Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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
D1037 : The super irreductibilty
Dattier   1
N 8 minutes ago by Dattier
Source: les dattes à Dattier
Let $P \in \mathbb Q[x]$, $P$ is super irreductible if $\forall Q \in \mathbb [x], \deg(Q)>0, P(Q(x))$ is irreductible.

Are there some $P \in \mathbb Q[x]$ super irreductible?
1 reply
Dattier
2 hours ago
Dattier
8 minutes ago
FE on Stems
mathscrazy   7
N 13 minutes ago by Thapakazi
Source: STEMS 2025 Category B4, C3
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in \mathbb{R}$, \[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]Proposed by Aritra Mondal
7 replies
mathscrazy
Dec 29, 2024
Thapakazi
13 minutes ago
Reflections and midpoints in triangle
TUAN2k8   2
N 15 minutes ago by Double07
Source: Own
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
2 replies
TUAN2k8
5 hours ago
Double07
15 minutes ago
Cool Functional Equation
Warideeb   0
24 minutes ago
Find all functions real to real such that
$f(xy+f(x))=xf(y)+f(x)$
for all reals $x,y$.
0 replies
Warideeb
24 minutes ago
0 replies
Simple inequality
sqing   59
N 27 minutes ago by ForeverSnow
Source: Shortlist BMO 2018, A1
Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:

$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant  \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$
59 replies
sqing
May 3, 2019
ForeverSnow
27 minutes ago
Inspired by 2025 Xinjiang
sqing   1
N 38 minutes ago by OGMATH
Source: Own
Let $ a,b >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(6+\frac {b}{a}\right) \left(2+\frac {a}{b}+\frac {b}{ a}\right)  \geq\frac {625}{ 12}$$$$  \left(1+\frac {a} { b}\right)\left(6+\frac {a}{b}\right) \left(2+\frac {a}{b}+\frac {b}{ a}\right)  \geq29+6\sqrt 6$$$$  \left(1+\frac {a} { b}\right)\left(2+\frac {b}{ a}\right) \left(2+\frac {a}{b}+\frac {b}{ a}\right)  \geq \frac{3(63+11\sqrt{33})}{16}  $$$$  \left(1+\frac {a} { b}\right)\left(2+\frac {a}{ b}\right) \left(2+\frac {a}{b}+\frac {b}{ a}\right)  \geq \frac{223+70\sqrt{10}}{27}  $$
1 reply
sqing
an hour ago
OGMATH
38 minutes ago
Integer-Valued FE comes again
lminsl   208
N 41 minutes ago by megahertz13
Source: IMO 2019 Problem 1
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$Proposed by Liam Baker, South Africa
208 replies
lminsl
Jul 16, 2019
megahertz13
41 minutes ago
2025 Beijing High School Mathematics Competition Q9
sqing   2
N 41 minutes ago by sqing
Source: China
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Find the maximum value of $a^2d.$
2 replies
sqing
2 hours ago
sqing
41 minutes ago
2025 Guangdong High School Mathematics Competition Q14
sqing   2
N 41 minutes ago by sqing
Source: China
Let $ x_1, x_2, x_3, x_4, x_5\geq 0 $ and $ x^2_1+x^2_2+x^2_3+ x^2_4+ x^2_5=4. $ Find the maximum value of
$\sum_{i=1}^5 \frac{1}{x_i+1} \sum_{i=1}^5 x_i .$
2 replies
sqing
2 hours ago
sqing
41 minutes ago
JBMO Shortlist 2023 G7
Orestis_Lignos   6
N an hour ago by MR.1
Source: JBMO Shortlist 2023, G7
Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at point $F$. Line $FE$ meets line $AD$ at point $G$, while line $FD$ meets line $BE$ at point $H$. Prove that lines $CF, AH$ and $BG$ pass through the same point.
6 replies
Orestis_Lignos
Jun 28, 2024
MR.1
an hour ago
Inspired by 2025 Xinjiang
sqing   0
an hour ago
Source: Own
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(2+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 12+8\sqrt 2 $$
0 replies
sqing
an hour ago
0 replies
2025 Xinjiang High School Mathematics Competition Q11
sqing   1
N an hour ago by sqing
Source: China
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(1+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 16 $$
1 reply
sqing
2 hours ago
sqing
an hour ago
Nice "if and only if" function problem
ICE_CNME_4   6
N an hour ago by BBNoDollar
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
6 replies
ICE_CNME_4
Yesterday at 7:23 PM
BBNoDollar
an hour ago
Inspired by 2025 Beijing
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
IMO ShortList 2002, algebra problem 1
orl   130
N Apr 25, 2025 by Ilikeminecraft
Source: IMO ShortList 2002, algebra problem 1
Find all functions $f$ from the reals to the reals such that

\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]

for all real $x,y$.
130 replies
orl
Sep 28, 2004
Ilikeminecraft
Apr 25, 2025
IMO ShortList 2002, algebra problem 1
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO ShortList 2002, algebra problem 1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Jndd
1417 posts
#130 • 1 Y
Y by radian_51
Setting $y=-f(x)$, we get \[f(f(x)-f(f(x))=f(0)=2x+f(f(-f(x))-x),\]so by varying $x$ across all reals, we get that $f$ is surjective. Now, using that $f$ is surjective, we must have some $t\in\mathbb{R}$ such that $f(t)=0$. Then, plugging in $x=t$ gives us \[f(f(t)+y)=f(y)=2t+f(f(y)-t),\]and letting $x=f(y)$, we have $x-2t=f(x-t)$, which can be written as $f(x)=x-t$. Replacing the constant $-t$ with $c$ and plugging it back into the original equation, we see that all such equations work, so our answer is $f(x)=x+c$ 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.
Markas
150 posts
#131 • 1 Y
Y by radian_51
Let y = -f(x). We get $f(0) = 2x + f(f(-f(x) - x)$ $\Rightarrow$ $f(f(-f(x) - x) = f(0) - 2x$ and now f(0) is fixed and we can vary x however we want $\Rightarrow$ f goes trough all values $\Rightarrow$ f is surjective. Now since f is surjective there $\exists$ t, for which $f(t) = 0$. Let x = t. We get $f(y) = 2t + f(f(y) - t)$. Let f(y) - t = u which goes trough all values $\Rightarrow$ $u - t = f(u)$ $\Rightarrow$ we get that $f(x) = x - t$, for every t. Now we have to check it so plugging in we get x - t + y - t = 2x + y - t - x - t, which is correct $\Rightarrow$ we are ready.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ezpotd
1287 posts
#132 • 1 Y
Y by radian_51
We claim the answer is all functions of the form $f(x) = x + c$, it is easy to verify all of these functions work.

Set $y =-f(x)$, then we get $f(0) - 2x = f(f(y) - x)$, varying $x$ shows that $f$ is surjective. Now we show $f$ is injective, assume otherwise and we have $f(a) = f(b)$ for unequal $a,b$, then set $y = c$ such that $f(c) = a + b$, then we get $ 2a + f(b)= f(f(b) + c)= f(f(a) + c)=2b + f(a) = 2b + f(b)$, giving $a = b$, contradiction. Now set $x = 0$, and undo a layer of $f$ via bijectivity, we get $f(0) + y = f(y)$, so $f(y) = y + c$ for some $c$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
balllightning37
389 posts
#133 • 1 Y
Y by radian_51
This is nice :)

The answer is $f(x)=x-c$, which trivially works.

If $y=-f(x)$, we get that $f(0)=2x+f(f(-f(x)-x))$, so $f$ is surjective.

Now, we take a constant $c$ such that $f(c)=0$. $P(c,y)$ gives $f(y)=2c+f(f(y)-c)$ which rearranges to $f(y)-c=f(f(y)-c)+c$. As $f$ is surjective, for all $x$ we can find $y$ such that $x=f(y)-c$. Thus, for all $x$, $f(x)=x-c$. It is easily verified that $c$ can be any real number, so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alexanderhamilton124
400 posts
#134 • 1 Y
Y by radian_51
$P(x, -f(x))$ gives $f(0) = 2x + f(f(-f(x)) - x)$ which means $f$ is surjective. Take $y$ such that $f(y) = x$, and we get $f(f(x) + y) = 2x + f(0)$. Say $f(a) = f(b)$, plugging this in into the new equation, we get $a = b$, so $f$ is injective. $P(0, y)$ gives $f(y + f(0)) = f(f(y)) \implies f(y) = y + f(0)$, so $f(x) = x + c$, which clearly works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anshu_Singh_Anahu
69 posts
#135 • 2 Y
Y by alexanderhamilton124, radian_51
alexanderhamilton124 wrote:
. Take $y$ such that $f(y) = x$, and we get $f(f(x) + y) = 2x + f(0)$. Say $f(a) = f(b)$, plugging this in into the new equation, we get $a = b$,

You cant say a=b becoz y is not fixed you taken y such that f(y) = x ,
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Saucepan_man02
1358 posts
#138 • 2 Y
Y by Sreepranad, radian_51
Note that $f$ is surjective. Let $f(a)=0$.
$P(a, y)$ gives: $f(y) = 2a+f(f(y)-a)$. Since $f$ is surjective, let $t=f(y)-a \in \mathbb R$. Thus: $f(t)=t-a$ for all $t \in \mathbb R$ for some constant $a$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alexanderhamilton124
400 posts
#139 • 2 Y
Y by Nobitasolvesproblems1979, radian_51
Anshu_Singh_Anahu wrote:
alexanderhamilton124 wrote:
. Take $y$ such that $f(y) = x$, and we get $f(f(x) + y) = 2x + f(0)$. Say $f(a) = f(b)$, plugging this in into the new equation, we get $a = b$,

You cant say a=b becoz y is not fixed you taken y such that f(y) = x ,

Thank you for spotting the error! Will try to fix today.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#140 • 1 Y
Y by radian_51
$y=-f(x)$ yields $f(x)$ is surjective. Thus let $k$ be such that $f(k)=0,$ then $$x=k \implies f(y)=2k+f(f(y)-k) \implies m-2k=f(m-k)$$for all real $m.$ Hence $f(x)=x-k,$ which after substitution into the assertion simply works.

Hence $f(x) \equiv x+c$ for some constant $c \in \mathbb R$ are the only solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ali123456
52 posts
#141 • 1 Y
Y by radian_51
$\textbf{Claim:}$ $f$ is surjective
$\textbf{Proof:}$ we notice that $P(x;-f(x)) \implies f(f(-x)-x)=f(0)-2x$
$\textbf{Claim:}$ $f$ is injective
$\textbf{Proof:}$ Let $x$ and $y$ such that $f(x)=f(y)$ Since f is surjective there exist $a$ such that $f(a)=x+y$ and so we conclude that $x=y$ by $P(x;a)$ and $P(y;a)$
$\textbf{Last step:}$ : $P(0;y) \implies f(f(0)+y)=f(f(y) \implies f(y)=y+f(0)$ Hence $f(y)=y+c$ with $c$ a real constant
This post has been edited 2 times. Last edited by ali123456, Feb 24, 2025, 11:07 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tony_stark0094
69 posts
#143 • 1 Y
Y by radian_51
dysfunctionalequations wrote:
Setting $ y=-f(x)$, we have $ f(0)=2x+f(f(-x)-x))$, so $ f(0)-2x=f(f(-x)-x)$, and $ f(0)-2x$ is a surjective function of $ x$, so $ f$ is surjective as well.

Then, by surjectivity, we can let $ c$ be a number for which $ f(c)=0$. Setting $ x=c$, we have $ f(y)=2c+f(f(y)-c)$. Since $ f(y)$ can take on all real values $ r$, $ r=2c+f(r-c)$ for all real numbers $ r$, implying $ f(r-c)=r-2c$. Letting $ s=r-c$, we find $ \boxed{f(s)=s-c}$ for some constant $ c$ for all real numbers $ s$. Also, it can be easily verified that all functions of this form satisfy this equation.

how can you let f(r)=r?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tony_stark0094
69 posts
#144 • 1 Y
Y by radian_51
djmathman wrote:
Here's a solution that I believe avoids injectivity and surjectivity altogether.

Let $x=0$ to get $f(f(0)+y)=f(f(y))$ and let $y=0$ to get $f(f(x))=2x+f(f(0)-x)$. Substituting our expression for $f(f(x))$ into the first equation yields \[f(f(0)+y)=f(f(y))=2y+f(f(0)-y).\]Now make the change of variables $m=f(0)+y$, $n=f(0)-y$. Then $2y=m-n$, so \[f(m)=m-n+f(n)\implies f(m)-m=f(n)-n.\]Let $g(x)=f(x)-x$. This implies $g(m)=g(n)$ for all pairs of real numbers $(m,n)$, so $g$ is constant. Thus $f(x)-x=c\implies f(x)=x+c$, where $c$ is a constant. Checking yields that all such functions work.

too good
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Marcus_Zhang
980 posts
#145 • 1 Y
Y by radian_51
Surjectivity.
This post has been edited 2 times. Last edited by Marcus_Zhang, Mar 16, 2025, 2:14 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lpieleanu
3002 posts
#146 • 2 Y
Y by megarnie, radian_51
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
658 posts
#147
Y by
I claim that $f(x) = x + c.$

Plug in $y = -f(x)$ to get that $f(f(-f(x)) - x) = f(0) - 2x,$ and thus $f$ is surjective. Let $P(x, y)$ denote our assertion.

Since $f$ is surjective, let us assume that $f(k) = 0.$ With $P(0, k),$ we get that $f(f(0) + k) = 0.$ With $P(-k, 0),$ we have that $f(f(-k)) = -2k.$ With $P(0, -k),$ we see that $f(f(0) - k) = -2k.$ With $P(k, 0),$ we get that $f(0) = 2k.$ Thus, we get that $f(k) = -2k,$ but by assumption, $f(k) = 0.$ Thus, $k = 0,$ and hence $f(0) = 0.$

With $P(0, x),$ we see that $f(x) = f(f(x)).$ Since $f(x)$ is surjective, we have that $f(x) = x.$
Z K Y
N Quick Reply
G
H
=
a