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
H not needed
dchenmathcounts   44
N 17 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
17 minutes ago
IZHO 2017 Functional equations
user01   51
N 38 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
38 minutes ago
chat gpt
fuv870   2
N 39 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
39 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 41 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
41 minutes ago
No more topics!
Functional equation
socrates   30
N Today at 8:37 AM by NicoN9
Source: Baltic Way 2014, Problem 4
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\] for all real numbers $x, y.$
30 replies
socrates
Nov 11, 2014
NicoN9
Today at 8:37 AM
Functional equation
G H J
G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2014, Problem 4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
socrates
2104 posts
#1 • 12 Y
Y by A-Thought-Of-God, FaThEr-SqUiRrEl, samrocksnature, mathematicsy, centslordm, megarnie, jhu08, tiendung2006, ImSh95, Adventure10, Mango247, ItsBesi
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\] for all real numbers $x, y.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Dadgarnia
164 posts
#2 • 20 Y
Y by socrates, LittelPerel, Abdollahpour, e_plus_pi, Wizard_32, Pluto1708, Beginner2004, A-Thought-Of-God, thczarif, FaThEr-SqUiRrEl, samrocksnature, mathematicsy, centslordm, jhu08, megarnie, Kanimet0, ImSh95, Adventure10, Mango247, Motaha313
socrates wrote:
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\] for all real numbers $x, y.$
Let $P(x,y)$ be the assertion $f(f(y))+f(x-y)=f(xf(y)-x)$. We get:
$P(0,0)\rightarrow f(f(0))=0$
$P(0,f(0)),P(f(0),f(0))\rightarrow f(0)=0$
$P(x,0)\rightarrow f(x)=f(-x)\Rightarrow P(x,-x),P(x,x)\rightarrow f(x)\equiv 0$
Hence $ f(x)\equiv 0 \,\,\, \forall x\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.
silouan
3952 posts
#3 • 10 Y
Y by socrates, john111111, starfire27, FaThEr-SqUiRrEl, samrocksnature, centslordm, jhu08, ImSh95, Adventure10, Mango247
Nice problem!!
Let $P(x,y)$ the given relation.
Let $y\in\mathbb{R}$ such that $f(y)\neq 2.$
Then $P(-\frac{y}{f(y)-2},y)$ gives $f(f(y))=0.$
So, if $f(y)\neq 2$ the initial gives: $f(x-y)=f(xf(y)-x)$
and for $x=0$ the last one gives $f(-y)=f(0)$
We conclude that either $f(y)=2$ or $f(-y)=f(0).$
But if $f(y)=2$ then $f(-y)=f(0)-f(2)$
So $f(y)=c_1$ for some $y$ and $f(y)=c_2$ for all the other and $c_1=2\neq 0.$
Take $y_0$ such that $f(y_0)=2.$ Then the initial gives: $f(2)+f(x-y_0)=f(x)$
Taking $x=2$ to the last one we get $f(2-y_0)=0$ so $c_2=0$ and $f(0)=f(2).$
So $f(y)=2$ for some $y$ and $f(y)=0$ for all the other.
Now take $x=y=y_0,$ then $f(2)+f(0)=f(y_0)=2.$ But $f(0)=f(2)$ so $f(0)=f(1)=1,$ this is a contradiction.
So $f(x)=0$ for all $x.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
utkarshgupta
2280 posts
#4 • 6 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, mannshah1211
Problem (Baltic Way 2014):
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\]for all real numbers $x, y.$


Solution :

Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$

$f \equiv 1$ isn't obviously a solution.
Thus there exists some $a \in \mathbb R$ such that $f(a) \neq 1$

$P(\frac{f(a)}{f(a)-1},a) \implies$
$$ f(\frac{f(a)}{f(a)-1}-a)=0$$
$\implies$ there exists a $k$ such that $f(k)=0$

$P(x,k) \implies f(0)+f(x-k)=f(-x)$
$P(0.k) \implies f(-k)=0$
$P(k,k) \implies f(0)=0$

$\implies f(0)=0$

$$P(0,y) \implies f(f(y)) + f(-y) = 0$$
$$P(\frac{-y}{f(y)-2},y) \implies f(y)=0$$for all $f(y) \neq 2$


Now let there exist some $t \in R$ such that $f(t)=2$
$f(f(t))= - f(t) = -2$
But we know that that for all $f(x) \neq 2$, we have $f(x)=0$

A contradiction.

$\implies \boxed{f \equiv 0}$ which indeed is a solution.
This post has been edited 1 time. Last edited by utkarshgupta, Nov 4, 2015, 1:26 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
E.A.K
52 posts
#6 • 6 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, Mango247
$x=y=0        f(f(0))=0$
$y=f(0)        f(0)+f(x-f(0))=f(-x)$
$x=0          f(0)+f(-f(0))=f(0)$
$ f(-f(0))=0$
$x=f(0)       2f(0)=f(-f(0))=0$
$ f(0)=0$
$y=0      f(x)=f(-x)$
$x=y       f(f(x))=f(xf(x)-x)$
$y=-x       f(f(x))+f(2x)=f(xf(x)-x)$
$ f(x)=0$
This post has been edited 4 times. Last edited by E.A.K, Apr 17, 2018, 6:27 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
e_plus_pi
756 posts
#7 • 8 Y
Y by nhusanboev, A-Thought-Of-God, FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, Mango247
Sweet and Simple :) .

Answer: We show that the only function satisfying the assertion is $\boxed{f(x) \equiv 0}$.

Proof: Let (as usual) $P(x,y)$ denote the assertion $f(f(y)) + f(x-y) = f(xf(y) - x)$.

Claim 1. $f(0) =0$.
$\longrightarrow$ $P(0,0) \rightarrow f^2(0) + f(0) = f(0) \implies f^2(0) =0$

Now let $f(0) =c $ for some $c \in \mathbb{R}$. Consider $P(0,c) \rightarrow f^2(c) + f(-c) = c \iff f(-c)=0$.

Also consider $P(c,c) \rightarrow f^2(c) + c = f( c f(c) -c) \implies 2c = 0 \iff c= 0$.


Claim 2. $f$ is even.
$\longrightarrow$ Consider $P(x,0) \rightarrow f(x) = f(-x)$

Claim 3. $f \equiv 0$.
$\longrightarrow$ Consider the two equations : $P(x,x)$ and $P(x,-x)$.

(1) $\rightarrow f^2(x) + f(2x) =f(xf(x) -x).$

(2) $\rightarrow f^2(x) = f(xf(x) -x).$

Compairing the two equations yields $ f(2x) = 0 \forall x\in \mathbb{R}$.
This post has been edited 1 time. Last edited by e_plus_pi, May 8, 2018, 6:34 AM
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
#8 • 6 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10, Mango247
Let $P(x,y)$ denote the assertion $f(f(y))+f(x-y)=f(xf(y)-x).$

We claim that the only solution is $\boxed{f(x)=0}.$

Clearly, this solution does work and we aim to prove that this is the only solution.

Claim 1: $f(f(0))=0.$
Note that $P(0,0)$ gives $f(f(0))+f(0)=f(0)$ implying the desired.

Let $f(0)=a$ so $f(a)=0$ so $f(f(0))=0$.

Claim 2: The only value of $a$ that works is $a=0$ so $f(0)=0.$
Let $P(0,a)$ so $a+f(-a)=f(0)=a\implies f(-a)=0.$ Let $P(a,a)$ so $2a=f(-a)$. Hence, $2a=f(-a)=0\implies a=0.$ Thus, $f(0)=0.$

Claim 3: $f$ is even.
Let $P(x,0)$ so $f(x)=f(-x)$ with the fact that $f(0)=0.$

Claim 4: $f(x+y)=f(x-y)$ for all real $x,y\in \mathbb{R}$.
Consider $P(x,y)$ and $P(x,-y)$ which give, respectively, \begin{align*} f(f(y))+f(x-y)=f(xf(y)-x)\\f(f(-y))+f(x+y)=f(xf(-y)-x). \end{align*}Rearranging and noting that $f$ is even gives $$f(x-y)=f(xf(y)-x)-f(f(y))=f(xf(-y)-x)-f(f(-y))=f(x+y).$$
Hence, we can set $x=y=\frac{x}{2}$ in $f(x-y)=f(x+y)$ so $f(x)=f(0)=0.$ $\blacksquare$
This post has been edited 1 time. Last edited by Jzhang21, Oct 22, 2018, 2:27 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
winnertakeover
1179 posts
#9 • 5 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95, Adventure10
Taking $x=y=0$ gives $f(f(0))=0$. Taking $y=f(0)$ implies $f(0)+f(x-f(0))=f(-x)$. Letting $x=f(0)/2$ in the previous equation yields $f(0)=0$. Putting this back into the equation it was derived from, we see $f(x)=f(-x)$ or that $f$ is even. From the original FE, we have $f(x-y)=f(xf(y)-x)-f(f(y)).$ Swapping the sign on $y$ and noting evenness, $f(x+y)=f(xf(y)-x)-f(f(y))$. Hence $f(x-y)=f(x+y)$ for all $x,y$, which shows the function is constant. Hence, there is only one solution: $\boxed{f(x)=0}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rafaello
1079 posts
#10 • 4 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95
Let $P(x,y)$ denote the assertion $f(f(y))+f(x-y)=f(xf(y)-x).$

$P(0,0)$: $f(f(0))+f(0)=f(0)\implies f(f(0))=0$.

$P(0,y)$: $f(f(y))+f(-y)=f(0)$.

$P(x,f(0))$: $f(f(f(0)))+f(x-f(0))=f(xf(f(0))-x) \implies f(0)+f(x-f(0))=f(-x)$.

Last two together give us that $f(f(x))=-f(x-f(0))$, which means that when $x=f(0)$, thus $f(f(f(0)))=-f(f(0)-f(0)) \implies f(0)=-f(0)\implies f(0)=0$
Hence, $-f(-x)=f(f(x))=-f(x)\implies f(x)=f(-x)$.

$P(x,y)$: $f(f(y))+f(x-y)=f(xf(y)-x)$
$P(y,x)$: $f(f(x))+f(y-x)=f(yf(x)-y)$
Since $f(x-y)=f(y-x)$, thus $f(y)-f(x)=f(f(x))-f(f(y))=f(xf(y)-x)-f(yf(x)-y)$, where the first equality comes from knowing that $-f(x)=f(f(x))$.
Now, set $x$ to be $0$ and we get that $f(y)-f(0)=f(0\cdot f(y)-0)-f(yf(0)-y)\implies f(y)=-f(-y)=-f(y) \implies f(y)=0\forall y \in \mathbb{R}$.
We claim $\boxed{f(y)=0\forall y \in \mathbb{R}}$ to be solution to this problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
A-Thought-Of-God
454 posts
#12 • 4 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95
Nice Problem. :)
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
quirtt
163 posts
#13 • 4 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95
Nice problem. :D

solution
This post has been edited 3 times. Last edited by quirtt, Feb 5, 2021, 5:37 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EulersTurban
386 posts
#14 • 4 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, ImSh95
The only solution is that $f \equiv 0$.

By setting $x=y=0$, we have that $f(f(0))=0$.
For simplicity let's say that $c=f(0)$.

Now let's plug into the assertion $y \rightarrow c$, then we have that $c+f(x-c)=f(-x)$.
Now let's plug into the assertion $y = 0$, then we must have that $f(x)=f(xc-x)$

Plug in $x=0$, then we must have that $f(f(y))=-f(-y)=-c-f(y-c)$
Set $y=0$ into the upper relation to get that $f(-c)=-c$

Into the original assertion plug in $y=-c$, then we must have that $-c+f(x+c)=f(-xc-x)$.
Into the upper relation plug in $x=0$ to get that $c=0$.

This implies that $f(x)=f(-x)$.

This easily implies that $f(x+y)=f(x-y)$ and then we just set $x=y$, to get that $f(2x)=0$.
Thus this implies that $f \equiv 0$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11091 posts
#15 • 5 Y
Y by FaThEr-SqUiRrEl, samrocksnature, centslordm, tony88, ImSh95
Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$.

$P(0,0)\Rightarrow f(f(0))=0$
$P(0,f(0))\Rightarrow f(-f(0))=0$
$P(f(0),f(0))\Rightarrow 2f(0)=f(-f(0))\Rightarrow f(0)=0$

$P(x,0)\Rightarrow f(x)=f(-x)$

$P(x,x)\Rightarrow f(f(x))=f(xf(x)-x)$
$P(x,-x)\Rightarrow f(f(x))+f(2x)=f(xf(x)-x)$
Comparing these, $\boxed{f(x)=0}$, which works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MrOreoJuice
594 posts
#16 • 2 Y
Y by centslordm, ImSh95
Same solution but anyways.
Answer: $f(x) = 0 \ \forall \ x$.
Proof:
Let $P(x,y)$ denote the given assertion.

Claim: $f(0) = 0.$
Proof:
$P(0,0) \to f(f(0)) + f(0) = f(0) \implies \boxed{f(f(0)) = 0}$.

$P(0 , f(0)) \to f(0) + f(-f(0)) = f(0) \implies \boxed{f(-f(0)) = 0}$.

$P(f(0) , f(0)) \to f(0) + f(0) = f(-f(0)) = 0 \implies \boxed{f(0) = 0}$.

Claim: $f$ is even.
Proof:
$P(x,0) \to f(f(0)) + f(x) = f(xf(0) - x) \implies \boxed{f(x) = f(-x)}$.

$P \left(\frac x2 , \frac x2 \right) \to f \left( f \left (\frac x2 \right) \right) + f(0) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right) \implies f \left( f \left (\frac x2 \right) \right) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right)$

$P \left(\frac x2 , -\frac x2 \right) \to f \left( f \left ( - \frac x2 \right) \right) + f(x) = f \left( \frac x2 f \left( - \frac x2 \right) - \frac x2 \right)$

$\implies f \left( f \left ( \frac x2 \right) \right) + f(x) = f \left( \frac x2 f \left( \frac x2 \right) - \frac x2 \right)$ (Using $f$ is even.)

$\implies f \left( f \left ( \frac x2 \right) \right) + f(x) = f \left( f \left ( \frac x2 \right) \right)$ (Equating $P \left(\frac x2 , \frac x2 \right)$ and $P \left(\frac x2 , -\frac x2 \right)$ .)

$\implies f(x) = 0$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hakN
428 posts
#17 • 3 Y
Y by centslordm, ImSh95, Mango247
Let $P(x,y)$ be the assertion.
$P(0,0)\implies f(f(0))=0$.
$P(x,0)\implies f(x)=f(x(f(0)-1))$.
If $c=f(0)-1 \neq 1$, then comparing $P(x,y)$ with $P(x,yc)$ gives $f(x)=f(x+y(1-c))$, which implies that $f$ is constant since $c\neq 1$. So only solution is $f(x)=0\ \forall x\in \mathbb{R}$ in this case.
If $f(0)=2$, then we have $f(2)=0$. From $P(x,2)$ we get $2+f(x-2)=f(-x)$, but plugging $x=4$ and $x=-2$ gives a contradiction. Thus our only solution is $\boxed{f(x)=0 \ \forall x\in \mathbb{R}}$, which clearly fits.
This post has been edited 1 time. Last edited by hakN, Sep 17, 2023, 5:58 PM
Reason: fixed error
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rafaello
1079 posts
#18 • 2 Y
Y by centslordm, ImSh95
Let $P(x,y)$ be the assertion of $f(f(y)) + f(x - y) = f(xf(y) - x)$.

We have $P(0,0)\implies f(f(0))=0$, thus $P(0,f(0))\implies f(-f(0))=0$ and therefore $P(f(0),f(0))\implies f(0)=0$.

Now, $P(x,0)\implies f(x)=f(-x)$ and $P(0,x)\implies f(f(x))=-f(-x)=-f(x)$.

This means that $$-f(x)=f(f(x))=f(-f(x))=f(f(f(x)))=-f(f(x))=f(x)\implies f\equiv 0,$$this obviously works.
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
#19 • 4 Y
Y by ImSh95, Mango247, Mango247, Mango247
Let $P(x,y)$ denote the assertion.
$P(0,0)\implies f(f(0))=0$
$P(1,0) \implies f(1)=0$
$P(0,1) \implies f(-1)=0$
$P(1,1) \implies f(0)=0$,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OGGY_666
62 posts
#20 • 1 Y
Y by ImSh95
Sprites wrote:
Let $P(x,y)$ denote the assertion.
$P(0,0)\implies f(f(0))=0$
$P(1,0) \implies f(1)=0$
$P(0,1) \implies f(-1)=0$
$P(1,1) \implies f(0)=0$,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$

Will you tell me how does $P(1,0) \implies f(1)=0 ?$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11091 posts
#21 • 1 Y
Y by ImSh95
Sprites wrote:
Let $P(x,y)$ denote the assertion.
$P(0,0)\implies f(f(0))=0$
$P(1,0) \implies f(1)=0$
$P(0,1) \implies f(-1)=0$
$P(1,1) \implies f(0)=0$,so by induction $\boxed{f(x)=0 \forall x\in \mathbb{R}}$

How do you induct to prove over $\mathbb R$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rahmanhabibi
12 posts
#22 • 1 Y
Y by ImSh95
hakN wrote:
Let $P(x,y)$ be the assertion.
$P(0,0)\implies f(f(0))=0$.
$P(x,0)\implies f(x)=f(xf(0))$.
Plugging $x=1$ in above, we get $f(1)=0$.
$P(0,x)\implies f(f(x))=f(0) - f(-x)$.
Plugging $x=1$ in above, we get $f(-1)=0$.
$P(x,1)\implies f(0) + f(x-1) = f(-x)$.
Plugging $x=1$ in above, we get $f(0)=0$.
So we get $\boxed{f(x)=0 \forall x\in \mathbb{R}}$ which clearly fits.

Can u explain how u received line 3
How f(x)=f(xf(0))$
This post has been edited 1 time. Last edited by Rahmanhabibi, Sep 12, 2021, 2:29 PM
Reason: Adding some thing
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
naruto_uzumaki_06
6 posts
#23 • 1 Y
Y by ImSh95
Let $P(x,y)$ be the assertion.
$P(0,0): f(f(0))=0.$ Let $f(0)=a \Rightarrow f(a)=0$
$P(a,a):f(-a)=2a$
$P(0,a): 3a=f(0)=a\Rightarrow f(0)=0$
$P(x,0):f(-x)=f(x) \forall x \in \mathbb{R}$
$P(x,-y):f(f(y))+f(x+y)=f(xf(y)-x)\Rightarrow f(x-y)=f(x+y) \forall x,y (1)$
In (1), let $x=y$, we get $f(x)=a \text{(a=const)}$
It is easy to show that $f(x)=0$ is the only function which works.
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
584 posts
#24 • 1 Y
Y by ImSh95
$$x=0 \implies f(0)=f(f(y))+f(-y)\rightarrow (1),x=y \implies f(0)=f(xf(x)-x)-f(f(x)) \implies f(f(0))=0$$$$x=y=f(0)\implies 2f(0)=f(-f(0)),y=f(0),x=0 \implies f(0)=f(f(f(0)))+f(-f(0))=f(0)+2f(0)=3f(0)\implies f(0)=0$$$$y=0 \implies f(x)=f(-x),f(0)=0 \implies f(f(x))=-f(x)\text{from (1)},f(x)=-(-f(x))=-f(f(x))=f(f(f(x)))=f(-f(x))\implies f(-f(x))=f(x)$$Finally,
$$x=f(x),y=0 \implies f(f(x))=f(-f(x)) \implies f(x)=-f(x) \implies f(x)=0$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZETA_in_olympiad
2211 posts
#25 • 1 Y
Y by ImSh95
Let $P(x,y)$ denote the given assertion.

$P(0,0)$ gives $f(f(0))=0.$ Further, $P(0,f(0))$ gives $f(-f(0))=0$ and finally $P(f(0),f(0))$ gives $f(0)=0.$ Then $P(x,0)$ implies $f$ is even. Subtracting $P(x,x)$ from $P(x,-x)$ implies $f(2x)=0.$ And it's easy to verify the identity zero function works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZETA_in_olympiad
2211 posts
#26 • 1 Y
Y by ImSh95
Rahmanhabibi wrote:
Can u explain how u received line 3
How f(x)=f(xf(0))$

It's wrong.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5531 posts
#27 • 3 Y
Y by ImSh95, Mango247, Mango247
The only solution is $\boxed{f\equiv 0}$, which works.


Let $P(x,y)$ denote the given assertion.

Let $f(0) = a$.
$P(0,0): f(a) = 0$.

$P(0,a): a + f(-a) = a$, so $f(-a) = 0$.

$P(a,a): 2a = f(-a)$, so $a=0$.

$P(x,0): f(x) = f(-x)$.

$P(0,x): f(f(x)) + f(x) = 0$.

$P(x+y,y): f(f(y)) + f(x) = f(xf(y)  - x)$.

$P(x+y,-y): f(f(y)) + f(x+2y) = f(xf(y) - x)$.

So $f(x) = f(x+2y)$, which implies $f$ is constant, so $f\equiv 0$.
This post has been edited 1 time. Last edited by megarnie, Sep 20, 2022, 4:36 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MagicalToaster53
159 posts
#28 • 1 Y
Y by ImSh95
I claim that the only solution is $f(x) \equiv 0$, which works.

Consider the following:
\begin{align*}
P(0, 0): &\rightarrow f(f(0)) = 0 \\
P(0, x): &\rightarrow f(f(x)) + f(-x) = f(0) \\
P(x, x): &\rightarrow f(f(x)) + f(0) = f(xf(x)-x) \\
P(x, 0): &\rightarrow f(x) = f(-x) \\
\\
P(0, f(0)): &\rightarrow f(-f(0)) = 0 \\
P(f(0), f(0)): &\rightarrow f(0) = 0 \\
\\
P(-x, x) \wedge P(x, 0): &\rightarrow f(f(x)) + f(-2x) = f(xf(x) - x) \\
P(-x, x) \wedge P(x, x): &\rightarrow f(-2x) = 0.
\end{align*}
Thus we have that as $-2x$ is surjective, $f(x) \equiv 0$ for all $x \in \mathbb{R}$. $\blacksquare$
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
#29 • 1 Y
Y by ImSh95
We claim that the only solution is $f\equiv0$, which clearly fits.
We refer the assertion $f(f(y))+f(x-y)=f(xf(y)-x)$ as $p(x,y)$. Note that if $f(x)$ is a constant function then $f\equiv0$. Assume that $f(x)$ is a non-constant function.
Note that $p(0,-y)$ gives $f(y)=f(f(-y))$.
Since the function is non-constant, there exists $y$ such that $f(y)\ne2$. Then, $p\left(\frac{y}{2-f(y)},y\right)$ gives $f(f(y))=0$. But $f(f(y))=f(-y)$. So, $f(-y)=0$. Now, $p\left(-\frac{y}{2},-y\right)$ gives $f(f(-y))=0=f(y)$. We conclude that either $f(y)\in\{0,2\}$. Note that $p(0,0)$ gives $f(f(0))=0=f(0)$. But $p(x,0)$ gives $f(x)=f(-x)$. Take $y$ such that $f(y)=2$. We have $2=f(y)=f(-y)=f(f(y))=f(2)$. Then, $p(x+2,2)$ gives $2+f(x)=f(x+2)$. It follows that $f(x)=0$ and $f(x+2)=2$. There is no restriction on $x$, so we can take any real and $f(x)=0$. So, $f\equiv0$, contradicting $f(2)=2$.
So, there are no solutions to the given equation other than $f\equiv0$. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tyx2019
21 posts
#30
Y by
Let $P(x,y)$ denote the assertion.

Note that if $f(y)\neq 2$, $f(f(y))=0$. Proof: Choose $x=-\frac{y}{f(y)-2}$.

Now clearly exists a number $l$ such that $f(l)\neq 2$, since $f(x)=2$ for all $x$ is not a solution. Hence $f(f(l))=0, f(f(f(l)))=f(0)=0$ and so $f(0)=0$. Now suppose exists $k$ such that $f(k)=2$,

$P(k,k): f(2)+f(0)=f(k)=2$ so $0=f(0)=f(2)-2$. So $f(2)=2$.

Now $P(x,0): 0+f(x)=f(-x)$, so $f(x)=f(-x)$.

$P(0,k): 2+f(-k)=0$. So $f(-k)=-2$ at the same time $f(k)=2$ and since $f(k)=f(-k)$, $2=-2$ which is not okay

So there is no $k$ such that $f(k)=2$. So $f(f(x))=0$ for all $x$.

$P(0,y): f(-y)=f(0)$ so $f$ is constant. Now checking only $f(x)=0$ works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ItsBesi
136 posts
#34
Y by
My solution is similar to #$6,7$ but I'm gonna post it anyway
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RedFireTruck
4207 posts
#35
Y by
Let $P(x, y)$ denote the assertion.

$P(0,0)$ gives $f(f(0))+f(0)=f(0)$ so $f(f(0))=0$.

$P(x, 0)$ gives $f(x)=f(x(f(0)-1))$.

$P(1, y)$ gives $f(f(y))+f(1-y)=f(f(y)-1)$ so plugging $y=f(0)$ gives $f(0)+f(1-f(0))=f(-1)$ but $f(x)=f(x(f(0)-1))$ so $f(0)+f(-1)=f(-1)$ so $f(0)=0$. Plugging this back into $f(x)=f(x(f(0)-1))$ gives $f(x)=f(-x)$.

$P(0, y)$ gives $f(f(y))+f(-y)=0$ so $f(f(y))=-f(y)$.

We claim that the only solution is $\boxed{f(x)=0}$. Assume FTSOC that there exists some $c\ne0$ in the image of $f$. Then, $f(c)=f(-c)=-c$ so $-c\ne 0$ is also in the image of $f$. However, this means that $f(-c)=f(c)=c$. Since $c\ne -c$, this is a contradiction, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NicoN9
78 posts
#36
Y by
The answer is $f(x)\equiv 0$, which obviously works. We claim that $f$ is constant. Let $P(x, y)$ denote the assertion.

Now we have \begin{align}
P(0, 0)\Longrightarrow f(f(0)) &= 0,\\
P(0, y)\Longrightarrow f(f(y)) + f(-y) &= f(0),\\
P(x, f(0))\Longrightarrow f(0) + f(x-f(0)) &= f(-x).
\end{align}But since , from ($2$) and ($3$), we have $f(f(x)) + f(-x) + f(x - f(0))=f(-x)$, thus \[
f(f(x))+f(x-f(0))=0.
\]Put $x=f(0)$ in above, we have $f(0) = 0$. Thus $f(x)=f(-x)$ from ($3$).

Now, comparing $P(x, y)$, and $P(x, -y)$ gives $f(x+y)=f(x-y)$. This implies $f$ is constant. because for any real $t$, Letting $x=y=\frac{t}{2}$ to get \[
f(t) = f(x+y)=f(x-y)=f(0).
\]And so we're done.
This post has been edited 1 time. Last edited by NicoN9, Today at 8:37 AM
Reason: typo fixed
Z K Y
N Quick Reply
G
H
=
a