Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Stay ahead of learning milestones! Enroll in a class over the summer!
algebra
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Divisibility on 101 integers
BR1F1SZ   3
N an hour ago by ClassyPeach
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
3 replies
1 viewing
BR1F1SZ
Aug 9, 2024
ClassyPeach
an hour ago
J
H
BMO 2021 problem 3
VicKmath7   19
N an hour ago by NuMBeRaToRiC
Source: Balkan MO 2021 P3
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Proposed by Serbia
19 replies
VicKmath7
Sep 8, 2021
NuMBeRaToRiC
an hour ago
J
H
USAMO 2002 Problem 4
MithsApprentice   89
N 2 hours ago by blueprimes
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y) \] for all pairs of real numbers $x$ and $y$.
89 replies
MithsApprentice
Sep 30, 2005
blueprimes
2 hours ago
J
H
pqr/uvw convert
Nguyenhuyen_AG   8
N 2 hours ago by Victoria_Discalceata1
Source: https://github.com/nguyenhuyenag/pqr_convert
Hi everyone,
As we know, the pqr/uvw method is a powerful and useful tool for proving inequalities. However, transforming an expression $f(a,b,c)$ into $f(p,q,r)$ or $f(u,v,w)$ can sometimes be quite complex. That's why I’ve written a program to assist with this process.
I hope you’ll find it helpful!

Download: pqr_convert

Screenshot:
IMAGE
IMAGE
8 replies
Nguyenhuyen_AG
Apr 19, 2025
Victoria_Discalceata1
2 hours ago
J
H
Inspired by hlminh
sqing   2
N 2 hours ago by SPQ
Source: Own
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that $$ |a-kb|+|b-kc|+|c-ka|\leq \sqrt{3k^2+2k+3}$$Where $ k\geq 0 . $
2 replies
sqing
Yesterday at 4:43 AM
SPQ
2 hours ago
J
H
A cyclic inequality
KhuongTrang   3
N 2 hours ago by KhuongTrang
Source: own-CRUX
IMAGE
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
3 replies
KhuongTrang
Monday at 4:18 PM
KhuongTrang
2 hours ago
J
H
Tiling rectangle with smaller rectangles.
MarkBcc168   60
N 3 hours ago by cursed_tangent1434
Source: IMO Shortlist 2017 C1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore
60 replies
MarkBcc168
Jul 10, 2018
cursed_tangent1434
3 hours ago
J
H
ALGEBRA INEQUALITY
Tony_stark0094   2
N 3 hours ago by Sedro
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
2 replies
Tony_stark0094
4 hours ago
Sedro
3 hours ago
J
H
Checking a summand property for integers sufficiently large.
DinDean   2
N 3 hours ago by DinDean
For any fixed integer $m\geqslant 2$, prove that there exists a positive integer $f(m)$, such that for any integer $n\geqslant f(m)$, $n$ can be expressed by a sum of positive integers $a_i$'s as
\[n=a_1+a_2+\dots+a_m,\]where $a_1\mid a_2$, $a_2\mid a_3$, $\dots$, $a_{m-1}\mid a_m$ and $1\leqslant a_1<a_2<\dots<a_m$.
2 replies
DinDean
Yesterday at 5:21 PM
DinDean
3 hours ago
J
H
Bunnies hopping around in circles
popcorn1   22
N 3 hours ago by awesomeming327.
Source: USA December TST for IMO 2023, Problem 1 and USA TST for EGMO 2023, Problem 1
There are $2022$ equally spaced points on a circular track $\gamma$ of circumference $2022$. The points are labeled $A_1, A_2, \ldots, A_{2022}$ in some order, each label used once. Initially, Bunbun the Bunny begins at $A_1$. She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}$, after which she hops back to $A_1$. When hopping from $P$ to $Q$, she always hops along the shorter of the two arcs $\widehat{PQ}$ of $\gamma$; if $\overline{PQ}$ is a diameter of $\gamma$, she moves along either semicircle.

Determine the maximal possible sum of the lengths of the $2022$ arcs which Bunbun traveled, over all possible labellings of the $2022$ points.

Kevin Cong
22 replies
popcorn1
Dec 12, 2022
awesomeming327.
3 hours ago
J
H
J
H
Functional equations
hanzo.ei   20
N Apr 17, 2025 by hanzo.ei
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
20 replies
hanzo.ei
Mar 29, 2025
hanzo.ei
Apr 17, 2025
J
Functional equations
G H J
Reveal topic tags
algebra
functional equation
L
G H BBookmark kLocked kLocked NReply
Source: Greekldiot
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hanzo.ei
20 posts
hanzo.ei
#1 h  Mar 29, 2025, 4:33 PM
Y by
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#2 h  Mar 29, 2025, 5:16 PM
Y by
Not my problem :D Havent made such a beautiful FE myself yet.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathzeus1024
826 posts
Mathzeus1024
#3 h  Mar 30, 2025, 10:12 AM
Y by
It works for $\textcolor{red}{f(x)=\frac{1}{x}}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#4 h  Mar 30, 2025, 10:23 AM
Y by
Yeah we established that in another post
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hanzo.ei
20 posts
hanzo.ei
#5 h  Mar 30, 2025, 10:42 AM
Y by
pco, can you solve it :omighty:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#6 h  Mar 30, 2025, 11:27 AM
Y by
$f$ seems to be an involution. I wonder if we are able to prove that. Then we can eliminate other solutions to the assertion very easily.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hanzo.ei
20 posts
hanzo.ei
#7 h  Apr 2, 2025, 3:28 PM
Y by
bump!!!!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MR.1
102 posts
MR.1
#8 h  Apr 3, 2025, 4:08 PM • 3 Y
Y by hanzo.ei, Akakri, giangtruong13
solved with GioOrnikapa if you guys want solution please give me $10$ likes :-D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#9 h  Apr 3, 2025, 4:17 PM
Y by
lol aint no way
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GioOrnikapa
76 posts
GioOrnikapa
#10 h  Apr 3, 2025, 4:33 PM • 1 Y
Y by MR.1
A lot of liars nowadays smh
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
truongphatt2668
323 posts
truongphatt2668
#11 h  Apr 3, 2025, 4:34 PM
Y by
hanzo.ei wrote:
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?
This post has been edited 3 times. Last edited by truongphatt2668, Apr 3, 2025, 4:52 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#12 h  Apr 3, 2025, 5:55 PM
Y by
truongphatt2668 wrote:
hanzo.ei wrote:
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

I noticed that there exists some homogenous-like function by isolating $y$ on the $RHS$. Can you post the claims you made with proof so that we can create a complete solution?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
truongphatt2668
323 posts
truongphatt2668
#13 h  Apr 4, 2025, 1:42 AM
Y by
GreekIdiot wrote:
truongphatt2668 wrote:
hanzo.ei wrote:
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

I noticed that there exists some homogenous-like function by isolating $y$ on the $RHS$. Can you post the claims you made with proof so that we can create a complete solution?

Just do it, and I will give a complete solution :D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#14 h  Apr 4, 2025, 3:29 PM
Y by
$f(x)>0$ for all $x \in \mathbb R_+$ so all $y \in \mathbb R_+$ can be written as $\dfrac {f(m)}{f(n)}$ for some $m,n \in \mathbb R_+$
Then there exists some homogenous-kinda function (lets call it $g$) such that $f(xf(y)+f(x))=y^{\ell +1} \cdot g(x)$ and also $f(x+yf(x))=y^{\ell} \cdot g(x)$ thats what I meant to say. Correct me if wrong lol. :oops:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11236 posts
jasperE3
#15 h  Apr 4, 2025, 4:36 PM
Y by
GreekIdiot wrote:
$f(x)>0$ for all $x \in \mathbb R_+$ so all $y \in \mathbb R_+$ can be written as $\dfrac {f(m)}{f(n)}$ for some $m,n \in \mathbb R_+$
Then there exists some homogenous-kinda function (lets call it $g$) such that $f(xf(y)+f(x))=y^{\ell +1} \cdot g(x)$ and also $f(x+yf(x))=y^{\ell} \cdot g(x)$ thats what I meant to say. Correct me if wrong lol. :oops:

What's a homogenous kinda function
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#16 h  Apr 4, 2025, 8:05 PM
Y by
I am not sure how to call it in english or even what it is. Hope you can understand what I am saying from the symbols :D Thats the important part anyways, not some random math definition.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11236 posts
jasperE3
#17 h  Apr 4, 2025, 8:07 PM
Y by
GreekIdiot wrote:
I am not sure how to call it in english or even what it is. Hope you can understand what I am saying from the symbols :D Thats the important part anyways, not some random math definition.

I can't, can you explain?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#18 h  Apr 4, 2025, 8:12 PM
Y by
So basically I am trying to define a second function, g, which exists and satisfies both relations above. Then proving g must be constant will help in proving that the only sol we have found so far is unique. Hope that clears things up.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11236 posts
jasperE3
#19 h  Apr 4, 2025, 8:15 PM
Y by
GreekIdiot wrote:
So basically I am trying to define a second function, g, which exists and satisfies both relations above. Then proving g must be constant will help in proving that the only sol we have found so far is unique. Hope that clears things up.

How is $g$ defined
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
174 posts
GreekIdiot
#20 h  Apr 4, 2025, 8:18 PM
Y by
$g(x) > 0$ is a must for all positive $x$. Then it could be any function but we may be able to narrow it down. Just brainstorming, nothing rigorous. This FE has been unsolved for some time, I doubt that I of all people will be the one to solve.
This post has been edited 3 times. Last edited by GreekIdiot, Apr 4, 2025, 8:28 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hanzo.ei
20 posts
hanzo.ei
#21 h  Apr 17, 2025, 2:09 PM
Y by
:blush: :blush:
Z K Y
N Quick Reply
G
H
=
a