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

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

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

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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
Trigo or Complex no.?
hzbrl   4
N 28 minutes ago by hzbrl
(a) Let $y=\cos \phi+\cos 2 \phi$, where $\phi=\frac{2 \pi}{5}$. Verify by direct substitution that $y$ satisfies the quadratic equation $2 y^2=3 y+2$ and deduce that the value of $y$ is $-\frac{1}{2}$.
(b) Let $\theta=\frac{2 \pi}{17}$. Show that $\sum_{k=0}^{16} \cos k \theta=0$
(c) If $z=\cos \theta+\cos 2 \theta+\cos 4 \theta+\cos 8 \theta$, show that the value of $z$ is $-(1-\sqrt{17}) / 4$.



I could solve (a) and (b). Can anyone help me with the 3rd part please?
4 replies
hzbrl
May 27, 2025
hzbrl
28 minutes ago
Circumscribed Quadrilateral
billzhao   17
N an hour ago by endless_abyss
Source: USAMO 2004, problem 1
Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
\[ 
\frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|. 
\]
When does equality hold?
17 replies
billzhao
Apr 29, 2004
endless_abyss
an hour ago
Own made functional equation
Primeniyazidayi   2
N 2 hours ago by JARP091
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
2 replies
Primeniyazidayi
May 26, 2025
JARP091
2 hours ago
IMO Shortlist 2008, Geometry problem 2
April   43
N 2 hours ago by ezpotd
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
43 replies
April
Jul 9, 2009
ezpotd
2 hours ago
Quadruple Binomial Coefficient Sum
P162008   3
N 3 hours ago by pineconee
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
3 replies
+1 w
P162008
Yesterday at 8:04 PM
pineconee
3 hours ago
In Cyclic Quadrilateral ABCD, find AB^2+BC^2-CD^2-AD^2
Darealzolt   0
3 hours ago
Source: KTOM April 2025 P8
Given Cyclic Quadrilateral \(ABCD\) with an area of \(2025\), with \(\angle ABC = 45^{\circ}\). If \( 2AC^2 = AB^2+BC^2+CD^2+DA^2\), Hence find the value of \(AB^2+BC^2-CD^2-DA^2\).
0 replies
Darealzolt
3 hours ago
0 replies
Plz give me the solution
Madunglecha   1
N 3 hours ago by top1vien
For given M
h(n) is defined as the number of which is relatively prime with M, and 1 or more and n or less.
As B is h(M)/M, prove that there are at least M/3 or more N such that satisfying the below inequality
|h(N)-BN| is under 1+sqrt(B×2^((the number of prime factor of M)-3))
1 reply
Madunglecha
6 hours ago
top1vien
3 hours ago
King's Constrained Walk
Hellowings   1
N 4 hours ago by Hellowings
Source: Own
Given an n x n chessboard, with a king starting at any square, the king's task is to visit each square in the board exactly once (essentially an open path); this king moves how a king in chess would.
However, we are allowed to place k numbers on the board of any value such that for each number A we placed on the board, the king must be in the position of that number A on its Ath square in its journey, with the starting square as its 1st square.
Suppose after we placed k numbers, there is one and only one way to complete the king's task (this includes placing the king in a starting square), find the minimum value of k set by n.

Didn't know I could post it here xd; I'm unsure how hard this question could be.
1 reply
Hellowings
6 hours ago
Hellowings
4 hours ago
Inspired by qrxz17
sqing   9
N 4 hours ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
9 replies
sqing
Yesterday at 8:50 AM
sqing
4 hours ago
Interesting inequality
sqing   0
5 hours ago
Source: Own
Let $  a, b,c>0,b+c\geq 3a$. Prove that
$$ \sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{1}{\sqrt 2}$$$$ \frac{3}{2}\sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{3}{2\sqrt 2}$$
0 replies
sqing
5 hours ago
0 replies
Inspired by m4thbl3nd3r
sqing   4
N 5 hours ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
4 replies
sqing
Yesterday at 3:43 AM
sqing
5 hours ago
Not so beautiful
m4thbl3nd3r   3
N 5 hours ago by m4thbl3nd3r
Let $a, b,c>0$ such that $b+c>a$. Prove that $$2 \sqrt[4]{\frac{a}{b+c-a}}\ge 2 +\frac{2a^2-b^2-c^2}{(a+b)(a+c)}.$$
3 replies
m4thbl3nd3r
Yesterday at 3:23 AM
m4thbl3nd3r
5 hours ago
2023 Putnam A1
giginori   29
N Yesterday at 10:52 PM by kidsbian
For a positive integer $n$, let $f_n(x)=\cos (x) \cos (2 x) \cos (3 x) \cdots \cos (n x)$. Find the smallest $n$ such that $\left|f_n^{\prime \prime}(0)\right|>2023$.
29 replies
giginori
Dec 3, 2023
kidsbian
Yesterday at 10:52 PM
A MATHEMATICA E BONITA
P162008   0
Yesterday at 7:54 PM
Source: Self made by my Elder brother
Let $K = \sum_{i=0}^{\infty} \sum_{j=0}^{\infty}\sum_{m=0}^{\infty}\sum_{l=0}^{\infty} \frac{1}{(i+j+m+l)!}$ where $i,j,k$ and $l \in W.$

Now, consider the ratio $Z$ defined as
$Z = \frac{\sum_{r=0}^{\lfloor k \rfloor} \sum_{i=0}^{\lfloor k \rfloor} (-1)^r \binom{\lfloor k \rfloor}{r}(\lfloor k \rfloor - r)^i}{\sum_{r=0}^{\lfloor k \rfloor + 1}(-1)^r\binom{\lfloor k \rfloor + 1}{r}(\lfloor k \rfloor + 1 - r)^{\lfloor k \rfloor + 1}}.$

The summation function $S(n)$ is given by
$S(n) = \sum_{j=1}^{n} \left(\binom{n}{j} (j!)\left(\sum_{b=0}^{j} \frac{(-1)^b}{b!}\right)\right)$

Let $p$ denotes the number of points of intersection between the curves
$x^2 + y^2 - \tan(e^x) - \frac{|x|}{\sin y} = 0, (x\sin (a))^y + (y - x\cos(a))^x = |a|.$

Define $A(m)$ as
$A(m) = p\left(\sum_{k=0}^{m} \binom{2m + 1}{k} ((2m + 1) - 2k) (-1)^k\right).$

The value of $X$ is
$X = \lim_{n \to \infty} \frac{\sqrt{n}}{e^n} \text{exp} \left(\int_{0}^{\infty} \lfloor ne^{-x} \rfloor \text{dx}\right).$

And, $8Y =$ Number of subsets of $\left(1,2,3,\cdots,100\right)$ whose sum of elements is divisible by $5.$

Finally, compute the value of $\frac{1}{Z} +S(4) + 1 + e^{A(20)} + X\sqrt{8\pi} + Y.$
0 replies
P162008
Yesterday at 7:54 PM
0 replies
Putnam 2000 A6
ahaanomegas   15
N Apr 6, 2025 by Levieee
Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0, a_1, \cdots $ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n \ge 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.
15 replies
ahaanomegas
Sep 6, 2011
Levieee
Apr 6, 2025
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ahaanomegas
6294 posts
#1 • 3 Y
Y by Rounak_iitr, Adventure10, Mango247
Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0, a_1, \cdots $ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n \ge 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.
Z K Y
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Kent Merryfield
18574 posts
#2 • 9 Y
Y by FinalSix, skrublord420, Centralorbit, math_and_me, Arkajit_Ganguly, Adventure10, Mango247, and 2 other users
If $a_1 = 0,$ then $a_n = 0\  \forall\, n$ and we are done. In all that follows, assume $a_1 \ne 0.$

Let $N$ be the smallest positive integer for which $a_N = 0.$ That there is such an N comes from the hypothesis of the problem that the sequence eventually reaches $0$ and from the well-ordering of the natural numbers. By our assumption, $N \ge 2.$

If for any $k < m, a_k = a_m,$ then for all $n > m,$ if we find that $s, k \le s < m,$ for which $n \equiv s \pmod {(m - k)},$ then $a_n = a_s.$ In other words, from that point on, the sequence $a_n$ is periodic of period $m - k.$ This follows from the fact that each $a_n$ depends only on its predecessor and an induction.

It now follows that we cannot have $a_k = a_m$ for any $0 < k < m < N,$ since if that did occur, the sequence would repeat only those values of $a_n$ for $k \le n < m,$ none of which are zero.

We observe that if $f$ is an integer-coefficient polynomial and $x$ and $y$ are integers, then $(x - y)\, |\, (f(x) - f(y)).$ This follows by induction and linearity from the fact that for each integer power $k, (x - y)\, |\, (x^k - y^k).$ This is clear for $k = 0$ and $1,$ and for larger $k,$ $(x_k - y_k) = (x - y)(x^{k-1} + x^{k-2}y + \cdots + y^{k-1}).$ Then, since $a_{n+2} - a_{n+1} = f(a_{n+1}) - f(a_n),$ we have that for all $n \ge 0,$ $(a_{n+1} - a_n)\, |\, (a_{n+2} - a_{n+1}).$ Each difference divides the next difference. None of these differences are zero (by the previous $a_k \ne a_m$ argument) but since the sequences are periodic, the differences are periodic. That means that no difference can ever be bigger (in absolute value) than any other difference: they are all, except possibly for sign, the same. Since the first difference is $a_1,$ all differences are $\pm a_1.$ That means that the sequence $a_n$ “walks” one step at a time, backwards or forwards, over the set of integer multiples of $a_1.$ This walking sequence starts at $0$ and must return to $0.$ However, from the $a_k \ne a_m$ argument, this sequence can never retrace any steps on the way from 0 to 0, and that means that it can never get two steps away from 0. Its first step is to $a_1;$ its next step cannot be to $2a_1$ so it must be back to $0.$ Hence, $a_2 = 0,$ which is what we were trying to prove.
Z K Y
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JSGandora
4216 posts
#3 • 3 Y
Y by Euler1728, tuanngoc298, Adventure10
Notice that $a_n=f^n(0)$ and $a_0=0$. Since $f^{m}(0)=f(f^{m-1}(0))=0$, we must have $f^{m-1}(0)$ to be a root of $f(x)$. Call this root $r$. Also call the constant term of $f$, $c_0$ and notice that $f(0)=c_0$.

There are two cases, where $c_0=0$ and where $c_0\neq 0$. If $c_0=0$ then $a_1=f(0)=c_0=0$ and we are done. If $c_0\neq 0$, we proceed as follows.

Notice that $c_0 | f(kc_0)$ where $k\in \mathbb{Z}$ since $f$ has integer coefficients. Then we have
\[c_0 | f(0)=c_0\]
so
\[
c_0 | f(0) | f(f(0)) | \cdots | f^{m-1}(0) = r
\]
Hence $c_0|r$. However, since $r$ is clearly an integer, by the rational root theorem, $r|c_0$ and thus we have $c_0=r$. Therefore, $a_2 = f(f(0))=f(c_0)=f(r)=0$.

Therefore, either $a_1=0$ or $a_2=0$ as desired.
Z K Y
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e_plus_pi
756 posts
#4 • 3 Y
Y by GreenKeeper, Adventure10, Mango247
Nice Problem.
Here is my solution:
$\longrightarrow$ Since $f(x)$ is an integer polynomial, we use the fact that $ m-n| f(m)-f(n) \forall m,n \in \mathbb{Z}$.

Let $ b_n =^{DEF} a_{n+1}-a_n$ . Then, $b_n | b_{n+1} \forall n$.

On the other hand, we are given that $a_0 = a_m = 0$ so $a_n$ is cyclic after m terms and so $b_0 = b_m$.

If $b_0 =0 $, then $a_i = a_j \forall i,j\in (0,1,2,\dots,m)$ and we are done. Otherwise $|b_0| = |b_1| =\dots $

So $ b_n = b_0$ or $b_n = - b_0$ and from here the conclusion follows easily.
This post has been edited 1 time. Last edited by e_plus_pi, Apr 26, 2018, 2:50 PM
Reason: Spacing
Z K Y
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Luta_Tnana
18 posts
#5 • 1 Y
Y by Adventure10
Thanks a lot
This post has been edited 1 time. Last edited by Luta_Tnana, Jul 1, 2019, 8:31 PM
Reason: I understood it later
Z K Y
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Luta_Tnana
18 posts
#6 • 1 Y
Y by Adventure10
Oh I get ya ! Sorry
Z K Y
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Ramanujan_math
375 posts
#7 • 1 Y
Y by Adventure10
ahaanomegas wrote:
Let $f(x)$ be a polynomial with integer coefficients. Define a sequence $a_0, a_1, \cdots $ of integers such that $a_0=0$ and $a_{n+1}=f(a_n)$ for all $n \ge 0$. Prove that if there exists a positive integer $m$ for which $a_m=0$ then either $a_1=0$ or $a_2=0$.

This is a question of ISI 2019
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Luta_Tnana
18 posts
#8 • 1 Y
Y by Adventure10
Ya it is !!
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Raunii
28 posts
#11
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JSGandora wrote:
Notice that $a_n=f^n(0)$ and $a_0=0$. Since $f^{m}(0)=f(f^{m-1}(0))=0$, we must have $f^{m-1}(0)$ to be a root of $f(x)$. Call this root $r$. Also call the constant term of $f$, $c_0$ and notice that $f(0)=c_0$.

There are two cases, where $c_0=0$ and where $c_0\neq 0$. If $c_0=0$ then $a_1=f(0)=c_0=0$ and we are done. If $c_0\neq 0$, we proceed as follows.

Notice that $c_0 | f(kc_0)$ where $k\in \mathbb{Z}$ since $f$ has integer coefficients. Then we have
\[c_0 | f(0)=c_0\]so
\[
c_0 | f(0) | f(f(0)) | \cdots | f^{m-1}(0) = r
\]Hence $c_0|r$. However, since $r$ is clearly an integer, by the rational root theorem, $r|c_0$ and thus we have $c_0=r$. Therefore, $a_2 = f(f(0))=f(c_0)=f(r)=0$.

Therefore, either $a_1=0$ or $a_2=0$ as desired.

it may happen that $c=-r$ then what? wont we have to take cases
This post has been edited 1 time. Last edited by Raunii, Mar 20, 2020, 4:03 PM
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peatlo17
77 posts
#12
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I agree with @above. I tried the same approach as @JSGandora with the rational root theorem, but was eventually stuck on the case $c = -r$.

I believe that the only way to resolve this is through the the differences / integer divisibility approach.
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lifeismathematics
1188 posts
#14
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define $d_{i}=a_{i+1}-a_{i}$

Now since $f(x) \in \mathbb{Z}[x]$ , we have $a_{i+1}-a_{i}|f(a_{i+1})-f(a_{i}) =a_{i+2}-a_{i+1}=d_{i+1}$

so we have $d_{i}|d_{i+1}$ $\forall$ $i \geqslant 0$

now we also have $a_{m}=0$ for some $m \in \mathbb{Z}^{+}$ , so $a_{m}=a_{0}$ , and $a_{m+1}=a_{1}$ , hence we have $d_{m}=d_{0}$

hence we have $|d_{0}|=|d_{1}|=|d_{2}|=\cdots =|d_{i}|=\Omega$ for $i \geqslant 0$

Now:

  • if $\Omega=0$

    so we have $a_{1}=0$.
  • if $\Omega \neq 0$

    so we have that the set $\{ a_{1} , a_{2} , \cdots , a_{m}\} \subseteq \mathbb{Z}^{+}_{0}$ , hence by well ordering principle this set has a minimum. say $a_{\ell}$ , so $a_{\ell-1}>a_{\ell}$ , $a_{\ell+1}>a_{\ell}$ , so we have $d_{\ell-1}=-d_{\ell} \implies a_{\ell-1}=a_{\ell+1}$ , now we can take $f$ over until we get $a_{2}=a_{0}=0$ , or if not then we have $f(a_{\ell-2})=f(a_{\ell})\implies a_{\ell-2}=a_{\ell}$ so we do this until we get $a_{2}=a_{0}=0$

Hence either $a_{1}=0$ or $a_{2}=0.$ $\blacksquare$.
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chakrabortyahan
385 posts
#15
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Lol isi 2019 problem...(not nearly a A6 though) in 2nd page there is a typo it would be $f(d)=2d,f(2d) = d$...and there was a third page with the line "Hence $a_2 = 0$ if $a_1\neq 0 $"
$\blacksquare\smiley$
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sanyalarnab
947 posts
#16
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Storage purposes only... one of my most favorite problems :-D
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PRMOisTheHardestExam
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#17
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lemma in imo 2006 p5
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Aiden-1089
302 posts
#18
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We prove the following generalisation:
If a polynomial $f(x) \in \mathbb{Z}[x]$ has a degree $n$ fixed point $a$, then $n=1$ or $2$.

Assume FTSOC that $n>2$.
Note that for all integers $x \neq y$, we have $x-y \mid f(x)-f(y)$.
Let $x_i=f^i(a)-f^{i-1}(a)$ for all positive integers $i$, so $x_{n+i}=x_{i}$. Then we have $x_2 \mid x_3 \mid \cdots \mid x_{n+2} = x_2$, from which we see that $\left| x_i \right|$ is a nonzero constant for all positive integers $i$.
Note that $x_1=x_2$ since $n>2$, and $\sum_{i=1}^n x_i =0$. Since it is not possible that $x_i$ is constant, let $k$ be the smallest positive integer such that $x_{k+2}=-x_1$.
Then we have $f^{i}(a) = ix_1 + a$ for all $1 \leq i \leq k+1$ and $f^{k+2}(a)= kx_1+a = f^k(a)$.
It follows that $f^{k+m}(a) = \begin{cases} f^k(a)=kx_1+a & \text{if } m \text{ is even} \\ f^{k+1}(a)=(k+1)x_1+a &\text{if } m \text{ is odd} \end{cases}$ for all nonnegative integers $m$, so $f^m(a) \neq a$ for all positive integers $m$, which contradicts the fact that $f^n(a)=a$.
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Levieee
244 posts
#19
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Define:= $b_{n}=a_{n+1}-a_{n}$
Lemma 1: if $f(x) \in \mathbb{Z}[x]$ then $x-y \mid f(x)-f(y)$ $\forall x,y \in \mathbb{Z}$

Coming back to the problem,
$f(a_{m})=a_{m+1}$
$f(0)=a_{m+1}$
we know $f(a_{0})=a_{1}$
$f(0)=a_{1}$
$a_{1}=a_{m+1}$
Similarly by induction we can show that $a_{m}=a_{m+k}$ $\forall k \in \mathbb{N}_{0}$
$\therefore$ $(a_{n})_{n\geq 1}$ is a cyclic sequence in $m$ terms

Since we know that the sequence $(a_{n})_{n \geq 1}$ is an integer sequence therefore Lemma 1 is applicable to all the terms in the sequence

Now,
$a_{n+1}-a_{n} \mid f(a_{n+1})-f(a_{n})$
or,$b_{n} \mid b_{n+1}$

from the fact that $a_{n}$ is a cyclic sequence in $m$ terms notice that $b_{m}$ is cyclic in $m$ terms,

$\therefore$ |$b_{0}$|=|$b_{m}$|

We know that $b_{n} \mid b_{n+1}$ and $(b_{n})_{n\geq 1}$ is a cyclic sequence in $m$ terms

therefore $|b_0| = |b_1| =\dots $

Case 1:

$b_0 =0 $

Then $a_i = a_j \forall i,j\in (0,1,2,\dots,m)$ and we are done


Case 2:

$|b_0| = |b_1| =\dots $ $= \alpha$

$ b_n =  b_0$ or $b_n = - b_0$

$a_{n+1}-a_{n}=a_{1}$ or $a_{n+1}-a_{n}=-a_{1}$
Hence, the conclusion follows $\mathbb{Q.E.D}$ $\blacksquare$
:whistling:
This post has been edited 2 times. Last edited by Levieee, Apr 6, 2025, 1:37 PM
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