Plan ahead for the next school year. Schedule your class today!

G
Topic
First Poster
Last Poster
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
and train with the best! Please note that early bird pricing ends August 19th!
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:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Wednesday, Jul 16 - Oct 29
Sunday, Aug 17 - Dec 14
Tuesday, Aug 26 - Dec 16
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10

Prealgebra 2 Self-Paced

Prealgebra 2
Friday, Jul 25 - Nov 21
Sunday, Aug 17 - Dec 14
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Tuesday, Jul 15 - Oct 28
Sunday, Aug 17 - Dec 14
Wednesday, Aug 27 - Dec 17
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Wednesday, Jul 2 - Sep 17
Sunday, Jul 27 - Oct 19
Monday, Aug 11 - Nov 3
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8

Introduction to Number Theory
Tuesday, Jul 15 - Sep 30
Wednesday, Aug 13 - Oct 29
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Friday, Jul 18 - Nov 14
Thursday, Aug 7 - Nov 20
Monday, Aug 18 - Dec 15
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30

Introduction to Geometry
Monday, Jul 14 - Jan 19
Wednesday, Aug 13 - Feb 11
Tuesday, Aug 26 - Feb 24
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
Friday, Aug 8 - Feb 20
Tuesday, Aug 26 - Feb 24
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4

Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24

Intermediate Number Theory
Wednesday, Sep 24 - Dec 17

Precalculus
Wednesday, Aug 6 - Jan 21
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31

Advanced: Grades 9-12

Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)

MATHCOUNTS/AMC 8 Advanced
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Tuesday, Aug 26 - Nov 11
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)

AMC 10 Problem Series
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 10 - Nov 2
Thursday, Aug 14 - Oct 30
Tuesday, Aug 19 - Nov 4
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 10 Final Fives
Friday, Aug 15 - Sep 12
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30

AMC 12 Problem Series
Wednesday, Aug 6 - Oct 22
Sunday, Aug 10 - Nov 2
Monday, Aug 18 - Nov 10
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28

AIME Problem Series A
Thursday, Oct 23 - Jan 29

AIME Problem Series B
Tuesday, Sep 2 - Nov 18

F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT


Programming

Introduction to Programming with Python
Thursday, Aug 14 - Oct 30
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3

Intermediate Programming with Python
Friday, Oct 3 - Jan 16

USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3

Physics

Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11

Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
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
GMA Problem
mihaig   0
11 minutes ago
Source: Own
Let $a_1,a_2, a_3,a_4\geq0$ be reals.
Prove
$$a_1+a_2+3a_3+3a_4+\sqrt{\left(a_{1}-a_{3}\right)^2+\left(a_{2}-a_{4}\right)^2}\geq\sum_{k=1}^{4}{\sqrt{a_k^2+a_{k+1}^2}},$$where $a_5=a_1.$ When do we have equality?
0 replies
mihaig
11 minutes ago
0 replies
Italian WinterCamps test07 Problem5
mattilgale   58
N 28 minutes ago by lpieleanu
Source: ISL 2006, A1, AIMO 2007, TST 1, P1
A sequence of real numbers $ a_{0},\ a_{1},\ a_{2},\dots$ is defined by the formula
\[ a_{i + 1} = \left\lfloor a_{i}\right\rfloor\cdot \left\langle a_{i}\right\rangle\qquad\text{for}\quad i\geq 0;
\]here $a_0$ is an arbitrary real number, $\lfloor a_i\rfloor$ denotes the greatest integer not exceeding $a_i$, and $\left\langle a_i\right\rangle=a_i-\lfloor a_i\rfloor$. Prove that $a_i=a_{i+2}$ for $i$ sufficiently large.

Proposed by Harmel Nestra, Estionia
58 replies
+1 w
mattilgale
Jan 29, 2007
lpieleanu
28 minutes ago
functional equation with two variables
Sayan   6
N 36 minutes ago by ParthivCalculus
Source: ISI(BS) 2005 #3
Let $f$ be a function defined on $\{(i,j): i,j \in \mathbb{N}\}$ satisfying

(i) $f(i,i+1)=\frac{1}{3}$ for all $i$

(ii) $f(i,j)=f(i,k)+f(k,j)-2f(i,k)f(k,j)$ for all $k$ such that $i <k<j$.

Find the value of $f(1,100)$.
6 replies
Sayan
May 20, 2012
ParthivCalculus
36 minutes ago
sorting sums (a_i +a_j) in ascending order, arithmetic progression when?
parmenides51   1
N an hour ago by ririgggg
Source: Dutch IMO TST 2016 day 3 p2
For distinct real numbers $a_1,a_2,...,a_n$, we calculate the $\frac{n(n-1)}{2}$ sums $a_i +a_j$ with $1 \le i < j \le  n$, and sort them in ascending order. Find all integers $n \ge  3$ for which there exist $a_1,a_2,...,a_n$, for which this sequence of $\frac{n(n-1)}{2}$ sums form an arithmetic progression (i.e. the dierence between consecutive terms is constant).
1 reply
parmenides51
Aug 30, 2019
ririgggg
an hour ago
IMO 2011 Problem 5
orl   87
N an hour ago by Jupiterballs
Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Proposed by Mahyar Sefidgaran, Iran
87 replies
orl
Jul 19, 2011
Jupiterballs
an hour ago
Numbers on cards (again!)
popcorn1   83
N an hour ago by eg4334
Source: IMO 2021 P1
Let $n \geqslant 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
83 replies
popcorn1
Jul 20, 2021
eg4334
an hour ago
(a^n + b^n + 1) is divisible by d for all positive integers n
parmenides51   3
N an hour ago by ririgggg
Source: Dutch IMO TST 2016 p2
Determine all pairs $(a, b)$ of integers having the following property:
there is an integer $d \ge  2$ such that $a^n + b^n + 1$ is divisible by $d$ for all positive integers $n$.
3 replies
parmenides51
Aug 30, 2019
ririgggg
an hour ago
An Inequality
Butterfly   0
2 hours ago

Let $x\ge y\ge z\ge 0$ and $x^2+y^2+z^2+xyz=4.$ Prove $x+y+z+(\sqrt{x}-\sqrt{z})^2\ge 3.$
0 replies
Butterfly
2 hours ago
0 replies
cubefree divisibility
DottedCaculator   64
N 2 hours ago by eg4334
Source: 2021 ISL N1
Find all positive integers $n\geq1$ such that there exists a pair $(a,b)$ of positive integers, such that $a^2+b+3$ is not divisible by the cube of any prime, and $$n=\frac{ab+3b+8}{a^2+b+3}.$$
64 replies
DottedCaculator
Jul 12, 2022
eg4334
2 hours ago
AOPS MO Introduce
MathMaxGreat   78
N 2 hours ago by Jackson0423
$AOPS MO$

Problems: post it as a private message to me or @jerryZYang, please post it in $LATEX$ and have answers

6 Problems for two rounds, easier than $IMO$

If you want to do the problems or be interested, reply ’+1’
Want to post a problem reply’+2’ and message me
Want to be in the problem selection committee, reply’+3’
78 replies
MathMaxGreat
Yesterday at 1:04 AM
Jackson0423
2 hours ago
Collinearity in Isosceles Trapezoid
utkarshgupta   16
N 2 hours ago by Rayvhs
Source: All Russian Olympiad 2017 Grade 9 Problem 2
$ABCD$ is an isosceles trapezoid with $BC || AD$. A circle $\omega$ passing through $B$ and $C$ intersects the side $AB$ and the diagonal $BD$ at points $X$ and $Y$ respectively. Tangent to $\omega$ at $C$ intersects the line $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear.
16 replies
utkarshgupta
Jul 5, 2017
Rayvhs
2 hours ago
Problem GEO
Math2030   0
2 hours ago
Problem
Let \(ABCD\) be a cyclic quadrilateral such that \(AD = BC\). Let \(I = AC \cap BD\), and let \(I_1, I_2\) be the incenters of triangles \(\triangle IAD\) and \(\triangle IBC\), respectively. Let \(X\) and \(Y\) be the midpoints of \(AB\) and \(CD\), respectively. Prove that the segment \(XY\) bisects the segment \(I_1I_2\).
0 replies
Math2030
2 hours ago
0 replies
IMO 2017 Problem 1
cjquines0   158
N 2 hours ago by lpieleanu
Source: IMO 2017 Problem 1
For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as
$$a_{n+1} = 
\begin{cases}
\sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\
a_n + 3 & \text{otherwise.}
\end{cases}
$$Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.

Proposed by Stephan Wagner, South Africa
158 replies
cjquines0
Jul 18, 2017
lpieleanu
2 hours ago
D1033 : A problem of probability for dominoes 3*1
Dattier   6
N 2 hours ago by maromex
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
6 replies
Dattier
May 15, 2025
maromex
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bobthesmartypants
4337 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$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $a-b|P(a)-P(b)$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bobthesmartypants
4337 posts
#15 • 1 Y
Y by Adventure10
Okay I guess I wasn't thinking. I forgot that $P(a)=b$ and $P(b)=c$. :|
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $P(c)=a$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $a,b,c.$ $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Pluto04
797 posts
#20 • 1 Y
Y by Adventure10
Indian MO 1986.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
vsamc
3803 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OlympusHero
17020 posts
#25 • 1 Y
Y by Mango247
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mogmog8
1080 posts
#26 • 2 Y
Y by centslordm, Mango247
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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?
$a-b|P(a)-P(b)=b-c$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ryanbear
1057 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)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megahertz13
3194 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.
Z K Y
N Quick Reply
G
H
=
a