We have your learning goals covered with Spring and Summer courses available. Enroll today!

G
Topic
First Poster
Last Poster
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/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
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
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
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
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, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
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
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
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
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
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
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
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

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
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
Sunday, Mar 23 - Aug 3
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Friday, Apr 11 - Jun 27
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21
Tuesday, Jun 10 - Aug 26

Calculus
Sunday, Mar 30 - Oct 5
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
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
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
Friday, Apr 11 - Jun 27
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
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
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

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
Monday, Mar 24 - Jun 16
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
Sunday, Mar 30 - Jun 22
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Tuesday, Mar 25 - Sep 2
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
Mar 2, 2025
0 replies
Inspired by Kazakhstan 2017
sqing   1
N a few seconds ago by sqing
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a^2+b^2+c^2=2. $ Prove that
$$\left(\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge \frac{8}{3}$$$$  \left( 2a^2+\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left( 2a^2+\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)  \ge 9\sqrt[3]{4}$$
1 reply
1 viewing
sqing
a minute ago
sqing
a few seconds ago
Number theory
Ecrin_eren   0
12 minutes ago
Show that there are no prime numbers satisfying the equation

(p + r)^q + (q + r)^p = (p + q)^r.

0 replies
Ecrin_eren
12 minutes ago
0 replies
Wait wasn't it the reciprocal in the paper?
Supercali   7
N 13 minutes ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
7 replies
Supercali
Jul 9, 2023
kes0716
13 minutes ago
Functional equation
Ecrin_eren   0
18 minutes ago
Find all functions f:R R satisfying the equation

f(f(x)y) + f(x+ f(y)) = x f(y) + f(x+y)

for all x,y real numbers

0 replies
Ecrin_eren
18 minutes ago
0 replies
Spheres and a point source of light
mofidy   3
N Yesterday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Yesterday at 9:22 PM
Real Analysis
rljmano   4
N Yesterday at 4:38 PM by alexheinis
In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?
4 replies
rljmano
Yesterday at 6:32 AM
alexheinis
Yesterday at 4:38 PM
find the isomorphism
nguyenalex   14
N Yesterday at 1:49 PM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Yesterday at 1:49 PM
find the convex sets satisfied
nguyenalex   2
N Yesterday at 9:54 AM by ILOVEMYFAMILY
In $\mathbb{R}^2$, let $B = \{(x, y) \mid x \geq 0\}$. Find all convex sets $C$ such that

\[\mathcal{E}(B \cup C) = B \cup C.\]
2 replies
nguyenalex
Yesterday at 8:57 AM
ILOVEMYFAMILY
Yesterday at 9:54 AM
A great result
steven_zhang123   10
N Yesterday at 9:35 AM by teomihai
Show that $\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4} $.
10 replies
steven_zhang123
Oct 31, 2024
teomihai
Yesterday at 9:35 AM
Weird family of sequences
AndreiVila   9
N Yesterday at 8:40 AM by Fibonacci_math
Source: Romanian District Olympiad 2025 12.3
[list=a]
[*] Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$. Show that $f(a)\cdot f(b)<0$.
[*] Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$$
9 replies
AndreiVila
Mar 8, 2025
Fibonacci_math
Yesterday at 8:40 AM
Monster Integral
Entrepreneur   1
N Yesterday at 7:41 AM by RezerdPrime
$$\color{blue}{\int_{0}^{\pi} \frac{\tan^{-1}\left(\frac{\ln(\sin(x))}{x}\right)dx}{\ln^2\left(x^2 + \ln^2(\sin(x))\right) + 4\arctan^2\left(\frac{\ln(\sin(x))}{x}\right)}= -\frac{\pi \tan^{-1}\left(\frac{2\ln(2)}{\pi}\right)}{\ln^2\left(\frac{\pi^2}{4} + \ln^2(2)\right) + 4\arctan^2\left(\frac{2\ln(2)}{\pi}\right)}.}$$
1 reply
Entrepreneur
Jan 10, 2025
RezerdPrime
Yesterday at 7:41 AM
convex closed set with a nonempty interior
ILOVEMYFAMILY   0
Yesterday at 6:11 AM
a) When $n = 2$, prove that a convex closed set with a nonempty interior that contains exactly one extreme point must contain a ray. (SOLVED)

b) When $n = 2$, find all convex closed sets with a nonempty interior that contain exactly one extreme point.
0 replies
ILOVEMYFAMILY
Yesterday at 6:11 AM
0 replies
Differentiation Marathon!
LawofCosine   178
N Yesterday at 5:26 AM by awzhang10
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
178 replies
LawofCosine
Feb 1, 2025
awzhang10
Yesterday at 5:26 AM
Integrals problems and inequality
tkd23112006   1
N Yesterday at 3:55 AM by removablesingularity
Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
1 reply
tkd23112006
Feb 16, 2025
removablesingularity
Yesterday at 3:55 AM
International Zhautykov Olympiad 2011 - Problem 2
wangsacl   14
N Jan 2, 2022 by Assassino9931
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy the equality,
\[f(x+f(y))=f(x-f(y))+4xf(y)\]for any $x,y\in\mathbb{R}$.
14 replies
wangsacl
Jan 16, 2011
Assassino9931
Jan 2, 2022
International Zhautykov Olympiad 2011 - Problem 2
G H J
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
wangsacl
182 posts
#1 • 2 Y
Y by Adventure10, Mango247
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy the equality,
\[f(x+f(y))=f(x-f(y))+4xf(y)\]for any $x,y\in\mathbb{R}$.
This post has been edited 2 times. Last edited by Amir Hossein, May 4, 2018, 2:50 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pco
23440 posts
#2 • 15 Y
Y by aahmeetface, Ankoganit, mhq, G.K-1.618, adilbek, Atpar, Jalil_Huseynov, qwertyboyfromalotoftime, Assassino9931, WinterSecret, Adventure10, Mango247, and 3 other users
wangsacl wrote:
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, then
\[f(x+f(y))=f(x-f(y))+4xf(y)\]
where $\mathbb{R}$ denote the set of real numbers.
Let $P(x,y)$ be the assertion $f(x+f(y))=f(x-f(y))+4xf(y)$

The function $f(x)=0$ $\forall x$ is a solution. Let us from now look for non all zero solutions.
Let $u$ such that $f(u)\ne 0$

$P(\frac x{8f(u)},u)$ $\implies$ $f(\frac x{8f(u)}+f(u))=f(\frac x{8f(u)}-f(u))+\frac x2$

And so $x=2f(\frac x{8f(u)}+f(u))-2f(\frac x{8f(u)}-f(u))$ and so any real $x$ may be written as $2f(y)-2f(z)$ for some $y,z$

$P(-f(z),z)$ $\implies$ $f(0)=f(-2f(z))-4f(z)^2$
$P(f(y)-2f(z),y)$ $\implies$ $f(2f(y)-2f(z))=f(-2f(z))+4f(y)^2-8f(y)f(z)$
Subtracting these two lines, we get : $f(2f(y)-2f(z))=(2f(y)-2f(z))^2+f(0)$

And so $f(x)=x^2+f(0)$ $\forall x$ which indeed is a solution.

Hence the two solutions :
$f(x)=0$ $\forall x$
$f(x)=x^2+a$ $\forall x$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mavropnevma
15142 posts
#3 • 2 Y
Y by Adventure10, Mango247
Notice the uncanny resemblance (if not utter equivalence) with Problem 3 at the 2007 Balkan Mathematical Olympiad.
See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=0d0afe7c258254c48c003fe03fe58e44#p827178.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Utkirstudios
56 posts
#4 • 1 Y
Y by Adventure10
Can anybody post other problems of first day?
Thanks!!!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lasha
204 posts
#5 • 2 Y
Y by Adventure10 and 1 other user
$f(x+f(y)) = f(x-f(y))+4xf(y) $. (1)
First note that $f(x)=0$ for any $x$ is solution to the given equation. Suppose there exist some $u$ such that $f(u)$ doesn't equal to $0$. Then, $4xf(u)$ can get equal to any real number, but by (1) it's the difference of two f-s. It means that any real can be expressed as a $f(a)-f(b)$. (2)
Denote $f(0)=c$. Than, setting $x=f(y)$ in (1), we get:
$f(2f(y))=(2f(y))^{2}+c $ (3).
Furthermore, setting $f(x)$ instead of $x$ in (1), $f(f(x)-f(y)=f(f(x)+f(y))-4f(x)f(y)$. So, as the right hand side is symmetric, $f(f(x)-f(y))=f(f(y)-f(x))$. Hence, $f$ is odd, because $f(x)-f(y)$ gets any real value by (2).
Denote by $A$ the set of all $a$, satisfying $f(a)=a^{2}+c$.
From (3), $2f(x)$ belong to $A$ and as $f$ is odd, also $-2f(x)$ belong to that set.
From (1), it's clear that if $a-b=2f(x)$ for some $x,a,b$, than $f(a)-f(b)=a^{2}-b^{2}$. So, if $a$ is in $A$, than so does $a+2f(x)$. From the last fact, as $-2f(a)$ is in$A$ for any $a$, than for any $b$, $-2f(a)+2f(b)=2(f(b)-f(a))$ is in $A$. But $f(b)-f(a)$ gets any real value, as well as $2(f(b)-f(a))$. So, $A=R$. It means, from the definition of $A$, that $f(x)=x^{2}+c$ holds for any $x$.
Finally, the only solutions are $f(x)=0$ and $f(x)=x^{2}+c$.
P.S. The problem is not easy, but the approach is quite well known :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MatjazL
36 posts
#6 • 2 Y
Y by Adventure10, Mango247
lasha wrote:
P.S. The problem is not easy, but the approach is quite well known :)

What about this approach is well known?

(I don't see what the main idea, that is said to be well known is?)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jjax
108 posts
#7 • 2 Y
Y by Adventure10, Mango247
The main idea is that the expression $f(a)-f(b)$ may take on any real value.
Such a step is certainly not easy to motivate. Indeed, it is the key step of IMO 1999 Q6 (so it's quite well known). However, after seeing a solution using this method, the method is easy to mimic for similar questions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
waver123
142 posts
#8 • 1 Y
Y by Adventure10
If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$.

Does that mean anything?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mavropnevma
15142 posts
#9 • 1 Y
Y by Adventure10
If you do that, you get $g(x+y^2 + g(y)) = g(x-y^2 - g(y))$. It remains for you to prove that this functional equation has as only solutions $g(x)=-x^2$ (leading to $f \equiv 0$), and $g(x) = c$ (constant, leading to $f(x) = x^2+c$).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lasha
204 posts
#10 • 2 Y
Y by Adventure10, Mango247
MatjazL wrote:
lasha wrote:
P.S. The problem is not easy, but the approach is quite well known :)

What about this approach is well known?

(I don't see what the main idea, that is said to be well known is?)



Sorry for a late response, I've just came across it :) Anyway, jjax anwered to you correctly. That approach has been used in IMO 1999/6 (But not only, I've solved several problems with that, just don't remember the sources).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
liimr
34 posts
#11 • 1 Y
Y by Adventure10
waver123 wrote:
If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$.

Does that mean anything?

solution with $g(x)$ is pretty good, you may see in
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=f14af484eddd9e2dc421aec7daa5ed21#p827178

for solution by Anto
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DanDumitrescu
196 posts
#12 • 3 Y
Y by Sharingan06, Adventure10, Mango247
I don't know if it's good but i think to create the function $g(x)=f(x)-x^2$ we can write $4xf(y)=(x+f(y))^2-(x-f(y))^2$ and we have that $g(x+f(y))=g(x-f(y)) $ and we can put y=0 and note that f(0)=a and we have that $f(k)=k^2+g(f(0))=k^2+f(a)-a^2$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
imn8128
13 posts
#13 • 2 Y
Y by Adventure10, Mango247
$P(x+f(y),z)$ yields $f(x+f(y)+f(z))=f(x+f(y)-f(z))+4(x+f(y))f(z)$
$P(x-f(y),z)$ yields $f(x-f(y)+f(z))=f(x-f(y)-f(z))+4(x-f(y))f(z)$
adding the two equations yields
$f(x+f(y)+f(z))-f(x+f(y)-f(z))+f(x-f(y)+f(z))-f(x-f(y)-f(z))=8xf(z)$
changing $y$ and $z$ yields
$f(x+f(y)+f(z))+f(x+f(y)-f(z))-f(x-f(y)+f(z))-f(x-f(y)-f(z))=8xf(y)$
subtracting the two equations yields
$f(x+f(y)-f(z))-f(x-(f(y)-f(z)))=4x(f(y)-f(z))$
since $f(a)-f(b)$ runs through the entire real numbers(when $f$ is nontrivial) we can write $$f(x+y)-f(x-y)=4xy$$thus $f(x)=x^2+c$, $f=0$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hungvuong
42 posts
#14
Y by
Let P(x,y) be the assertion f(x-f(y)) = f(x) -f(y)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Assassino9931
1188 posts
#15
Y by
liimr wrote:
waver123 wrote:
If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$.

Does that mean anything?

solution with $g(x)$ is pretty good, you may see in
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=f14af484eddd9e2dc421aec7daa5ed21#p827178

for solution by Anto

@liimr Are you sure that this solution can be used for this problem? We can get $g$ to be periodic but then in the $x=y+T$ equation $g(y)$ does not cancel and so we cannot work with $y = (-z-2T^2)/4T$ (or a similar expression) here.
This post has been edited 1 time. Last edited by Assassino9931, Jan 2, 2022, 5:41 PM
Z K Y
N Quick Reply
G
H
=
a