Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

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, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
D1027 : Super Schoof
Dattier   1
N 7 minutes ago by Dattier
Source: les dattes à Dattier
Let $p>11$ a prime number with $a=\text{card}\{(x,y) \in \mathbb Z/ p \mathbb Z: y^2=x^3+1\}$ and $b=\dfrac 1 {((p-1)/2)! \times ((p-1)/3)! \times ((p-1)/6)!} \mod p$ when $p \mod 3=1$.



Is it true that if $p \mod 3=1$ then $a \in \{b,p-b, \min\{b,p-b\}+p\}$ else $A=p$.
1 reply
Dattier
3 hours ago
Dattier
7 minutes ago
minimizing sum
gggzul   0
9 minutes ago
Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$. Find
$$min\{12x-4y-3z\}.$$
0 replies
gggzul
9 minutes ago
0 replies
Prove that 4p-3 is a square - Iran NMO 2005 - Problem1
sororak   23
N 14 minutes ago by reni_wee
Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.
23 replies
sororak
Sep 21, 2010
reni_wee
14 minutes ago
IMO 2009, Problem 2
orl   142
N 21 minutes ago by pi271828
Source: IMO 2009, Problem 2
Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$

Proposed by Sergei Berlov, Russia
142 replies
orl
Jul 15, 2009
pi271828
21 minutes ago
(a-1)(b-1)(c-1) is a divisor of abc-1
ehsan2004   22
N an hour ago by reni_wee
Source: IMO 1992, Day 1, Problem 1
Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that \[ (a-1)(b-1)(c-1)  \] is a divisor of $abc-1.$
22 replies
ehsan2004
Jan 22, 2005
reni_wee
an hour ago
Balkan MO 2025 p1
Mamadi   0
an hour ago
Source: Balkan MO 2025
An integer \( n > 1 \) is called good if there exists a permutation \( a_1, a_2, \dots, a_n \) of the numbers \( 1, 2, 3, \dots, n \), such that:

\( a_i \) and \( a_{i+1} \) have different parities for every \( 1 \le i \le n - 1 \)

the sum \( a_1 + a_2 + \dots + a_k \) is a quadratic residue modulo \( n \) for every \( 1 \le k \le n \)

Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.

Remark: Here an integer \( z \) is considered a quadratic residue modulo \( n \) if there exists an integer \( y \) such that \( y^2 \equiv z \pmod{n} \).
0 replies
1 viewing
Mamadi
an hour ago
0 replies
Number theory
MathsII-enjoy   3
N an hour ago by KevinYang2.71
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
3 replies
MathsII-enjoy
Yesterday at 3:22 PM
KevinYang2.71
an hour ago
Inspired by Bet667
sqing   1
N an hour ago by ytChen
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
1 reply
sqing
5 hours ago
ytChen
an hour ago
4-var inequality
sqing   1
N 2 hours ago by arqady
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
1 reply
sqing
5 hours ago
arqady
2 hours ago
Extremaly hard inequality
blug   1
N 2 hours ago by arqady
Source: Polish Math Olympiad Training Camp 2024
Let $a, b, c$ be non-negative real numbers. Prove that
$$a+b+c+\sqrt{a^2+b^2+c^2-ab-bc-ca}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+c^2}.$$Looking for an algebraic solution!
1 reply
blug
2 hours ago
arqady
2 hours ago
Common external tangents of two circles
a1267ab   56
N 2 hours ago by wassupevery1
Source: USA Winter TST for IMO 2020, Problem 2, by Merlijn Staps
Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.

Suppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear.

Merlijn Staps
56 replies
a1267ab
Dec 16, 2019
wassupevery1
2 hours ago
That's Vietnamese geo!
wassupevery1   8
N 2 hours ago by cj13609517288
Source: 2025 Vietnam National Olympiad - Problem 3
Let $ABC$ be an acute, scalene triangle with circumcenter $O$, circumcircle $(O)$, orthocenter $H$. Line $AH$ meets $(O)$ again at $D \neq A$. Let $E, F$ be the midpoint of segments $AB, AC$ respectively. The line through $H$ and perpendicular to $HF$ meets line $BC$ at $K$.
a) Line $DK$ meets $(O)$ again at $Y \neq D$. Prove that the intersection of line $BY$ and the perpendicular bisector of $BK$ lies on the circumcircle of triangle $OFY$.
b) The line through $H$ and perpendicular to $HE$ meets line $BC$ at $L$. Line $DL$ meets $(O)$ again at $Z \neq D$. Let $M$ be the intersection of lines $BZ, OE$; $N$ be the intersection of lines $CY, OF$; $P$ be the intersection of lines $BY, CZ$. Let $T$ be the intersection of lines $YZ, MN$ and $d$ be the line through $T$ and perpendicular to $OA$. Prove that $d$ bisects $AP$.
8 replies
wassupevery1
Dec 25, 2024
cj13609517288
2 hours ago
Equation has no integer solution.
Learner94   33
N 2 hours ago by bjump
Source: INMO 2013
Let $a,b,c,d \in \mathbb{N}$ such that $a \ge b \ge c \ge d $. Show that the equation $x^4 - ax^3 - bx^2 - cx -d = 0$ has no integer solution.
33 replies
Learner94
Feb 3, 2013
bjump
2 hours ago
Problem 1 — Symmetric Squares, Symmetric Products
RockmanEX3   8
N 2 hours ago by Baimukh
Source: 46th Austrian Mathematical Olympiad National Competition Part 1 Problem 1
Let $a$, $b$, $c$, $d$ be positive numbers. Prove that

$$(a^2 + b^2 + c^2 + d^2)^2 \ge (a+b)(b+c)(c+d)(d+a)$$
When does equality hold?

(Georg Anegg)
8 replies
RockmanEX3
Jul 14, 2018
Baimukh
2 hours ago
Easy Number Theory
math_comb01   37
N Apr 23, 2025 by John_Mgr
Source: INMO 2024/3
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
37 replies
math_comb01
Jan 21, 2024
John_Mgr
Apr 23, 2025
Easy Number Theory
G H J
G H BBookmark kLocked kLocked NReply
Source: INMO 2024/3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_comb01
662 posts
#1 • 3 Y
Y by GeoKing, Rounak_iitr, polynomialian
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
This post has been edited 3 times. Last edited by math_comb01, Jan 22, 2024, 8:22 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mannshah1211
651 posts
#2 • 3 Y
Y by ATGY, GeoKing, megarnie
If $p$ divides one of them, it divides all of them. Assume that it divides none of them. Then, there exists an inverse for all of them in modulo $p$ and then it's some algmanip to get $b \equiv c \pmod{p}$ a contradiction
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ATGY
2502 posts
#3
Y by
mannshah1211 wrote:
If $p$ divides one of them, it divides all of them. Assume that it divides none of them. Then, there exists an inverse for all of them in modulo $p$ and then it's some algmanip to get $b \equiv c \pmod{p}$ a contradiction

Did it the same for the first part but used sum of cubes to arrive at the result (more lengthy)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
LuciferMichelson
18 posts
#4 • 1 Y
Y by Aliosman
isn't that this question toooo easy for INMO p3
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
starchan
1608 posts
#5 • 1 Y
Y by kamatadu
LuciferMichelson wrote:
isn't that this question toooo easy for INMO p3

On what sample set are you basing your opinion on difficulty?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#6 • 1 Y
Y by Raj_singh1432
Proposed by Navilarekallu Tejaswi
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
181 posts
#7 • 1 Y
Y by S_14159
If $p$ divides one of them, it divides all of them. Assume that it divides none of them. Then just do some things to get $b \equiv c \pmod{p}$.
Which gives a contradiction
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Pluto1708
1107 posts
#8 • 4 Y
Y by ATGY, GeoKing, kamatadu, Rounak_iitr
math_comb01 wrote:
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$.
Prove that $p$ divides each of $a,b,c$.
Clearly if $p$ divides any of $a,b,c$ we are done, so from now onwards we assume none of them are divisible by $p$.Define $x,y$ such that
$x = \dfrac{a}{b} \mod p \Longleftrightarrow x^{2023}\equiv -1\mod p$ and $y = \dfrac{b}{c}\mod p \Longleftrightarrow y^{2024}\equiv -1\mod p$.Then notice
\[\dfrac{1}{xy} = \dfrac{c}{a}\mod p\Longleftrightarrow (x\cdot y)^{2025}\equiv -1\mod p\]Claim : $y\equiv 1\mod p$ that is $b\equiv c \mod p$
Based on previous equations $$-1 = (x\cdot y)^{2023\cdot 2025} = x^{2023\cdot 2025}\cdot y^{2023\cdot 2025} = -y^{2024^2-1}=-y^{-1}\mod p\Longleftrightarrow y\equiv 1\mod p $$
Now plugging back in the original equation we get $1=y^{2024}=-1\mod p\implies p\mid 2$ a contradiction to the fact that $p$ was an odd-prime.$\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HoRI_DA_GRe8
597 posts
#9
Y by
Yeah basically belows solution I got.
This post has been edited 1 time. Last edited by HoRI_DA_GRe8, Jan 22, 2024, 1:05 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kamatadu
480 posts
#10 • 4 Y
Y by GeoKing, polynomialian, iamahana008, kkloveMinecraft
Firstly note that if even one of them is divisible by $p$, all of them are. So assume on the contrary that none of them are divisible by $p$. So assume none of them are.

Then we get that,
\begin{align*}
    \left(\dfrac{a}{b}\right)^{2023} &\equiv -1 \pmod{p}\\
    \left(\dfrac{b}{c}\right)^{2024} &\equiv -1 \pmod{p}\\
    \left(\dfrac{c}{a}\right)^{2025} &\equiv -1 \pmod{p}
.\end{align*}
We multiply all three of them to get $a^2 \equiv -bc \pmod{p}$.

Now note that we have $a^{2023} \equiv -b^{2023} \pmod{p}$ and $a^{2025} \equiv -c^{2025} \pmod{p}$. Multiplying these, we get,
\[ a^{2023 + 2025} \equiv b^{2023}c^{2025} \implies (a^2)^{2024} \equiv c^2 \cdot (bc)^{2023} \implies bc \cdot (bc)^{2023} \equiv c^2 \cdot (bc)^{2023} \implies b \equiv c \pmod{p}. \]
But then,
\[ p\mid b^{2024} + c^{2024} \equiv 2b^{2024} \implies p \mid 2 \]which is a contradiction and we are done. :yoda:
This post has been edited 1 time. Last edited by kamatadu, Jan 21, 2024, 2:34 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5606 posts
#11 • 2 Y
Y by Anchovy, HoRI_DA_GRe8
Solved with Anchovy.

I assume the problem means that $p$ divides all of $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$
If $p$ divides one of $a,b,c$, clearly $p$ divides all of $a,b,c$, so assume $p$ divides none of them.

Let $x = \frac ab, y = \frac bc, z = \frac ca$. The condition $x^{2023} \equiv -1\pmod p, y^{2024} \equiv 1\pmod p, z^{2025} \equiv -1\pmod p$ and we also have $xyz = 1$, so multiplying the three equations gives $(xyz)^{2023} \cdot y z^2 \equiv 1\pmod p$, so $yz^2 \equiv 1\pmod p$. This implies that $y\equiv -\frac{1}{z^2}$, so $(-1/z^2)^{2024} \equiv 1$, so $z^{4048} \equiv 1$. But $(z^{2025})^2 = z^{4050}\equiv 1$, so dividing the two gives $z^2 \equiv 1$, so $z\in \{-1,1\}$ mod p. Since $z^{2025} \equiv -1$, we have $z\equiv -1$. But then since $yz^2 \equiv 1$, $y\equiv 1\pmod p$, so $b\equiv c \pmod p$. Now, this means that $2b^{2024}$ is a multiple of $p$, contradiction since $p$ doesn't divide $b$ and $p > 2$.
This post has been edited 1 time. Last edited by megarnie, Jan 21, 2024, 2:42 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SatisfiedMagma
458 posts
#13 • 1 Y
Y by ATGY
Nah bro, I ain't even gonna comment on this one.

Solution: Observe that $p$ divides either of $a,b,c$, then it divides everything and we're done. Henceforth, assume that $p \nmid a,b,c$. We will now work modulo $p$ with special powers of existence of inverses of $a,b,c$.

From $p \mid a^{2023}+b^{2023}$, we get that $b^{2024} \equiv -a^{2023}b \pmod{p}$. Swing this into $p \mid b^{2024}+c^{2024}$ to get $c^{2025} \equiv a^{2023}bc \pmod{p}$. At last, putting this into $p \mid a^{2025} + c^{2025}$, we get
\[a^{2025} + a^{2023}bc \equiv 0 \iff a^2 + bc \equiv 0 \pmod{p}.\]From here, we wish to eliminate $b$ completely modulo $p$. This can be achieved via substituting $b \equiv -a^2/c$. Putting this in $p \mid a^{2023}+b^{2023}$ we get
\[a^{2023} - \frac{a^{4046}}{c^{2023}} \equiv 0 \iff a^{2023} \equiv c^{2023} \pmod{p}.\]This yields
\[a^{2025} \equiv a^2c^{2023} \equiv -c^{2025} \pmod{p} \implies a^2 + c^2 \equiv 0 \pmod{p}.\]Finally, on combining this with $a^2 \equiv -bc$, we get
\[p \mid c(c-b) \iff c \equiv b \pmod{p}\]since $\gcd(c,p) = \gcd(b,p) = 1$. Upon putting $c \equiv b$ in say $p \mid b^{2024}+c^{2024}$, we get immediately get $2b \equiv 0 \pmod{p}$ which is a contradiction since $p$ is odd and we're done. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Samujjal101
2799 posts
#14
Y by
............
This post has been edited 1 time. Last edited by Samujjal101, Jan 21, 2024, 4:53 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Samujjal101
2799 posts
#15
Y by
Samujjal101 wrote:
Just see that if gcd(a,b)=1 then the given conditions are not possible. So, a|b which means b=ak for all integers k. Now p divides (a^2023 + k^2023.a^2023) so p either divides a^2023 or (1+ k^2023) => p divides a^2023 =>p divides a

..............
This post has been edited 1 time. Last edited by Samujjal101, Jan 21, 2024, 4:53 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
taptya17
29 posts
#16 • 1 Y
Y by Om245
If $p$ divides any one of them, it divides the other two as well.

Assume for the sake of contradiction that $p$ doesn't divide any of the three. Since $p$ is an odd prime, it has a primitive root, say $g$.
Let $a=g^x,b=g^y,c=g^z$.

Claim. $g^m+g^n=0(p)\implies m-n=k(2k)$ where $p=2k+1$.
Proof. $g^{m-n}=-1(p)\implies m-n=k(2k)\implies m-n=2kq+k=k(2q+1)$.
Using the claim and given information,
$$0=(x-y)+(y-z)+(z-x)=\frac{k(2q_1+1)}{2023}+\frac{k(2q_2+1)}{2024}+\frac{k(2q_3+1)}{2025}$$Cancelling $k$ and the denominator and taking mod 2 leads to a contradiction. Done!
This post has been edited 2 times. Last edited by taptya17, Jan 21, 2024, 4:01 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
djmathman
7938 posts
#17 • 2 Y
Y by starchan, Rounak_iitr
math_comb01 wrote:
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$.
This sentence seems to be missing something. Are all three integers divisible by $p$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mannshah1211
651 posts
#18 • 3 Y
Y by GeoKing, sanyalarnab, Erratum
djmathman wrote:
math_comb01 wrote:
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$.
This sentence seems to be missing something. Are all three integers divisible by $p$?

Yes. It's supposed to say that all three are divisible by $p$, but I suppose a lot of us got tunnel vision and didn't see the error because we already saw it in the test lol :rotfl:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
maths_arka
6 posts
#19 • 1 Y
Y by sanyalarnab
First see that if $p|a$ or $p|b$ or $p|c$ then $p$ divides all of them.

Now suppose that $p{\not|}a$ $p{\not|}b$ $p{\not|}c$.
Then the desired contradiction will come from the fact that $p$ is odd prime, i.e we would prove $p=2$.

Since $p{\not|}a$ $p{\not|}b$ $p{\not|}c$ we could take inverses.
Reducing everything $(\textrm{mod}\ p)$ we get
$$(ab^{-1})^{2023}\equiv-1 ({mod}\,p) \cdots(1)$$$$(bc^{-1})^{2024}\equiv-1 ({mod}\,p)\cdots(2) $$$$(ca^{-1})^{2025}\equiv-1 ({mod}\,p)\cdots(3) $$Multiplying equations $1$, $2$ and $3$ we get
$$a^{-2}bc\equiv-1 ({mod}\,p)\cdots(4)$$.
$$\Rightarrow (ab^{-1})^{-1}ca^{-1}\equiv-1 ({mod}\,p) \cdots(5)$$i.e
$$\Rightarrow ca^{-1}\equiv-(ab^{-1}) ({mod}\,p) \cdots(6)$$Now from $3$ we get
$$(ab^{-1})^{2025}\equiv1 ({mod}\,p) \cdots(7)$$and from 1 we get
$$(ab^{-1})^{4046}\equiv1 ({mod}\,p) \cdots(8)$$Let the order of $ab^{-1}$ mod $p$ be k.
Thus $k|gcd(4046,2025)$ i.e $k|1$.
Therefore we have
$$ab^{-1}\equiv1 ({mod}\,p)$$or
$$a\equiv b ({mod}\,p)$$Putting this information in equation $1$ we get
$$1\equiv -1 ({mod}\,p)$$or $p=2$
which is a contradiction.
$\blacksquare$ :-D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mannshah1211
651 posts
#20
Y by
Well, I'll post my solution just for the sake of it, I guess.

If $p$ divides one of $a, b, c,$ then it divides all of them, so henceforth assume it doesn't divide any of them. Thus, there exists a valid inverse in modulo $p$ for each of $a, b, c.$ Then, we have $\left(\frac{a}{b}\right)^{2023} \equiv -1 \pmod{p}, \left(\frac{b}{c}\right)^{2024} \equiv -1\pmod{p}, \left(\frac{c}{a}\right)^{2025}\equiv -1\pmod{p},$ and thus, multiplying all of them together, we have $bc \equiv -a^2 \pmod{p}.$ Thus, $b^{1012}c^{1012} \equiv a^{2024} \pmod{p},$ which gives $ab^{1012}c^{1012} + c^{2025} \equiv 0 \pmod{p} \implies ab^{1012} \equiv -c^{1013} \pmod{p}.$ Thus, $a \equiv \frac{-c^{1013}}{b^{1012}} \pmod{p},$ which by putting in the first equation gives $\frac{b^{1013 \cdot 2023} - c^{1013 \cdot 2023}}{b^{1012 \cdot 2023}} \equiv 0 \pmod{p} \implies b^{1013 \cdot 2023} - c^{1013 \cdot 2023} \equiv 0 \pmod{p}.$ From the second equation, we get $b^{4048} - c^{4048} \equiv 0 \pmod{p},$ so $p \mid b^{\gcd(4048, 1013 \cdot 2023)} - c^{\gcd(4048, 1013 \cdot 2023)} = b - c,$ so $p \mid 2b^{2024},$ a contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Safal
169 posts
#21
Y by
My attempt,
Click to reveal hidden text
This post has been edited 8 times. Last edited by Safal, Jan 23, 2024, 12:46 PM
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
#22 • 1 Y
Y by Rounak_iitr
we work with $a^{r}+b^{r} \equiv 0 \pmod p , b^{r+1}+c^{r+1} \equiv 0 \pmod p , a^{r+2}+c^{r+2} \equiv 0 \pmod p$ for any $r \in \mathbb{Z}^{+}$ and it follows for $r=2023$.

$\mathsf{Claim 1:-}$ If $p$ divides any one of $a,b,c$ then $p|a,b,c$.

$\mathsf{Proof:-}$ W.L.O.G $p|a \implies p|a^{r}$ , now since $a^{r}+b^{r} \equiv 0 \pmod p \implies p|b^{r} \implies p|b$ and also $b^{r+1}+c^{r+1} \equiv 0 \pmod p \implies p|c^{r+1} \implies p|c$. $\square$

From $\mathsf{Claim 1}$ it suffices to disprove the fact that that $p \nmid a, p \nmid b , p\nmid c$ ,so FTSOC we assume that is the case.

$\mathsf{Claim 2:-}$ $p$ does not divide any of $a-b , b-c,c-a$.

$\mathsf{Proof:-}$ FTSOC $p|a-b$ , then $a \equiv b \pmod p \implies a^{r} \equiv b^{r} \pmod p \implies 2a^{r} \equiv 0 \pmod p$, now since $\gcd(p,2)=\gcd(p,a)=1$ , it gives a contradiction. $\rightarrow \leftarrow$.

Now ,
$\bullet a^{r}+b^{r} \equiv 0 \pmod p$

$\bullet b^{r+1}+c^{r+1} \equiv 0 \pmod p$

$\bullet a^{r+2}+c^{r+2} \equiv 0 \pmod p$


so $a^{r} \equiv -b^{r}   \pmod p$ and $a^{r+2} \equiv -c^{r+2} \pmod p \implies a^2c^{r+1} \equiv -bc^{r+2} \pmod p$ and since $\gcd(c,p)=1 \implies a^2 \equiv -bc \pmod p , b^2 \equiv -ca \pmod p , c^2 \equiv -ab \pmod p \qquad (\star)$

now substract the subsequent cogruences to get $a^2-b^2 \equiv c(a-b) \pmod p$ similarly symmetrically other in $b$ and $c$ , but from $\mathsf{Claim 2}$ we have $p \nmid a-b , b-c , c-a$ , which implies $a+b \equiv c \pmod p , b+c \equiv a \pmod p , c+a \equiv b \pmod p \qquad (\dagger)$

Now from $(\star)$ we have $(a+b)^2+(b+c)^2+(c+a)^2 \equiv 0 \pmod p$ and from $(\dagger)$ we get this is equivalent to $a^2+b^2+c^2 \equiv 0 \pmod p \implies ab+bc+ca \equiv 0 \pmod p \implies a+b+c \equiv 0 \pmod p \implies 2a,2b , 2c \equiv 0 \pmod p$ , but this is not possible as $\gcd(2,p)=\gcd(a,p)=\gcd(b,p)=\gcd(c,p)=1$.

Hence we get a contradiction $\rightarrow \leftarrow$.

Hence $p$ must divide $a,b,c$. $\blacksquare$

A cute NT for sure! :blush:
This post has been edited 2 times. Last edited by lifeismathematics, Jan 23, 2024, 9:15 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Master_of_Aops
71 posts
#24
Y by
Isn’t this a one-liner:
For $a^x \equiv -b^x$, raise both sides to power $yz$ to get terms $a^{xyz}, b^{xyz}, c^{xyz} (mod p)$ and done
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tintarn
9042 posts
#25
Y by
Master_of_Aops wrote:
... to get terms $a^{xyz}, b^{xyz}, c^{xyz} (mod p)$ and done
How exactly are you done from here?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Master_of_Aops
71 posts
#26
Y by
You get $a^{xyz} \equiv b^{xyz}, b^{xyz} \equiv -c^{xyz},  a^{xyz} \equiv c^{xyz}(mod p)$ so all are divisible by $p$
This post has been edited 1 time. Last edited by Master_of_Aops, Feb 1, 2024, 5:49 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
idkk
118 posts
#27
Y by
if p doesnt divide any of them

$(\frac{a}{b})^{2023} \equiv -1(mod p)$

$(\frac{b}{c})^{2024} \equiv -1(mod p)$

$(\frac{a}{c})^{2025} \equiv -1(mod p)$

$x^{2023} \equiv -1(mod p)$
$y^{2024} \equiv -1(mod p)$
$x^{2025}y^{2025} \equiv -1(mod p)$

so $x^2y \equiv -1(mod p)$

so $y \equiv \frac{-1}{x^2} (mod p)$

so $x^{2025} \equiv 1 (mod p)$

also $x^{2*2023} \equiv 1(mod p)$

let $k$ be the order of $x$ mod p so

$k | gcd(2025,2*2023) \implies k|1$

$x \equiv 1(mod p)$

then $1 \equiv -1(mod p)$

not possible as p is odd.

so p divides one of them then the answer is easy to conclude
This post has been edited 1 time. Last edited by idkk, Feb 3, 2024, 4:23 AM
Reason: .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Pyramix
419 posts
#28 • 1 Y
Y by GeoKing
We first prove that if $p$ divides any one of $a,b,c$, then it divides all the three numbers. Suppose $p\mid a$. Then, $p\mid a^{2023}$ but $p\mid a^{2023}+b^{2023}$, which means that $p\mid b^{2023}$, forcing $p\mid b$. Also, $p\mid c^{2025}+a^{2025}$ and $p\mid a^{2025},$ which means $p\mid c^{2025}$, forcing $p\mid c$.

Similarly, if $p\mid b$, then $p\mid b^{2023}$ and $p\mid a^{2023}+b^{2023}$, which means $p\mid a^{2023}$, forcing $p\mid a$. Since $p\mid a$, we also have that $p\mid c$ from the earlier proof.

Finally, if $p\mid c$, then $p\mid c^{2024}$ and $p\mid b^{2024}+c^{2024}$, which means $p\mid b^{2024}$, forcing $p\mid b$, which in turn forces $p\mid c$. So, if $p$ divides any one of $a,b,c$, then it divides all the three numbers.

Assume that $p\nmid a,b,c$. For integer $a$ and prime $p$, we define the order of $a\pmod{p}$ to be the smallest positive integer $k$ such that $a^k\equiv1\pmod{p}$.

We prove a Lemma.

$\textbf{Lemma.}$ If $k$ is order of $a$, $\pmod{p}$ and $a^n\equiv1\pmod{p}$, then $k\mid n$.

$\emph{Proof.}$ Suppose it was possible that $k\nmid n$. Note that if $n<k$, then it will contradict the minimality of $k$. So, $n>k$. Hence, there exists $q>0$ and $0<r<k$ such that $n=kq+r$. Now, $a^{k}\equiv1\pmod{p}$, which means $a^{kq}\equiv1\pmod{p}$, so $a^n\equiv a^{kq+r}\equiv a^r\equiv1\pmod{p}$. But since $0<r<k$ and $a^r\equiv1\pmod{p}$, the minimality of $k$ gives us a contradiction. Hence, $k\nmid n$ is impossible, and $r=0$ is forced. The proof is complete. $\blacksquare$

We have the equations \[\left(\frac{a}{b}\right)^{4046}\equiv1\pmod{p} \ \ \ \ (1)\]\[\left(\frac{b}{c}\right)^{4048}\equiv1\pmod{p} \ \ \ \ (2)\]\[\left(\frac{c}{a}\right)^{4050}\equiv1\pmod{p} \ \ \ \ (3)\]Multiplying these equations $(1),(2),(3),$ we obtain $a^4\equiv b^2c^2\pmod{p}$. So, $a^2\equiv\pm bc\pmod{p}$.

Note, the equation $(1)$ is $\left(\frac{a}{b}\right)^{4046}\equiv1\pmod{p}$. So, \[\left(\frac{\left(bc\right)^{2023}}{b^{4046}}\right)\equiv\pm1\pmod{p},\]which gives \[\left(\frac{b}{c}\right)^{4046}\equiv1\pmod{p} \ \ \ \ (4)\]It follows that if $k$ is the order of $\left(\frac{b}{c}\right)$, $\pmod{p}$, then from equation $(1)$, we get $k\mid4048$, and from equation $(4)$, we get $k\mid4046$, using the $\textbf{Lemma}$ in each case.

This forces $k\mid2$, so $k=1,2$. Hence, we have that $\left(\frac{b}{c}\right)^{2}\equiv1\pmod{p}$. So, $b^2\equiv c^2\pmod{p}$.

Taking to the power $1012$, we get $b^{2024}\equiv c^{2024}\pmod{p}$, and since $p\mid b^{2024}+c^{2024}$, we have $p\mid2b^{2024}$, which forces $p\mid b$, as $p$ is an odd prime. However, if $p\mid b$ then $p\mid a$, and $p\mid c$ are forced by our initial arguments.

The proof is complete. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1005 posts
#29
Y by
In contest solution:

Observe that if $p$ divides any one of $a, b, c$, then it divides all of them. Assume this to not be the case. Then we get the system of congruences $$a^{2023} \equiv -b^{2023} \pmod{p} \cdots(A)$$$$b^{2024} \equiv -c^{2024}  \pmod{p} \cdots(B)$$$$c^{2025} \equiv -a^{2025}  \pmod{p} \cdots(C)$$Multiplying $(A)$ and $(B)$, $$b \equiv c\left(\frac{c}{a}\right)^{2023}  \pmod{p}$$$$\implies b^{2025} \equiv c^{2025}\left(\frac{c}{a}\right)^{2023 \cdot 2025} \pmod{p}$$but from $(C)$, $$\left(\frac{c}{a}\right)^{2025} \equiv -1 \pmod{p}$$Substituting in, $$b^{2025} \equiv c^{2025}(-1)^{2023} \equiv -c^{2025} \pmod{p}$$Dividing $(B)$ from this new equation, $$b \equiv c \pmod{p}$$Therefore $$b^{2024} \equiv -b^{2024} \pmod{p}$$$$\implies 2b^{2024} \equiv 0 \pmod{p}$$$$\implies b^{2024} \equiv 0 \pmod{p}$$$$\implies p \mid b, $$a contradiction. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Shreyasharma
682 posts
#30
Y by
We are given,
\begin{align*}
a^{2023} + b^{2023} \equiv b^{2024} + c^{2024} \equiv c^{2025} + a^{2025} \equiv 0 \pmod{p}
\end{align*}Now assume that $a, b, c \not\equiv 0 \pmod{p}$ as if one of the three is $0$ modulo $p$, all of them must be. Then we find that,
\begin{align*}
a^{2023}b \equiv c^{2024} \pmod{p}
\end{align*}However then from this we may conclude that,
\begin{align*}
a^{2023}bc \equiv -a^{2025} \pmod{p}
\end{align*}Dividing as $a \not\equiv 0$ modulo $p$, we find that,
\begin{align*}
-bc \equiv a^2 \pmod{p}
\end{align*}Now rewriting our conditions we have,
\begin{align*}
b^{1012} &\equiv c^{1011}a \pmod{p}\\
b^{2024} &\equiv -c^{2024} \pmod{p}\\
c^{1013} &\equiv -b^{1012}a \pmod{p}
\end{align*}Squaring the first equation we have,
\begin{align*}
b^{2024} &\equiv c^{2022}a^2 \pmod{p}\\
-c^{2024} &\equiv c^{2022}a^2 \pmod{p}\\
-c^2 &\equiv a^2 \pmod{p}\\
-c^2 &\equiv -bc \pmod{p}\\
b &\equiv c \pmod{p}
\end{align*}From the first equation we then find $a \equiv b \bmod p$. Thus we have,
\begin{align*}
a \equiv b \equiv c \pmod{p}
\end{align*}Thus we have from the original equations,
\begin{align*}
2a^{2023} \equiv 2a^{2024} \equiv 2a^{2025} \equiv 0 \pmod{p}
\end{align*}from which we easily see $a \equiv 0 \bmod{p}$ and hence are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SilverBlaze_SY
66 posts
#31
Y by
The solution I originally did in the contest:
First, FTSOC, we assume that $p \nmid a$, $p \nmid b$, $p \nmid c$.
$$p \mid a^{2023}+b^{2023}...(i)$$$$p \mid b^{2024}+c^{2024}...(ii)$$$$p \mid a^{2025}+c^{2025}...(iii)$$From $(i)$, we get that: $p \mid a^{2025}+a^2b^{2023}$
Combining this with $(iii)$, $p \mid a^2b^{2023}-c^{2025}$

As $c^{2024} \equiv -b^{2024}$, $p \mid a^2b^{2023}+b^{2024}c \implies p \mid b^{2023}(a^2+bc)$
$\implies \boxed{p \mid a^2+bc}...(A)$
$\implies a^2 \equiv -bc \pmod{p}$

From $(iii)$, we have: $p \mid a^{2025}+c^{2025} \implies p \mid a.(bc)^{1012}+c^{2025}$
$\implies p \mid c^{1012}(ab^{1012}+c^{1013})$
$\implies p \mid a^2b^{2024}-c^{2026}$
Combining with $(ii)$, we get: $p \mid a^2b^{2024}+b^{2024}c^2 \implies p \mid b^{2024}(a^2+c^2)$
$\implies \boxed{p \mid a^2+c^2}...(B)$

Therefore, from $(A)$ and $(B)$, we have: $p \mid c(b-c) \implies b \equiv c \pmod{p}$
But then, in $(ii)$, $p\mid 2b^{2024} \implies p \mid 2$, which is a contradiction!

Then, $p$ must divide one of $a,b,c$. If $p$ divides any one of the numbers, $p$ must divide all the numbers individually, and we're done! :wow:
This post has been edited 1 time. Last edited by SilverBlaze_SY, Apr 27, 2024, 7:17 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shafikbara48593762
17 posts
#32 • 3 Y
Y by E.Sultan, allaith.sh, BR1F1SZ
$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$
Let $n$=2023×2024×2025
$$\implies a^{n} \equiv b^{n} \pmod{p} $$$$\implies b^{n} \equiv -c^{n} \pmod{p} $$$$\implies a^{n} \equiv c^{n} \pmod{p} $$And from this we get
$$c^{n} \equiv -c^{n} \pmod{p} $$and we are done
This post has been edited 2 times. Last edited by shafikbara48593762, May 9, 2024, 2:20 PM
Reason: Nicer
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
E.Sultan
7 posts
#33
Y by
shafikbara48593762 wrote:
$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$
Let $n$=2023×2024×2025
$$\implies a^{n} \equiv b^{n} \pmod{p} $$$$\implies b^{n} \equiv -c^{n} \pmod{p} $$$$\implies a^{n} \equiv c^{n} \pmod{p} $$And from this we get
$$c^{n} \equiv -c^{n} \pmod{p} $$and we are done

what a solution!!!
I was surprised by how organized and simple this solution was, it is a MASTERPIECE!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shafikbara48593762
17 posts
#34
Y by
E.Sultan wrote:
shafikbara48593762 wrote:
$$a^{2023} \equiv -b^{2023} \pmod{p} $$$$b^{2024} \equiv -c^{2024} \pmod{p} $$$$c^{2025} \equiv -a^{2025} \pmod{p} $$
Let $n$=2023×2024×2025
$$\implies a^{n} \equiv b^{n} \pmod{p} $$$$\implies b^{n} \equiv -c^{n} \pmod{p} $$$$\implies a^{n} \equiv c^{n} \pmod{p} $$And from this we get
$$c^{n} \equiv -c^{n} \pmod{p} $$and we are done

what a solution!!!
I was surprised by how organized and simple this solution was, it is a MASTERPIECE!

Thx bro
Such an easy problem for INMO P3
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
peppapig_
281 posts
#35
Y by
Note that if $p\mid a$, then
\[a^{2023}+b^{2023}\equiv 0 \mod p,\]\[\iff b^{2023} \equiv 0 \mod p,\]\[\iff p\mid b,\]and similarly we can find that $p\mid c$. In general, using a similar proof, it can be shown that if $p\mid abc$, then $p$ divides each of $a$, $b$, and $c$. FTSOC, assume that $p$ divides none of $a$, $b$, or $c$.

Now, notice that
\[a^{2023}+b^{2023}\equiv 0 \mod p,\]\[\iff a^{2023}\equiv -b^{2023} \mod p,\]\[\iff \left(\frac{a}{b}\right)^{2023}\equiv -1 \mod p,\]\[\iff ord_p\left(\frac{a}{b}\right) \mid 4046 \text{ and } 2\mid ord_p\left(\frac{a}{b}\right),\]since we had that $\left(\frac{a}{b}\right)^{2023}\equiv -1 \mod p$. Similarly, from the second equation, we get that
\[ord_p\left(\frac{b}{c}\right) \mid 4048 \text{ and } 16\mid ord_p\left(\frac{b}{c}\right).\]
Now, since $p$ is prime, it is well-known that it must have a primitive root. Let said primitive root be $g$ and define $m$ and $n$ to be the unique integers such that $\frac{a}{b}\equiv g^m\mod p$ and $\frac{b}{c}\equiv g^n\mod p$, where $1\leq m,n\leq p-1$. Now, notice that if
\[2\mid ord_p\left(\frac{a}{b}\right) \text{ and } ord_p\left(\frac{a}{b}\right) \mid 4046,\]we get that
\[\nu_2\left(ord_p\left(\frac{a}{b}\right)\right)=1,\]which in turn gives that
\[\nu_2(m)=\nu_2(p-1)-1,\]since the order of $g$ mod $p$ is $p-1$. Similarly, if
\[16\mid ord_p\left(\frac{b}{c}\right) \text{ and } ord_p\left(\frac{b}{c}\right) \mid 4048,\]then we have that
\[\nu_2(n)=\nu_2(p-1)-4.\]
Now, notice that
\[\frac{a}{c}\equiv \left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\equiv g^{m+n} \mod p,\]which means that
\[\nu_2\left(ord_p\left(\frac{a}{c}\right)\right)=\nu_2(p-1)-\nu_2(m+n).\]However, since
\[\nu_2(p-1)-1=\nu_2(m)>\nu_2(n)=\nu_2(p-1)-4,\]we get that
\[\nu_2(m+n)=\nu_2(p-1)-4,\]which gives that
\[\nu_2\left(ord_p\left(\frac{a}{c}\right)\right)=\nu_2(p-1)-(\nu_2(p-1)-4)=4.\]
However, we had that
\[c^{2025}+a^{2025}\equiv 0 \mod p,\]\[\iff a^{2025}\equiv -c^{2025} \mod p,\]\[\iff \left(\frac{a}{c}\right)^{2025}\equiv -1 \mod p,\]\[\iff ord_p\left(\frac{a}{c}\right) \mid 4050,\]a contradiction, since $16$ does not divide $4050$. Therefore $p$ must divide each of $a$, $b$, and $c$, as desired, finishing the problem.
This post has been edited 3 times. Last edited by peppapig_, Jun 7, 2024, 8:18 PM
Reason: Typo corrections
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alexanderhamilton124
391 posts
#36
Y by
If \( p \) divides one of \( a, b, c \), WLOG \( a \), we have \( p \mid b^{2023} \implies p \mid b \), \( p \mid c^{2025} \implies p \mid c \), so then \( p \) divides all of them.

Assume, for the sake of contradiction, \( p \nmid a, b, c \). We have the following modular congruences:
\[ \left(\frac{a}{b}\right)^{4046} \equiv 1 \pmod{p} \]\[ \left(\frac{b}{c}\right)^{4048} \equiv 1 \pmod{p} \]\[ \left(\frac{c}{a}\right)^{4050} \equiv 1 \pmod{p} \]Firstly, observe that \( p \mid a^{2023} + b^{2023} \implies p \mid a^{2025} + a^2b^{2023} \). Since \( p \mid a^{2025} + c^{2025} \), we have \( p \mid c^{2025} - a^2b^{2023} \). Further, \( p \mid b^{2024} + c^{2024} \implies p \mid c^{2025} + b^{2024}c \). From this, we conclude that:
\[ p \mid b^{2024}c + a^2b^{2023} = b^{2023}(bc + a^2) \implies a^2 \equiv -bc \pmod{p} \]since \( p \nmid b^{2023} \).

We have \( a^{4046} \equiv -(bc)^{2023} \), which, replacing in our first congruence, gives \( c^{2023} \equiv -b^{2023} \pmod{p} \implies c^{4046} \equiv b^{4046} \pmod{p} \). Replacing this in our second congruence, we have
\[ \frac{b^{4046}}{c^{4046}} \cdot \frac{b^2}{c^2} \equiv 1 \pmod{p} \implies b^2 \equiv c^2 \pmod{p} \implies b^{2024} \equiv c^{2024} \pmod{p} \]Therefore, \( b^{2024} + c^{2024} \equiv 2b^{2024} \equiv 0 \pmod{p} \). Since \( p \) is odd, we have \( p \mid b^{2024} \implies p \mid b \), which gives \( p \mid a, c \) as well, so we are done.

Edit: I would like to see a solution that doesn't use the even parity of 2024.
This post has been edited 1 time. Last edited by alexanderhamilton124, Aug 4, 2024, 5:18 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Siddharthmaybe
106 posts
#39
Y by
Raise everything to the power of 2023x2024x2025 mod p, then p divides a^2023x2024x2025 + or - c^2023x2024x2025 which gives p | a and the rest follows
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
little-fermat
147 posts
#40
Y by
I have discussed this question in my INMO 2024 video on my channel "little fermat". Here is the Video
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anudeep
172 posts
#41
Y by
Showing that any one of $a,b,c$ is divisible by $p$ would suffice. Assume all of them are invertible on $\mathbb{Z}_p$. As $b^{2024}\equiv -ba^{2023}$ we may say,
$$c^{2024}+b^{2024}\equiv c^{2024}-ba^{2023}\quad\text{and}\quad c^{2025}\equiv cba^{2023}.$$We then have $bca^{2023}+a^{2025}\equiv 0$. Since $a$ is invertible we are left with the key result,
$$a^2\equiv -bc.$$Using the above fact, $a(-bc)^{1012}\equiv-c^{2025}$ which deduces to $ab^{1012}\equiv -c^{1013}$ and as $a^{2025}\equiv cb^{2024}$ we can easily show that $b^{1012}\equiv ac^{1011}$. Now we are left with,
$$ab^{1012}\equiv a^2c^{1011}\equiv -c^{1013}\quad\implies\quad a^2\equiv -c^2.$$But we had already seen $a^2\equiv -bc$ and is absurd, contradicting our assumption, hence none of them are invertible as desired.$\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
John_Mgr
69 posts
#42
Y by
If any of the $a,b,c$ is divisible by $p$, them all of them are .
FOTSOC, assume that $p\nmid a, b, c$.
$a^{2023}\equiv -b^{2023}\pmod{p} $, $c^{2024}\equiv -b^{2024} \pmod{p} $ and $c^{2025}\equiv -a^{2025}\pmod{p} $
then, $c^{2025}\equiv c\cdot -b^{2024}\equiv -a^{2025}\equiv a^2\cdot b^{2023} \implies b^{2023}(c+ab)\equiv 0\pmod{p} $
$p\nmid b$. so $ab+c \equiv 0\pmod{p} $, $ab\equiv -c\pmod{p}  \rightarrow a^{2023}\cdot b^{2023}\equiv-c^{2023}\pmod{p} \implies (b^2)^{2023}\equiv c^{2023}\pmod{p} \implies b^2 \equiv c\pmod{p} $. Also $ab+c \equiv 0(modp)\implies ab+b^2 \equiv 0\pmod{p} \equiv b(a+b) (modp) $. so, $a\equiv -b\pmod{p} $
$a^{2025}+c^{2025} \equiv -b^{2025}+(b^2)^{2025}\equiv b^{2025}(-1+b^{2025})\equiv 0(modp) \implies b\equiv 1\pmod{p} $
then $b^{2024}+c^{2024}\equiv 0\pmod{p}  \implies b^{2024}+(b^2)^{2024}\equiv 2\equiv 0\pmod{p} $, Contradiction!!
So, $p$ divides and one of them which leads to $p$ divides of them i.e $p\mid a,b,c$.
Z K Y
N Quick Reply
G
H
=
a