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

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

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
2 var inquality
Iveela   20
N a minute ago by TigerOnion
Source: Izho 2025 P1
Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$. Prove that \[(a-b)^2 + a + b \geq 2\]
20 replies
Iveela
Jan 14, 2025
TigerOnion
a minute ago
Four concyclic points
jayme   2
N 13 minutes ago by whwlqkd
Source: own?
Dear Mathlinkers,

1. ABC an A-isoceles triangle
2. (O) the circumcircle
3. D a point on the segment AC
4. E le second point d’intersection de (O) avec (BD)
5. Te the tangent to (O) at E
6. Z the point of intersection of the parallel to BC through D and Te
7. Y the second point of intersection of CZ and (O).

Question : E, D, Y et Z are cocyclic.

Sincerely
Jean-Louis
2 replies
jayme
an hour ago
whwlqkd
13 minutes ago
Root of polynomial
Eeightqx   0
13 minutes ago
Source: 2025 China South East Mathematical Olympiad Grade10 P5
Suppose $\alpha,\,\beta,\,\gamma\ne1$ are roots of $f(x)=x^3+ax^2+bx+a+2\in\mathbb R[x]$ where $2a+b\ge24$. Prove that
$$\dfrac1{\sqrt[3]{\alpha-1}}+\dfrac1{\sqrt[3]{\beta-1}}+\dfrac1{\sqrt[3]{\gamma-1}}\le0.$$
0 replies
Eeightqx
13 minutes ago
0 replies
van der Waerden Theorem
steven_zhang123   0
21 minutes ago
Source: 2025 CSMO Grade 11 P8
Do there exist sets \( A \) and \( B \) such that \( A \cap B = \emptyset \), \( A \cup B = \mathbb{N} \), and for every positive integer \( d \), there exists an integer \( l > 2 \) such that neither \( A \) nor \( B \) contains an arithmetic progression of common difference \( d \) and length \( l \)? Prove your answer.
0 replies
1 viewing
steven_zhang123
21 minutes ago
0 replies
can you pls give me the reason?
youochange   2
N 25 minutes ago by maromex

Evaluate the integral:
\[
\int_0^{2\pi} \frac{4\sin x + 8\cos x}{5\sin x + 4\cos x} \, dx
\]
Click to reveal hidden text
2 replies
youochange
an hour ago
maromex
25 minutes ago
Density of a Recurrence-Defined Sequence
steven_zhang123   0
27 minutes ago
Source: 2025 CSMO Grade 11 P7
Given a positive real number \( c \), consider a sequence of positive integers \( \{a_n\} \) with \( a_1 \equiv 3 \pmod{4} \) and \( a_{n+1} = a_n + 4 \left\lfloor c \cdot \sqrt[3]{a_n} \right\rfloor \) for \( n \geq 1 \). Prove that there exist a real number \( r \) and $2025$ distinct positive integers \( k \) such that \( \left\lfloor 10^{2^kr} \right\rfloor \) is a term of \( \{a_n\} \) for each such \( k \). (Here \( \lfloor x \rfloor \) is the greatest integer \( \leq x \).)
0 replies
steven_zhang123
27 minutes ago
0 replies
Some of my less-seen proposals
navid   2
N 27 minutes ago by ayeen_izady
Dear friends,

Since 2003, I have had several nice days in AOPS-- i.e., Mathlinks; as some of you may remember. I decided to share you some of my less-seen proposals. Some of them may be considered as some early ethudes; several of them already appeared on some competitions or journals. I hope you like them and this be a good starting point for working on them. Please take a look at the following link.

https://drive.google.com/file/d/1bntcjZAHZ-WN1lfGbNbz0uyFvhBTMEhz/view?usp=sharing

Best regards,
Navid.
2 replies
navid
Yesterday at 7:08 PM
ayeen_izady
27 minutes ago
Compatible Fillings of a Toroidal Grid
steven_zhang123   0
28 minutes ago
Source: 2025 CSMO Grade 11 P6
Let \( m \) be a positive integer. Initially, each cell of a \( 3 \times 3 \) grid is filled with a number from \( \{1, 2, \dots, 2m\} \) (repetitions allowed). The cell at row \( i \) and column \( j \) is denoted \( (i, j) \), and rows/columns are indexed modulo 3 (e.g., cell \( (4, 2) \) is \( (1, 2) \)). An operation consists of choosing integers \( i, j \), and adding 1 to the cells \( (i, j) \), \( (i+1, j) \), \( (i, j+1) \), \( (i+1, j+1) \). An initial filling is "compatible" if finitely many operations can make all cells equal. Find the number of compatible fillings in terms of \( m \).
0 replies
steven_zhang123
28 minutes ago
0 replies
Concyclic points
Eeightqx   0
30 minutes ago
Source: 2025 China South East Mathematical Olympiad Grade10 P6
Given $\triangle$$ABC$. $E,\,F$ lies on $AC,\,AB$ resp., such that $B,\,C,\,E,\,F$ concyclic. $EF$ cuts $\odot(ABC)$ at $X,\,Y$. $BE$ cuts $CF$ at $P$. $XP,\,YP$ cut $\odot(BPC)$ again at $M,\,N$ resp. Prove that $X,\,Y,\,M,\,N$ concyclic and $A$ is the center of the circle.
0 replies
Eeightqx
30 minutes ago
0 replies
Geometric Equality in a Triangle
steven_zhang123   0
31 minutes ago
Source: 2025 CSMO Grade 11 P5
In triangle \( ABC \), points \( E \) and \( F \) are taken on sides \( AC \) and \( AB \) respectively, such that \( EF \parallel BC \). The angle bisector of \(\angle BAC\) intersects \( BC \) at \( D \). Point \( P \) lies on the extension of line \( DA \) beyond \( A \), satisfying \(\angle PBE = \angle ABC\) and \(\angle PCF = \angle ACB\). Prove that \( AP = AD \).
0 replies
steven_zhang123
31 minutes ago
0 replies
The result before GMA Problem 575
mihaig   0
an hour ago
Source: Farfuridi
Given $n\geq2.$ Prove that $\sqrt{\left[\frac{n^2}4\right]}$ is the least constant $K$ such that
$$\sum_{i=1}^{n}{a_i}+K\left(\sqrt{a_1}-\sqrt{a_n}\right)^2\geq\sqrt{n\sum_{i=1}^{n}{a_i^2}}$$holds true for all $a_1\geq\cdots\geq a_n\geq0,$ with $a_1>0.$
0 replies
mihaig
an hour ago
0 replies
Digit Sum Equality in Multiple Bases
steven_zhang123   1
N an hour ago by sarjinius
Source: 2025 Hope League Test 2 P4
Let \( S_q(a) \) denote the sum of the digits of the positive integer \( a \) in base \( q \), where \( q \) is an integer greater than or equal to 2. Find the largest positive integer \( k \) such that for any infinite subset \( S \) of positive integers, there exist integers \( 1 < q_1 < q_2 < \cdots < q_{100} \) and \( k \) distinct positive integers \( a_1, a_2, \cdots, a_k \) in \( S \) satisfying \( S_{q_i}(a_1) = S_{q_i}(a_2) = \cdots = S_{q_i}(a_k) \) for each \( i = 1, 2, \cdots, 100 \).
Proposed by Wu Zhuo
1 reply
steven_zhang123
Jul 24, 2025
sarjinius
an hour ago
Polynomial with prime
Eeightqx   0
an hour ago
Source: 2025 China South East Mathematical Olympiad Grade10 P8
Find all positive integer pairs $(p,\,m)$ where $p$ is a prime such that the polynomial
$$f(x)=x^{2m}-px^m+p^2$$can be written as the product of two rational coefficient polynomial whose degree less than $2m$.
0 replies
1 viewing
Eeightqx
an hour ago
0 replies
A symmetric inequality in n variables (9)
Nguyenhuyen_AG   1
N 2 hours ago by Nguyenhuyen_AG
Let $a_1,a_2,\ldots,a_n \, (n \geqslant 1)$ be positive real numbers. Prove that
\[(n - 1)^2 \displaystyle \sum_{i=1}^{n} \frac{a_i}{\displaystyle \sum_{j=1}^{n} a_j - a_i} + 4n \cdot \frac{\left( \displaystyle \sum_{i=1}^{n} a_i \right)^2 - \displaystyle \sum_{i=1}^{n} a_i^2}{\left(\displaystyle \sum_{i=1}^{n} a_i \right)^2} \geqslant (n - 1)(n + 4).\]Case $k=3:$ https://artofproblemsolving.com/community/c6h395973
1 reply
Nguyenhuyen_AG
2 hours ago
Nguyenhuyen_AG
2 hours ago
Integer Polynomial with P(a)=b, P(b)=c, and P(c)=a
Brut3Forc3   30
N Jun 6, 2025 by megahertz13
Source: 1974 USAMO Problem 1
Let $ a,b,$ and $ c$ denote three distinct integers, and let $ P$ denote a polynomial having integer coefficients. Show that it is impossible that $ P(a) = b, P(b) = c,$ and $ P(c) = a$.
30 replies
Brut3Forc3
Mar 13, 2010
megahertz13
Jun 6, 2025
Integer Polynomial with P(a)=b, P(b)=c, and P(c)=a
G H J
G H BBookmark kLocked kLocked NReply
Source: 1974 USAMO Problem 1
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Brut3Forc3
1948 posts
#1 • 2 Y
Y by Mathlover_1, Adventure10
Let $ a,b,$ and $ c$ denote three distinct integers, and let $ P$ denote a polynomial having integer coefficients. Show that it is impossible that $ P(a) = b, P(b) = c,$ and $ P(c) = a$.
Z K Y
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Agr_94_Math
881 posts
#2 • 2 Y
Y by Adventure10, Mango247
$ a,b,c \in \mathbb{Z}$, and $ a,b,c$ are distinct.
Since we have $ P[\mathbb{Z}]$, suppose we have by the given conditions to be exisitng such that
$ a-b | b-c$
$ b-c | c-a$
$ c-a | a-b$.
As we are dealing over integers,
$ 2b\ge c+a$
$ 2c\ge a+b$
$ 2a\ge b+c$
THis however implies equaltiy holds everywhere and therefore contradiction.
Z K Y
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Zhero
2043 posts
#3 • 8 Y
Y by cycliccircle, TheThor, Adventure10, and 5 other users
I don't think that's directly implied from the divisibilities you've arrived at. What if $ a - b > 0$ and $ b - c < 0$?

What I did
Z K Y
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KWTLEO
5 posts
#4 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Agr_94_Math wrote:
$ a,b,c \in \mathbb{Z}$, and $ a,b,c$ are distinct.
Since we have $ P[\mathbb{Z}]$, suppose we have by the given conditions to be exisitng such that
$ a - b | b - c$
$ b - c | c - a$
$ c - a | a - b$.
you can arrive the result immediately by using the fact that
If $ x|y$ and $ y|x$, then $ |x|=|y|$.
Z K Y
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yugrey
2326 posts
#5 • 2 Y
Y by rd123, Adventure10
Well, thinking about the constant term (let it be z), b=z mod a. Similarly, c= z mod b, a=z mod c. b=z mod a=z mod (z mod(a)) =z mod z-na for some n, which means c=na, and c is a multiple of a. Similarly, a is a multiple of b, and b is in turn a multiple of c, so it follows that they are all equal, as a=1b, b=1c, and c=1a, as otherwise all the numbers would be unequal to themselves.

Q.E.D.

P.S. Oh my god, did I just do a USAMO problem? I didn't even qualify for AIME!! :D :lol: :) But they're much harder nowadays. Also, by = I meant congruence, sorry, and I obviously used the parentheses for order. I think, however, that it's a good solution. Edit: If there is something flawed with my solution, or if it is correct, tell me. If it is correct, I'd love to add this brief solution to the AoPS wiki, but I don't know latex. :blush:
Z K Y
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yugrey
2326 posts
#6 • 2 Y
Y by Adventure10, Mango247
Oh, there is a flaw. A number that is zmod z-na isn't necessarily equal to z-na.
Z K Y
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facis
642 posts
#8 • 2 Y
Y by Adventure10, Mango247
funny. A very similar problem was just used this year for the University of Illinois Undergrad Contest. It was pretty much this problem plus also the case of $P(a)=b$ and $P(b)=a$.
Z K Y
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hrithikguy
1791 posts
#9 • 3 Y
Y by MRenkhtur, Adventure10, Mango247
Suppose that $a < b < c$. Then $|P(a) - P(c)| = |b-a|< |a-c|$, a contradiction.
Next, suppose that $b < a < c$. Then $|P(b) - P(c)| = |c - a| < |c-b|$, also a contradiction.

Note that these are both contradictions because $|a-c|$ divides $|P(a) - P(c)|$ and $|c-b|$ divides $|P(b) - P(c)|$.
Z K Y
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cobbler
2180 posts
#10 • 2 Y
Y by Adventure10 and 1 other user
Using the well-known fact that $(a-b)\mid P(a)-P(b)$ we get
  • $\frac{P(a)-P(b)}{a-b}=\frac{b-c}{a-b}=k_1$
  • $\frac{P(b)-P(c)}{b-c}=\frac{c-a}{b-c}=k_2$
  • $\frac{P(c)-P(a)}{c-a}=\frac{a-b}{c-a}=k_3$

where the $k_i$ are integral.

Multiplying the first and last equations we have that \[\left(\frac{P(a)-P(b)}{a-b}\right)\left(\frac{P(c)-P(a)}{c-a}\right)=\frac{b-c}{c-a}=k_1k_3.\] But $\frac{b-c}{c-a}=\frac{1}{k_2}$, so the only way for $k_2$ to be an integer is if $|k_2|=1$. Similarly we get that $|k_1|=|k_3|=1$. If one of $k_1,k_2,k_3$ is -1, WLOG say $k_1$, then $\tfrac{b-c}{a-b}=-1\Leftrightarrow{b-c=-a+b}\Leftrightarrow{a=c}$, contradiction. Thus, $k_1,k_2,k_3$ are all equal to 1, and plugging these in it follows that $2a=b-c$, $2b=c-a$ and $2c=a-b$. Solving this system yields $a=b=c=0$, but $a,b,c$ are distinct integers by definition, a contradiction.
Z K Y
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sayantanchakraborty
505 posts
#11 • 2 Y
Y by Adventure10, Mango247
These conditions imply that [a-b]<=[b-c]<=[c-a]<=[a-b],where [a]=modulus of a, so equality holds everywhere.
Thus the fact that a,b,c are distinct implies (a-b)=(b-c)=(c-a).Solving we get a=b=c. Contradiction!
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bobthesmartypants
4336 posts
#12 • 1 Y
Y by Adventure10
Agr_94_Math wrote:
$ a,b,c \in \mathbb{Z}$, and $ a,b,c$ are distinct.
Since we have $ P[\mathbb{Z}]$, suppose we have by the given conditions to be exisitng such that
$ a-b | b-c$
$ b-c | c-a$
$ c-a | a-b$.
As we are dealing over integers,
$ 2b\ge c+a$
$ 2c\ge a+b$
$ 2a\ge b+c$
THis however implies equaltiy holds everywhere and therefore contradiction.

How did he immediately arrive at the conclusion:

$ a-b | b-c$
$ b-c | c-a$
$ c-a | a-b$
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Abe27342
146 posts
#13 • 2 Y
Y by Adventure10, Mango247
It's well-known that $a-b|P(a)-P(b)$ for integers $a$ and $b$ and a polynomial $P$ with integral coefficients. To see why, write out the expression $P(a)-P(b)$ and then refactor it so you get a lot of terms of the form $a^k-b^k$ for some $k$. Then $a-b$ divides all of these terms, and $P(a)-P(b)$ is a linear combination of terms of this form, so $a-b|P(a)-P(b)$.
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tastymath75025
3223 posts
#14 • 2 Y
Y by Adventure10, Mango247
Here's what I did:
Since $\frac{a-b}{b-c}, \frac{b-c}{c-a}, \frac{c-a}{a-b}$ are all integers and their product is equal to $1$, all three of them are each either $1$ or $-1$.
At least one of the fractions equals 1, so WLOG $\frac{a-b}{b-c}=1$. THen either both of the remaining fractions equal 1 or both equal -1, so we examine the two cases. Both cases give us that $a=b=c$, contradiction.
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bobthesmartypants
4336 posts
#15 • 1 Y
Y by Adventure10
Okay I guess I wasn't thinking. I forgot that $P(a)=b$ and $P(b)=c$. :|
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Wave-Particle
3690 posts
#16 • 3 Y
Y by luka2002, Adventure10, Mango247
Here is a solution I wrote up a while ago:

Lemma 1: If $P(x)$ is a polynomial with integer coefficients then $a-b|P(a)-P(b)$ where $a,b$ are integers.

Proof: Let $P(x)=a_n*x^n+a_{n-1}x^{n-1}...+a_1x+a_0$. We have $P(a)=a_n*a^n+a_{n-1}a^{n-1}...+a_1a+a_0$ and $P(b)=a_nb^n+a_{n-1}b^{n-1}...a_1b+a_0$. So $P(a)-P(b)=(a^n-b^n)a_n+(a^{n-1}-b^{n-1})a_{n-1}+...+(a-b)a_1$. Notice that $a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+...b^{k-1})$ so it is easy to see $a-b|P(a)-P(b)$.

We will prove this by contradiction. By Lemma 1 we have $a-b|P(a)-P(b)$, $b-c|P(b)-P(c)$, and $c-a|P(c)-P(a)$. We know what $P(a), P(b),$ and $P(c)$ are as they were given in the problem so we substitute them in. Let $\frac{b-c}{a-b}=k$, $\frac{c-a}{b-c}=m$, and $\frac{a-b}{c-a}=t$ where $k,m,t$ are integers. Note that $kmt=1$ and because $k,m,t$ are integers it follows either all of them are $1$ or two of them are $-1$ and the other is $1$. Let's split this into two cases:

Case 1: $k=m=t=1$.

From here we have $b-c=a-b$, $c-a=b-c$, and $a-b=c-a$. This gives us that $a=b=c$ but the problem said $a,b,c$ are distinct so this case is ruled out.

Case 2: WLOG let $k=m=-1$ and $t=1$.

This means that $a=b=c$ again, meaning that it is impossible for $P(a)=b$, $P(b)=c$, and $P(c)=a$.
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strategos21
491 posts
#17 • 2 Y
Y by Adventure10, Mango247
solution
This post has been edited 1 time. Last edited by strategos21, Jun 25, 2018, 3:27 PM
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Jzhang21
308 posts
#18 • 2 Y
Y by Adventure10, Mango247
Assume for the sake of contradiction that is possible for unique integers $a,b,c$. Let $P(x)=d_1x^n+d_2x^{n-1}+\cdots+d_n.$ Note that $$b=P(a)=d_1a^n+d_2a^{n-1}+\cdots+d_n$$$$c=P(b)=d_1b^n+d_2b^{n-1}+\cdots+d_n$$$$a=P(c)=d_1c^n+d_2c^{n-1}+\cdots+d_n$$Subtracting the second from the first, third from second, and first from third gives
$$b-c=P(a)-P(b)=d_1(a^n-b^n)+d_2(a^{n-1}-b^{n-1})+\cdots+ d_{n-1}(a-b)$$$$c-a=P(b)-P(c)=d_1(b^n-c^n)+d_2(b^{n-1}-c^{n-1})+\cdots +d_{n-1}(b-c)$$$$a-b=P(c)-P(a)=d_1(c^n-a^n)+d_2(c^{n-1}-a^{n-1})+\cdots +d_{n-1}(c-a)$$By the RHS, note that $a-b\mid P(a)-P(b)$ so $\lvert{a-b}\rvert\leq\lvert{b-c}\rvert.$ Similarly, $\lvert{b-c}\rvert\leq\lvert{c-a}\rvert$ and $\lvert{c-a}\rvert\leq\lvert{a-b}\rvert.$ Hence, $\lvert{a-b}\rvert\leq\lvert{b-c}\rvert\leq\lvert{c-a}\rvert\leq\lvert{a-b}\rvert$ so $\lvert{a-b}\rvert=\lvert{b-c}\rvert=\lvert{c-a}\rvert.$ Assume WLOG that $a>b>c$ so $a-b=a-c$ and $b-c=c-a.$ From the first equation, we get $b=c$ and substituting this in the second gives $c=a.$ Hence, $a=b=c$, contradicting the uniqueness of $a,b,c.$ $\blacksquare$
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amar_04
1917 posts
#19 • 1 Y
Y by Adventure10
A bit different finishing.......

Like others did, I also used the well known fact that $x-y|P(x)-P(y)$. So, by this fact we get $$a-b|b-c$$$$b-c|c-a$$$$c-a|a-b$$. Now we will consider this into two cases when $a-b=b-c$, $b-c=c-a$ and and the other case will be when $a-b<b-c$, $b-c<c-a$ and $c-a<a-b$. So, by adding all the inequalities we get $a+b+c<a+b+c$ which is impossible. Now we will consider when $a-b=b-c$, $b-c=c-a$ and $c-a=a-b$. From this we get $a=b=c$, which is clearly not possible, as the question says that $a,b,c$ are distinct integers. Hence, not possible.
This post has been edited 3 times. Last edited by amar_04, Aug 18, 2019, 5:53 AM
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Pluto04
797 posts
#20 • 1 Y
Y by Adventure10
Indian MO 1986.
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AopsUser101
1750 posts
#21 • 3 Y
Y by vsamc, v4913, Adventure10
Let $P(x) = d_nx^n+d_{n-1}x^{n-1}+...+d_0$. We have that $$P(m)-P(n) = d_n(m^n-n^n)+d_{n-1}(m^{n-1}-n^{n-1})+...+d_1(m-n)$$, which is clearly divisible by $m-n$. Hence, $$(a-b)|(b-c)$$$$(b-c)|(c-a)$$$$(c-a)|(a-b).$$Therefore, $|a-b|=|b-c|=|c-a|$. Letting $x=a-b$, $y=b-c$, and $x+y=c-a$, we have that $|x|=|y|=|x+y|$. $x=-y \implies a=c$, so we know that $x=y$. In other words, $|y|=|2y| \implies y = 0$. If $y=0$, then $b=c$, which is impossible. $\blacksquare$
This post has been edited 2 times. Last edited by AopsUser101, Feb 3, 2020, 12:38 AM
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Math-Shinai
396 posts
#22 • 1 Y
Y by Adventure10
I don't see the point of commenting on an ancient post that is well-known... especially when the solution that you just posted is basically the same as everyone else's
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vsamc
3812 posts
#23 • 2 Y
Y by Mango247, Mango247
@above well writing your solution up helps you become better at math because you can reflect on your solution.

my first ever USAMO solve :)
My Solution
This post has been edited 1 time. Last edited by vsamc, Jun 4, 2020, 4:32 PM
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bryanguo
1032 posts
#24 • 4 Y
Y by v4913, Mango247, Mango247, Mango247
My solution is basically @sayantanchakraborty's, but I feel like the explanation of mine is slightly more concise:

Solu
This post has been edited 1 time. Last edited by bryanguo, Jun 4, 2020, 10:18 PM
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OlympusHero
17020 posts
#25 • 1 Y
Y by Mango247
Solution
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Mogmog8
1080 posts
#26 • 2 Y
Y by centslordm, Mango247
Solution
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Sprites
478 posts
#27
Y by
FTSOC,assume that we can have $P(a)=b,P(b)=c,P(c)=a$ simultaneously.
Then $$a-c|P(P(c))-P(P(a))=a-b$$and $$a-b|P(P(b))-P(P(a))=a-c$$implying $b=c$, contradiction.
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lifeismathematics
1188 posts
#28
Y by
div and done
This post has been edited 1 time. Last edited by lifeismathematics, Nov 21, 2022, 1:22 PM
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huashiliao2020
1292 posts
#29
Y by
Agr_94_Math wrote:
$ a,b,c \in \mathbb{Z}$, and $ a,b,c$ are distinct.
Since we have $ P[\mathbb{Z}]$, suppose we have by the given conditions to be exisitng such that
$ a-b | b-c$
$ b-c | c-a$
$ c-a | a-b$.
As we are dealing over integers,
$ 2b\ge c+a$
$ 2c\ge a+b$
$ 2a\ge b+c$
THis however implies equaltiy holds everywhere and therefore contradiction.

hmm, I don't understand how you got to a-b|b-c. Can someone explain the omitted part?
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pco
23515 posts
#30
Y by
huashiliao2020 wrote:
hmm, I don't understand how you got to a-b|b-c. Can someone explain the omitted part?
$a-b|P(a)-P(b)=b-c$
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ryanbear
1058 posts
#31
Y by
$a-b|P(a)-P(b)$ --> $a-b|b-c$
Similarly, $b-c|c-a$, $c-a|a-b$
So this means $|a-b|=|b-c|=|c-a|$
WLOG $a<b<c$
$b-a=c-b=c-a$
$a=b$ (contradiction)
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megahertz13
3206 posts
#32
Y by
First, observe that \[P(a)-P(b)=P(b+(a-b))-P(b)\]is always a multiple of $b-a$. From this we obtain $a-b|b-c$, $b-c|c-a$, and $c-a|a-b$. Thus, \[|a-b|=|b-c|=|c-a|.\]Clearly $a-b=-(b-c)$ cannot hold, as that would imply $a=c$. Therefore, $a-b=b-c\implies c+a=2b$. Similarly, we get $a+b=2c$. Subtracting this gives $c-b=2b-2c\implies c=b$, a contradiction.
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