1/2 \implies a_k=-5/4$

I think we need to consider f(f(x)) where f(x)=x^2+x-1,
to show $-1<x<0 \implies -1<f(f(x))<x<0$.
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MexicOMM
292 posts
#4 • 5 Y
Y by tensor, samrocksnature, Adventure10, Mango247, ehuseyinyigit
Hint: sum all the equations.
2 Hint: sum 1 to all equations, and then multiply this ones.
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tensor
433 posts
#5 • 3 Y
Y by samrocksnature, Adventure10, Mango247
aerile wrote:
tensor wrote:
Suppse in the way of contradiction $\exists k$ such that $a_{k-1}>a_k$
$\implies a_k=a_{k-1}^2+a_{k-1}+1>a_k^2+a_k-1\implies |a_k|<1$
$a_{k-1}>a_k$ does not imply $|a_k|<1$, for example, $a_{k-1}=-1/2 \implies a_k=-5/4$

I think we need to consider f(f(x)) where f(x)=x^2+x-1,
to show $-1<x<0 \implies -1<f(f(x))<x<0$.

Aerile @... OH REALLY?? but unfortunately ur counterexample doesn't seem to work

$a_{k-1}^2+a_{k-1}-1=\frac 14-\frac 12+1\neq \frac 54=a_k$
This post has been edited 1 time. Last edited by tensor, Jun 22, 2014, 6:45 PM
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aerile
8 posts
#6 • 3 Y
Y by samrocksnature, Adventure10, Mango247
tensor wrote:
$a_{k-1}^2+a_{k-1}-1=\frac 14-\frac 12+1\neq \frac 54=a_k$ :D
You are changing the constant -1 into +1
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tensor
433 posts
#7 • 3 Y
Y by samrocksnature, Adventure10, Mango247
oho noo i should learn how to see my typos thanks friend :wallbash: :wallbash_red:
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aerile
8 posts
#8 • 2 Y
Y by samrocksnature, Adventure10
MexicOMM wrote:
Hint: sum all the equations.
2 Hint: sum 1 to all equations, and then multiply this ones.
That's a nice way. I think the Third Hint will be the AM-GM inequality:
$(\frac{a_1^2+a_2^2+...+a_n^2}{n})^n \geq (a_1a_2...a_n)^2$
with equality if and only if $a_1^2=a_2^2=...=a_n^2$
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randomusername
1059 posts
#9 • 7 Y
Y by aerile, abbosjon2002, pavel kozlov, samrocksnature, Adventure10, Mango247, MarioLuigi8972
Let's compile a complete solution.
juckter wrote:
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 = a_2 +1\\
  a_2^2 + a_2 = a_3 +1\\
  \hspace*{3.3em} \vdots \\
 a_{n}^2 + a_n = a_1 +1\]

Adding these $n$ equations we get $a_1^2+a_2^2+\dots+a_n^2=n$. (1)
Multiplying these equations we arrive at $\prod_{i=1}^n (a_i^2+a_i)=\prod_{i=1}^n(a_i+1)$, so we get
$(a_1a_2\dots a_n-1)(a_1+1)(a_2+1)\dots (a_n+1)=0$. (2)
Now in (2) we may have $a_i=-1$ for some $i$. Then we get $a_{i+1}=a_i^2+a_i-1=1-1-1=-1$, $a_{i+2}=-1$, and so on. It is easy to see that $a_1=a_2=\dots=a_n=-1$ is in fact a solution of the system.
Next consider the more general case $a_1\dots a_n=1$. Using the AM-GM inequality, we get
$1=\frac{a_1^2+\dots+a_n^2}n\ge \root{n}\of{a_1^2a_2^2\dots a_n^2}=1$,
where equality only holds if $a_1^2=a_2^2=\dots=a_n^2=A$. Put $a=\sqrt A$, then all the $a_i$ are $\pm a$.
The system becomes ($a_{n+1}:=a_1$)
$a_{i+1}-a_i=A-1$, $i=1,2,\dots,n$.
Adding these for $i=1,\dots,n$ gives $n(A-1)=0$, $A=1$, $a=1$. This means $a_1=\dots=a_n=+1$ or $=-1$. Both are solutions.
Therefore there are two solutions:
$a_1=\dots=a_n=1$ and $a_1=\dots=a_n=-1$.
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aerile
8 posts
#10 • 4 Y
Y by tensor, samrocksnature, Adventure10, Mango247
tensor wrote:
Proof 1:This proof is quite flawed but i think it can be repaired plz give ur suggestions
Actually, $a_{k-1}>a_{k}$ implies $a_{k-1}>a_{k-1}^2+a_{k-1}-1$.
Solving this quadric inequality, the thing we have is $-1<a_{k-1}<1$.

$|a_k|<1$ holds if $0<a_{k-1}<1$, so the remaining is $-1<a_{k-1}\leq0$.
A repairment for that case is using f(f(x)).

Let $f(x)=x^2+x-1$, then $f(f(x))=...=x^4+2x^3-x-1$.
Assume with the supposition $-1<x\leq0$,
[1] $f(f(x))<x$ and [2] $-1<f(f(x))\leq0$ are shown.

Since $a_{k+1}=f(f(a_{k-1}))$, [1] implies $a_{k+1}<a_{k-1}$ and [2] says $a_{k+1}$ also satisfies the supposition, so we can repeat to use [1] to show the chain:
$...a_{k+3}<a_{k+1}<a_{k-1}$, which also gives a contradiction.

Proof for [1] and [2]
[1]: $x-f(f(x))=x-(x^4+2x^3-x-1)=...=(1-x)(1+x)^3,$
which is positive when $-1<x\leq0$.
[2]: From [1] we have $f(f(x))<x\leq0$. For the other side, $f(f(x))-(-1)=x^4+2x^3-x=x(x+1)(x^2+x-1)$,
which will be also found out to be positive with the supposition.
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codyj
723 posts
#11 • 2 Y
Y by samrocksnature, Adventure10
Notice that if we find the right value of $a_1$, our system of equations is less of a system of equations than a system of definitions! Let $f(x)=x^2+x-1$ and $f^k(x)=f(f(\dots f(x)\dots))$ where $f$ is composed $k$ times. We seek all real solutions $x$ to $x=f^n(x)$.

We can prove inductively the following results: For $x>1$, $f^n(x)>x$. For $-1<x<1$, $f^n(x)<x$. For $x<-1$, $f^n(x)>x$. Therefore, the only solutions may occur at $x=\pm1$. Indeed, these are solutions as $f^n(x)=x$. Therefore, the only solutions are $(1,1,\dots,1,1)$ and $(-1,-1,\dots,-1,-1)$.
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AopsUser101
1750 posts
#12 • 3 Y
Y by v4913, samrocksnature, Mango247
Summing, we see that $\sum _{i = 1} ^ n a_i^2 = n$. We write each equation as $a_i(a_i + 1) = 1 + a_{i + 1}$ for all integer $i \in [1,n-1]$ and $a_n(a_n+1) = a_1$; multiplying all of them yields that $\prod_{i = 1}^n a_i = 1$. By the AM-GM inequality, we know that:
$$\sum_{i = 1}^n a_i^2 \ge n\sqrt[n]{\prod_{i = 1}^n a_i^2}=n$$Since equality clearly holds, $a_1^2 = a_2^2 = ... = a_n^2 = 1$. Plugging in $a_1 = 1$ into the first equation, we see that $a_1 = a_2$ and it follows that $a_1 = a_2 = ... = a_n$. Therefore, we see that the only possible solutions are $a_1 = a_2 = .... = a_n = \pm 1$, which we can check do indeed work.
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pad
1671 posts
#13 • 1 Y
Y by samrocksnature
We have
\[ a_{k+1}+1 = a_k(a_k+1).\]Taking the cyclic product over all $k$, we get $a_1\cdots a_n=1$. Taking the cyclic sum of the original sum, we get $a_1^2+\cdots+a_n^2 = n$. By AM-GM, $a_1^2+\cdots+a_n^2 \ge n (a_1\cdots a_n)^{2/n} = n$, and since equality is achieved, $|a_1|=\cdots=|a_n|=1$.

If $a_i=1$ for some $i$, then $a_{i+1}=1$, and so on for all $a$'s, and similarly if $a_i=-1$ then all are -1. So either all are 1 or all are -1.
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wolfpack
1274 posts
#14 • 1 Y
Y by samrocksnature
This solution felt natural and much more easier to find without any high level manipulations.

Solution
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juckter
323 posts
#15 • 2 Y
Y by Emathmaster, samrocksnature
Doesn't seem obviously correct, for values close to $-1$ the sequence can oscillate wildly between things greater than $-1$ and things smaller than $-1$.
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Jay17
268 posts
#16 • 1 Y
Y by samrocksnature
sol
This post has been edited 1 time. Last edited by Jay17, Sep 13, 2020, 9:52 PM
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Zorger74
760 posts
#17 • 1 Y
Y by samrocksnature
Solved with andyxpandy99.

Solution
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JustKeepRunning
2958 posts
#19
Y by
Very nice problem! Similar flavor to Shortlist 2018 A2, but more intuitive.

The answers are $\{a_i\}=\{1\}$ and $\{a_i\}=\{-1\}.$ These work. Notice that if any one of the $a_i$ is equal to $1$ or $-1,$ the rest must be as well, so from now on, we will assume that no $a_i$ is equal to $1$ or $-1$. Furthermore, there can also be no $a_i=0,$ as that would make $a_{i+1}$(indices taken $\pmod n$) equal to $-1,$ a contradiction.

Summing, we get that $\sum a_i^2 = n,$ and factoring, we get that $a_1(a_1+1)=a_2+1$. Taking the product of cyclic permutations, we get that $\prod a_i(a_i+1)=\prod (a_i+1),$ and since $a_i\neq 0,-1\forall 1\leq i\leq n,$ we have that $\prod a_i = 1$. However, by AM-GM, we have that $\frac{\sum a_i^2}{n}\geq 1,$ and because equality holds, we get that the entire sequence is equal up to sign. From here, it is just a matter of checking cases!
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Camilaeon
17 posts
#20
Y by
Summing every equation gives that the sum of the squares of the sequence is exactly n (this means that the quadratic mean is 1). Adding 1 to both sides and multiplying gives that, as long as no number in the sequence is - 1, the product is also 1. However, Q(x) >= A(x) >= G(x) this means that A(x) = 1. Thus the sum of every number in the sequence is n, but this means that the sums of the squares equals the regular sum of every term. This implies that a_i=1 for all i unless it is - 1 for some i.
Going into this second case, if a term is - 1 then the next is also - 1, thus a_n is - 1. This means that a_1 is also - 1 and the whole sequence is - 1.
This post has been edited 1 time. Last edited by Camilaeon, Aug 31, 2021, 8:10 AM
Reason: Correction in the proof
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blackbluecar
303 posts
#21 • 1 Y
Y by centslordm
Let $P(x)=x^2+x-1$. Notice that $a_1=P^n(a_1)$, but by IMO 2006/5 $P^n$ has at most two fixed points which in this case are at $1$ and $-1$. Thus, either $a_1=1$ or $a_1=-1$ for which it follows that all terms are equal.
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juckter
323 posts
#22
Y by
blackbluecar wrote:
Let $P(x)=x^2+x-1$. Notice that $a_1=P^n(a_1)$, but by IMO 2006/5 $P^n$ has at most two fixed points which in this case are at $1$ and $-1$. Thus, either $a_1=1$ or $a_1=-1$ for which it follows that all terms are equal.

This solution doesn't work as stated since IMO 2006/5 says that $P^n$ has at most two integer fixed points.
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blackbluecar
303 posts
#23
Y by
@above oops yes you are right, thanks for pointing that out. Back to the drawing board :blush:
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IAmTheHazard
5001 posts
#24
Y by
The answer is $(1,\ldots,1)$ and $(-1,\ldots,-1)$, both of which clearly work. Take indices modulo $n$, so $a_{k+1}=a_k^2+a_k-1$ (when we say the sequence is periodic, this is by considering the infinite extension of the sequence in periodic manner). We now present two solutions to show that no other $n$-tuples work.

Solution 1 (very bad!!!): If $a_i=-1$ then all of them are $-1$ as well, hence suppose none of them are $-1$. Add one to both sides of the equation to get $a_{k+1}+1=a_k(a_k+1)$. Since no $a_i$ equals $-1$, we have $a_1\ldots a_n=1$. On the other hand, summing gives $a_1^2+\cdots+a_n^2=n$. But by AM-GM we have
$$n=a_1^2+\cdots+a_n^2\geq n(a_1\ldots a_n)^{2/n}=n,$$hence $a_1=\cdots=a_n$, which means all of them are either $1$ or $-1$, so we extract the given answer. $\blacksquare$

Solution 2 (very good!!!) By summing all the equations we have $a_1^2+\cdots+a_n^2=n$, so there must exist some $i$ with $|a_i| \geq 1$. It is clear that if $a_i>1$, $a_{i+1}>a_i$, but this is absurd as it makes a periodic sequence strictly increasing, so we either have $a_i=1$ for all $i$, $a_i=-1$ for all $i$, or there exists some $i$ such that $a_i<-1$. We will show that the third case is impossible. Indeed, suppose $a_i=-1-a$ for $x>0$. Since we have $x^2+x-1 \geq -5/4$ for all $x \in \mathbb{R}$, we must have $a_i \geq -5/4$ for all $i$, hence $a \in (0,1/4]$. Now we can calculate $a_{i+1}=-1+(a^2+a)$, and
$$a_{i+2}=-1-(a-(2a^3+a^4)).$$We can check that $a>2a^3+a^4$ for $a \in (0,1/4]$ (factor it as $a(a+1)(a^2+a-1)<0$), so $a_{i+2}+1$ is also negative but also strictly greater than $a_i+1$. Thus, we have $a_{i+2}>a_i$ for all $i$, but this is again impossible as $(a_i)$ is periodic. From this, we either have $a_i=1~\forall i$ or $a_i=-1~\forall i$the desired solution. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Apr 17, 2022, 2:28 PM
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jasperE3
11320 posts
#25
Y by
Take indices$\pmod n$. We have the equations $a_i^2+a_i=a_{i+1}+1$ for each $i$.
If $a_i=-1$ for any $i$, then:
$$a_{i+1}=a_i^2+a_i-1=-1$$so by induction $(a_1,a_2,\ldots,a_n)=\boxed{(-1,-1,\ldots,-1)}$ which is indeed a solution. Otherwise, suppose that $a_i\ne-1$ for all $i$.

Summing the equations $a_i^2+a_i=a_{i+1}+1$ for $1\le i\le n$ gives:
$$\sum_{i=1}^na_i^2+\sum_{i=1}^na_i=\sum_{i=1}^na_{i+1}+n$$so
$$\sum_{i=1}^na_i^2=n.$$Multiplying these equations gives:
$$\prod_{i=1}^na_i(a_i+1)=\prod_{i=1}^n(a_{i+1}+1)$$so
$$\prod_{i=1}^na_i=1.$$
But by QM-AM, we have:
$$1=\sqrt{\frac1n\sum_{i=1}^na_i^2}\ge\sqrt[n]{\prod_{i=1}^na_i}=1,$$so equality holds and all of the $a_i$ are equal. Rearranging the first equation gives that $a_1^2=1$, and since $a_1\ne-1$ we have $(a_1,a_2,\ldots,a_n)=\boxed{(1,1,\ldots,1)}$ which is indeed a solution.
This post has been edited 1 time. Last edited by jasperE3, Apr 16, 2022, 2:57 PM
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IAmTheHazard
5001 posts
#26
Y by
jasperE3 wrote:
if $a_i\le-1$ then $a_2\le-1$

If you mean $a_{i+1}$ instead of $a_2$, this is false—take $a_i=-5/4$, so $a_{i+1}=-11/16>-1$. In fact, if $a_i \leq -1$, we have $a_{i+1} \geq -1$
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cj13609517288
1916 posts
#27
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Note: The version of the problem I was given asks for the number of solutions, but the solution works out the same either way.

Note that if $a_k=-1$ for any $k$, then everything becomes $-1$. We will add in this case later, so assume that $a_k+1\ne 0$ for now.

First, sum over all $k$ to get
\[a_1^2+a_2^2+\dots+a_n^2=n.\]
Now rewrite our original equation as
\[\frac{a_{k+1}+1}{a_k+1}=a_k\]and multiply over all $k$ to get
\[a_1a_2\dots a_n=1.\]
Thus
\[\sqrt{\frac{a_1^2+a_2^2+\dots+a_n^2}{n}}=\sqrt[n]{a_1a_2\dots a_n}.\]This implies QM-GM for the variables $|a_1|,|a_2|,\dots,|a_n|$ and only has the equality case of everything being equal.

If $x$ goes to $-x$, then $x^2+x-1=-x$ so $x=-1\pm\sqrt2$. We can manually check that for each of these cases, $-x$ doesn't go to either of $x$ or $-x$. Thus the sequences are indeed constant.

Plugging back in, the sequence must be all $1$'s or all $-1$'s, so the answer is $\boxed{2}$.
This post has been edited 8 times. Last edited by cj13609517288, Apr 19, 2023, 6:29 PM
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YaoAOPS
1541 posts
#28
Y by
If we sum cyclically, we get \[ a_{1}^2 + a_{2}^2 + \dots + a_{n}^2 = n\]Also, note that
\[ a_{k+1} + 1 = a_k(a_k + 1) \]As such, \[ a_1a_2\dots a_n(a_i + 1) = (a_i + 1) \]If $a_i = -1$ for some $i$, then all $a_i$ equal $-1$. Else, assume $a_1a_2\dots a_n = 1$ and no $a_i = -1$. By AM-GM, \[ \frac{a_1^2 + a_2^2 + \dots + a_n^2}{n} \ge \sqrt[n]{a_1^2a_2^2\dots a_n^2} = 1 \]Equality can only hold if $a_1^2 = a_2^2 = \dots = a_n^2 = 1$, so thus $a_1 = a_2 = \dots = a_n = 1$.
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Mathandski
757 posts
#29
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Subjective Rating (MOHs) $       $
Attachments:
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AshAuktober
1005 posts
#30
Y by
Firstly add all the equations to get $\sum_i a_i^2 = n$
Now, rewrite the equations as $$a_i(a_i + 1) = a_{i+1} + 1.$$$i = 1, 2, \cdots. n$.
Then we have two cases:
Case 1: Some $a_i$ is $-1$.
Then evidently, $$a_{i+1} + 1 = 0 \implies a_{i+1} = -1$$and so on, so $$a_i = -1 \forall i.$$Case 2: No $a_i$ is $-1$.
Take the cyclic product of all the equations to get $$\prod_{i} a_i \cdot \prod_i (a_i + 1) = \prod_i (a_i + 1).$$But $\prod_i (a_i+1) \ne 0$, so $$\prod_{i} a_i  = 1 \implies \prod_i a_i^2 = 1.$$Now AM-GM gives us $$1 = \frac{\sum_i a_i^2}{n} \ge \left(\prod_i a_i^2\right)^{1/n} = 1.$$Since equality holds, $a_1^2 = \cdots = a_n^2 = 1$. But no $a_i$ is $-1$, so all $a_i$ are $1$, yielding $a_i = 1 \forall i$.
Since $\boxed{(a_1, \cdots, a_n) = (-1, \cdots, -1)}$ and $\boxed{(a_1, \cdots, a_n) = (1, \cdots, 1)}$ both clearly work, we're done. $\square$
This post has been edited 2 times. Last edited by AshAuktober, Dec 13, 2024, 12:11 PM
Reason: typo
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Ihatecombin
60 posts
#31
Y by
Claim 1:
\(a_{k+2} = a_{k}^4 + 2a_{k}^3 - a_{k} - 1\)
Proof:
The problem condition is
\[a_{k+1} = a_{k}^2 + a_{k} - 1\]We can substitute this value of \(a_{k+1}\) into the problem condition to obtain
\[a_{k+2} = a_{k+1}^2 + a_{k+1} - 1 = {(a_{k}^2 + a_{k} - 1)}^2 + (a_{k}^2 + a_{k} - 1) -1 = a_{k}^4 + 2a_{k}^3 - a_{k} - 1\]QED

Claim 2:
If \(-1 < a_{k} < 1\), then \(|-1-a_{k}| > |-1-a_{k+2}|\), in other words \(a_{k+2}\) will get closer to \(-1\).
Proof:
We can represent \(a_{k}\) as \(-1+\delta\) for some positive \(\delta < 2\). Substituting this value into the equation for \(a_{k+2}\) gives us
\[a_{k+2} = {(-1+\delta)}^4 + 2{(-1+\delta)}^3 - (-1+\delta) -1 = \delta^4 - 2\delta^3 + \delta - 1\]Our goal is to show that
\[\delta > |-1 - (\delta^4 - 2\delta^3 + \delta - 1)|\]Simplifying, we have
\[|-1 - (\delta^4 - 2\delta^3 + \delta - 1)| = |\delta^4 -2\delta^3 + \delta| = \delta \cdot |(\delta-1)(\delta^2-\delta-1)|\]Thus it suffices to show
\[|(\delta-1)(\delta^2-\delta-1)| < 1\]Which can be bashed. QED

Claim 3:
If \(\frac{-1-\sqrt{5}}{2} < x < - 1\), then \(|-1-a_{k}| > |-1-a_{k+2}|\), in other words \(a_{k+2}\) will get closer to \(-1\).
Proof:
We substitute \(a_{k} = -1 + \delta\) like before, however this time for negative delta (I don't want to reexpand the equation lol). Again it suffices to show
\[-\delta > |\delta(\delta-1)(\delta^2-\delta-1)|\]However notice that
\[-\delta > |\delta(\delta-1)(\delta^2-\delta-1)| \iff 1 > |(\delta-1)(\delta^2-\delta-1)|\]for \(\frac{1-\sqrt{5}}{2}< \delta < 0\). Which is true since \(\frac{1-\sqrt{5}}{2}\) is a root of \((\delta^2-\delta-1)\). QED
Claim 4:
If \(-2< a_{k} < \frac{-1-\sqrt{5}}{2}\), then \(1 > a_{k+2} > - 1\)
Proof:
More boring subs. QED
Claim 5:
If there is an \(a_{k} > 1\), then
\[a_{k+1} > a_{k}\]Also if there is an \(a_{k} < -2\), then \(a_{k+1} > 1\)
Proof:
Easy substitution into the original equation. QED

Let \(n\) denote the \(n\)-th composition of \(x^4+2x^3-x-1\), notice that
\[a_{1} = f^{n}(a_{1})\]Claims 1 until 5 imply that \(a_{1} = \{1,-1\}\). Thus the only possible sequences are \(a_{i} = 1\) for all \(i\) or \(a_{i} = -1\) for all \(i\).
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Primeniyazidayi
98 posts
#32
Y by
Nice problem:
If any of them is equal to -1,then all of them are -1.If not,then multiplying the equations gives us that $a_1a_2...a_n=1$.But we have that $\sum_{i=1}^n a_i^2= n$ and defining the sequence $b_i = |a_i|$ gives us $b_i =1$ via AM-GM.This gives us all the results.
This post has been edited 3 times. Last edited by Primeniyazidayi, Apr 3, 2025, 11:39 AM
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blueprimes
353 posts
#33
Y by
We claim the only solutions are $(a_1, a_2, \dots, a_n) = (-1, -1, \dots, -1), (1, 1, \dots, 1)$ which clearly work.

Now note that any $a_k = -1$ is sufficient enough to imply the solution set $(-1, -1, \dots, -1)$, so assume otherwise. We can rearrange the relation $a_{k + 1} = a_k^2 + a_k - 1$ (indices $\pmod{n}$) to $a_k^2 = a_{k + 1} - a_k + 1$ which summing over all $k$ telescopes to $a_1^2 + a_2^2 + \dots + a_n^2 = n$. Moreover, we also have $\dfrac{a_{k + 1}  + 1}{a_k + 1} = a_k$ which multiplying over all $k$ gives $a_1 a_2 \dots a_n = 1$.

So $\dfrac{|a_1|^2 + |a_2|^2 + \dots + |a_n|^2}{n} = \sqrt[n]{|a_1|^2 |a_2|^2 \dots |a_n|^2}$ which is the equality case of AM-GM. Then $|a_1| = |a_2| = \dots = |a_n|$ easily gives $(1, 1, \dots, 1)$ as a solution set, done.
This post has been edited 2 times. Last edited by blueprimes, Apr 18, 2025, 3:40 AM
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Ilikeminecraft
627 posts
#34
Y by
Addingi all the equations, we get that $\sum a_n^2 = n.$ We also have that $a_{k + 1} + 1 = a_k^2 + a_k,$ and so $\prod a_n = 1.$ Thus, by AM-GM, we have that $a_k = \pm 1.$ If $a_0 = 1,$ then the rest are all 1. If $a_0 = -1,$ the rest are -1.
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Maximilian113
575 posts
#36
Y by
Observe that $$a_k=\frac{a_{k+1}+1}{a_k+1}$$so multiplying these yields $\prod a_i = 1.$ Meanwhile $$a_k^2-1 = a_{k+1}-a_k \implies \sum a_i^2 = n.$$But by the Power Mean Inequality $$\sqrt{\frac{\sum a_i^2}{n}} \geq \sqrt[n]{\prod a_i} \iff 1=1$$so equality in fact holds and all $a_i$ are equal. $x^2+x-1=x \iff x=\pm 1$ so the only solutions are $$(a_1, a_2, \dots, a_n) = (1, 1, \dots, 1), (-1, -1, \dots, -1).$$
this felt like 2015 ISL A1
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torch
1013 posts
#37
Y by
The answer is $\boxed{(a_1, a_2, \dots, a_n)=\pm(1, 1, \dots, 1)}$. Summing the condition over $k$ gives $a_1^2+a_2^2+\dots+a_n^2 = n$. The condition can also be manipulated to $\frac{a_{k+1}+1}{a_k+1}=a_k$, which gives $a_1a_2\dots a_n=1$ by multiplying over $k$. This gives the equality case of AM-GM on $(a_1^2, a_2^2, \dots, a_n^2),$ which implies $a_1=a_2=\dots=a_n=\pm 1$. Note that if $a_k=1$, then $a_{k+1}=1$ as well, and likewise, if $a_k=-1$, then $a_{k+1}=-1$. The result follows by induction.
Maximilian113 wrote:
this felt like 2015 ISL A1
well they're both in alg manip
This post has been edited 2 times. Last edited by torch, Apr 29, 2025, 12:30 AM
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shendrew7
796 posts
#38
Y by
We claim our only solution are $\boxed{(-1, \ldots, -1), (1, \ldots, 1)}$. We aim to show that there are no other solutions.

Consider dividing the parabola $y=x^2+x-1$ into four regions $A$, $B$, $C$, $D$ with y-coordinates $\ge 1$, $[0,1]$, $[-1,0]$, $\leq -1$. Then we can see that the mapping $x \rightarrow x^2+x-1$ maps
\[A \mapsto A, B \mapsto \text{down}, C \mapsto D, D \mapsto C.\]
We see that the only cycle of length greater than 1 we could have is the involution $C \mapsto D \mapsto C$. Thus the only other solutions we could have are from
\[(x^2+x-1)^2+(x^2+x-1)-1=x \implies (x-1)(x+1)^3 = 0.\]
Thus there are no other solutions. $\blacksquare$
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