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
JBMO Shortlist 2022 N1
Lukaluce   9
N 8 minutes ago by MathIQ.
Source: JBMO Shortlist 2022
Determine all pairs $(k, n)$ of positive integers that satisfy
$$1! + 2! + ... + k! = 1 + 2 + ... + n.$$
9 replies
Lukaluce
Jun 26, 2023
MathIQ.
8 minutes ago
fraction sum
miiirz30   5
N 14 minutes ago by MathIQ.
Source: 2025 Euler Olympiad, Round 1
Evaluate the following sum:
$$ \frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + \ldots + \frac{1}{1 + 2 + 3 + 4 + \dots + 2025} $$
Proposed by Prudencio Guerrero Fernández
5 replies
miiirz30
Mar 31, 2025
MathIQ.
14 minutes ago
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   8
N an hour ago by MathLuis
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
8 replies
OgnjenTesic
Today at 4:02 PM
MathLuis
an hour ago
Primes and sets
mathisreaI   41
N an hour ago by Tinoba-is-emotional
Source: IMO 2022 Problem 3
Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.
41 replies
mathisreaI
Jul 13, 2022
Tinoba-is-emotional
an hour ago
Minimum times maximum
y-is-the-best-_   64
N an hour ago by ezpotd
Source: IMO 2019 SL A2
Let $u_1, u_2, \dots, u_{2019}$ be real numbers satisfying \[u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.\]Let $a=\min \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$ and $b=\max \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$. Prove that
\[
a b \leqslant-\frac{1}{2019}.
\]
64 replies
y-is-the-best-_
Sep 22, 2020
ezpotd
an hour ago
Prove $x+y$ is a composite number.
mt0204   1
N 2 hours ago by sharknavy75
Let $x, y \in \mathbb{N}^*$ such that $1000 x^{2023}+2024 y^{2023}$ is divisible by $x+y$ and $x+y>2$. Prove that $x+y$ is a composite number.
1 reply
mt0204
Today at 3:59 PM
sharknavy75
2 hours ago
Serbian selection contest for the IMO 2025 - P1
OgnjenTesic   2
N 2 hours ago by MathLuis
Source: Serbian selection contest for the IMO 2025
Let \( p \geq 7 \) be a prime number and \( m \in \mathbb{N} \). Prove that
\[\left| p^m - (p - 2)! \right| > p^2.\]Proposed by Miloš Milićev
2 replies
OgnjenTesic
Today at 4:01 PM
MathLuis
2 hours ago
JBMO Shortlist 2021 N1
Lukaluce   15
N 2 hours ago by LeYohan
Source: JBMO Shortlist 2021
Find all positive integers $a, b, c$ such that $ab + 1$, $bc + 1$, and $ca + 1$ are all equal to
factorials of some positive integers.

Proposed by Nikola Velov, Macedonia
15 replies
Lukaluce
Jul 2, 2022
LeYohan
2 hours ago
a+b+c+d divides abc+bcd+cda+dab
v_Enhance   51
N 2 hours ago by BossLu99
Source: USA Team Selection Test for IMO 2021, Problem 1
Determine all integers $s \ge 4$ for which there exist positive integers $a$, $b$, $c$, $d$ such that $s = a+b+c+d$ and $s$ divides $abc+abd+acd+bcd$.

Proposed by Ankan Bhattacharya and Michael Ren
51 replies
v_Enhance
Mar 1, 2021
BossLu99
2 hours ago
three discs of radius 1 cannot cover entirely a square surface of side 2
parmenides51   1
N 3 hours ago by Blast_S1
Source: 2014 Romania NMO VIII p4
Prove that three discs of radius $1$ cannot cover entirely a square surface of side $2$, but they can cover more than $99.75\%$ of it.
1 reply
parmenides51
Aug 15, 2024
Blast_S1
3 hours ago
Floor sequence
va2010   88
N 3 hours ago by heheman
Source: 2015 ISL N1
Determine all positive integers $M$ such that the sequence $a_0, a_1, a_2, \cdots$ defined by \[ a_0 = M + \frac{1}{2}   \qquad  \textrm{and} \qquad    a_{k+1} = a_k\lfloor a_k \rfloor   \quad \textrm{for} \, k = 0, 1, 2, \cdots \]contains at least one integer term.
88 replies
va2010
Jul 7, 2016
heheman
3 hours ago
2025 Caucasus MO Juniors P6
BR1F1SZ   2
N 3 hours ago by IEatProblemsForBreakfast
Source: Caucasus MO
A point $P$ is chosen inside a convex quadrilateral $ABCD$. Could it happen that$$PA = AB, \quad PB = BC, \quad PC = CD \quad \text{and} \quad PD = DA?$$
2 replies
BR1F1SZ
Mar 26, 2025
IEatProblemsForBreakfast
3 hours ago
Interesting functions with iterations over integers
miiirz30   5
N 3 hours ago by OGMATH
Source: 2025 Euler Olympiad, Round 2
For any subset $S \subseteq \mathbb{Z}^+$, a function $f : S \to S$ is called interesting if the following two conditions hold:

1. There is no element $a \in S$ such that $f(a) = a$.
2. For every $a \in S$, we have $f^{f(a) + 1}(a) = a$ (where $f^{k}$ denotes the $k$-th iteration of $f$).

Prove that:
a) There exist infinitely many interesting functions $f : \mathbb{Z}^+ \to \mathbb{Z}^+$.

b) There exist infinitely many positive integers $n$ for which there is no interesting function
$$
f : \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}.
$$
Proposed by Giorgi Kekenadze, Georgia
5 replies
1 viewing
miiirz30
Today at 11:07 AM
OGMATH
3 hours ago
Inequality with one variable rational functions
liliput   14
N 3 hours ago by IEatProblemsForBreakfast
Source: 2022 Junior Macedonian Mathematical Olympiad P2
Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=3$. Prove the inequality
$$\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}.$$
Proposed by Anastasija Trajanova
14 replies
liliput
Jun 7, 2022
IEatProblemsForBreakfast
3 hours ago
FE over R
IAmTheHazard   20
N Today at 3:28 PM by shanelin-sigma
Source: ELMO Shortlist 2024/A3
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu
20 replies
IAmTheHazard
Jun 22, 2024
shanelin-sigma
Today at 3:28 PM
FE over R
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO Shortlist 2024/A3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5003 posts
#1 • 3 Y
Y by ItsBesi, ihatemath123, pho1234
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu
This post has been edited 1 time. Last edited by IAmTheHazard, Jun 22, 2024, 3:41 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MarkBcc168
1595 posts
#2
Y by
Solution
This post has been edited 1 time. Last edited by MarkBcc168, Jun 22, 2024, 3:42 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
VicKmath7
1390 posts
#3
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
fractals
3028 posts
#4
Y by
Let $P(x,y)$ denote the relation $f(x+f(y))+xy=f(x)f(y)+f(x)+y$.
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
CyclicISLscelesTrapezoid
372 posts
#5 • 1 Y
Y by pho1234
The solutions are $f(x) \equiv x$ and $f(x) \equiv -x$, which work.

Let $P(x,y)$ denote the assertion that
\[f(x+f(y))+xy=f(x)f(y)+f(x)+y.\]$P(0,x)$ gives $f(f(x))=f(0)f(x)+f(0)+x$. Notice that $P(f(x),y)$ gives
\[f(f(x)+f(y))+yf(x)=f(f(x))f(y)+f(f(x))+y,\]and subtracting this by its symmetric variant with $x$ and $y$ swapped gives
\[yf(x)-xf(y)=f(f(x))f(y)+f(f(x))+y-f(f(y))f(x)-f(f(y))-x.\]Using $P(0,x)$ to rewrite all the nested $f$'s, the RHS simplifies to $xf(y)-yf(x)$, so we have $xf(y)=yf(x)$. Plugging in $y=1$, we get $f(x)=xf(1)$. If $f(1)=c$, then the functional equation simplifies to $c(x+cy)+xy=c^2xy+cx+y$, which is only true for $c=1,-1$, 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.
KevinYang2.71
428 posts
#6
Y by
Solved with Orthogonal..

We claim the only functions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$. It is easy to check that these work.

Let $P(x,y)$ denote the given assertion. $P(0,x)$ gives $f(f(x))=f(0)f(x)+f(0)+x$. Comparing $P(f(x),1)$ and $P(f(1),x)$ gives $f(x)\equiv f(1)x$. We can show that $f(1)=\pm 1$, as desired. $\square$
This post has been edited 2 times. Last edited by KevinYang2.71, Jun 23, 2024, 2:26 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5610 posts
#7 • 1 Y
Y by KevinYang2.71
The only solutions are $f(x) = x$ and $f(x) = -x$, which work. Let $P(x,y)$ denote the given assertion.

Case 1: $f(0) = 0$
Then $P(0,x): f(f(x)) = x$, so $f$ is an involution.

$P(1, f(x)): f(x + 1)= x f(1) + f(1)$, so $f$ is linear, meaning $f(x) = cx$ for some constant $c$. Since $f$ is an involution, either $c = 1$ or $c = -1$.

Case 2: $f(0) \ne 0$.
Claim: $f$ is injective.
Proof: If $f(a)= f(b)$, then $P(0,a)$ compared with $P(0,b)$ gives $a = b$. $\square$

Claim: $x + f(x)$ is injective.
Proof: Suppose $a + f(a) = b + f(b)$. Then $P(x,x): f(x + f(x)) - (x + f(x)) = (f(x) - x)(x + f(x))$, so comparing $x = a$ and $x = b$ gives that $(a + f(a)) (f(a) - a)) = (a + f(a)) (f(b) - b)$. If $a + f(a) = 0$, then $P(a,a): f(0) = 0$, absurd. Hence $a + f(a) \ne 0$, so $f(a) - a = f(b)  - b$. Hence \[(f(a) - a) + (f(a) + a) = (f(b) - b) + (f(b) + b) \implies f(a) = f(b) \implies a = b\]$\square$

Now, the equation gives \[ f(x + f(y)) - f(x) - y = f(x) f(y) - xy,\]so swapping $x,y$ here gives $f(x + f(y)) - f(x) - y = f(y + f(x)) - f(y) - x$, so $f(x + f(y))  + (x + f(y)) = f(y + f(x)) + (y + f(x))$, which by our earlier claim implies $x + f(y) = y + f(x)\implies f(x) - x = f(y) - y$, so $f(x) = x + c$ for some constant $c$.

$P(0,0): f(f(0)) = f(0)^2 + f(0)$, so $2c = c^2 + c\implies c\in \{0,1\}$. Checking, we see that $c = 1$ fails, so $c = 0$, but this is absurd since $f(0) \ne 0$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
P2nisic
406 posts
#8
Y by
IAmTheHazard wrote:
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu

$f(x)f(y)f(z)=f(x+f(y))f(z)+xyf(z)-f(z)f(x)-yf(z)=f(x+f(y)+f(z))+(x+f(y))f(z)-f(x+f(y))-z+xyf(z)-f(x+f(z))-xz+f(x)+z-yf(z)$

Now by the symmetry of $y,z$ we get that:

$f(x+f(y)+f(z))+(x+f(y))f(z)-f(x+f(y))-z+xyf(z)-f(x+f(z))-xz+f(x)+z-yf(z)=$
$f(x+f(z)+f(y))+(x+f(z))f(y)-f(x+f(z))-y+xzf(y)-f(x+f(y))-xy+f(x)+y-zf(y)$
$\Rightarrow xf(z)+xyf(z)-xz-yf(z)=xf(y)+xzf(y)-xy-zf(y)$

In this for $x=0$ we get that:
$-yf(z)=-zf(y)\Rightarrow \frac{f(z)}{z}=\frac{f(y)}{y}\Rightarrow f(x)=cx$

Now in the start we have that:
$cx+c^2y+xy=c^2xy+cx+y\Rightarrow c^2y=y\Rightarrow c=+-1$
So we get that: $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$.



In the same way we can solve the problem if $f : \mathbb{R^+}\to\mathbb{R^+}$
This post has been edited 1 time. Last edited by P2nisic, Jun 23, 2024, 11:20 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
X.Allaberdiyev
104 posts
#9
Y by
Cute one.
$P(0,x)$ -> $f(f(x))=f(x)f(0)+f(0)+x$ $(1)$
By looking at $P(f(x),y)$ and $P(f(y),x)$ we have $f(f(x)+f(y))=f(f(x))f(y)+f(f(x))+y-f(x)y=f(f(y))f(x)+f(f(y))+x-f(y)x$, and by using $(1)$ we observe that $xf(y)=yf(x)$, which means that $f(x)=cx$. And by plugging it into equation one can prove that only solutions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ItsBesi
146 posts
#13
Y by
I am not sure about the surjectivety part
$\textbf{Answer:}$ $f(x)=\pm x \forall x \in \mathbb{R}$

$\textbf{Solution:}$

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

$\textbf{Claim:}$ $f$-bijective

$\textbf{Proof:}$

$P(0,x) \implies f(f(x))=f(x)f(0)+f(0)+x \implies f$-is surjective

f-injective

Since $f$ is both injective and surjective we get that $f$-is bijective. $\square$


$\textbf{Claim:}$ $f(0)=0$
$\textbf{Proof:}$

Since $f$-is surjective there exists an $\alpha$ such that $f(\alpha)=0 \iff \exists \alpha \in \mathbb{R} : f(\alpha)=0$

$P(\alpha,0) \implies f(\alpha+f(0))=0=f(\alpha) \implies f(\alpha+f(0))=f(\alpha) \stackrel{f-injective}{\implies} \alpha+f(0)=\alpha \implies f(0)=0.$ $\square$

$\textbf{Claim:}$ $f(x)= \pm x \forall \in \mathbb{R}$

$\textbf{Proof:}$
$P(0,x) \implies f(f(x))=x$ $...(*)$

$P(x,x) \implies f(x+f(x))+x^2=f(x)^2+f(x)+x \implies f(x+f(x))-x-f(x)=f(x)^2-x^2$
$...(3)$

$P(f(x),f(x)) \stackrel{(*)}{\implies} f(x+f(x))+f(x)^2=x^2+x+f(x) \implies f(x+f(x))-x-f(x)=x^2-f(x)^2$ $...(4)$

$(3)-(4) \implies f(x)^2-x^2=x^2-f(x)^2 \implies f(x)^2=x^2 \implies f(x)= \pm x \forall \in \mathbb{R}$ $\blacksquare$

$\textbf{POINTWISE TRAP!}$
This post has been edited 1 time. Last edited by ItsBesi, Jul 3, 2024, 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.
omar1tun
34 posts
#14
Y by
This solution might be one of the lamest ever :
$P(x,y):f(x+f(y))+xy=f(x)f(y)+f(x)+y$
We will just simplify $P(f(y),1)$ to find the answer :

$ \boxed{f(f(y)+f(1)) +f(y)-1= f(f(y))(f(1)+1)}$

Now : $P(0,y)$ , gives $f(f(y))=f(0)(f(y)+1)+y$
So for the right side we get that : $\boxed{f(f(y))(f(1)+1)=f(y).f(0)(f(1)+1)+f(0)(f(1)+1)+y(f(1)+1)}$
For the left side using $P(f(1),y)$ , we get that :
$f(f(y)+f(1))=f(f(1))(f(y)+1)+y(1-f(1))=[f(0)(f(1)+1)+1](f(y)+1)+y(1-f(1))$

So :$ \boxed{f(f(y)+f(1))+f(y)-1=f(0)(f(1)+1).f(y)+f(0)(f(1)+1)+f(y)+1+f(y)-1+y(1-f(1))}$
Identifying both sides will be left with :
$2f(y)+y(1-f(1)) = y(f(1)+1)$ , thus $f(y)=yf(1)$ , the rest is obvious .
This post has been edited 2 times. Last edited by omar1tun, Aug 1, 2024, 5:23 PM
Reason: .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
omar1tun
34 posts
#15
Y by
For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x)
This post has been edited 1 time. Last edited by omar1tun, Aug 1, 2024, 5:28 PM
Reason: .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
omar1tun
34 posts
#16
Y by
For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x) an example is $f(x)=(x+1)^2$ ,will get on the right : $ (x+1)^2+(x+1)$ , which doesn't take all values in $R$ such as $-4$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihatemath123
3449 posts
#17
Y by
The solutions are $f(x) = x$ and $f(x)=-x$, which are both easy to verify.

Claim: $f(0) = 0$.
Proof: Assume that $f(0) \neq 0$ FTSOC. Taking $P(x,0)$ gives us
\[f(x+f(0)) = f(x) \cdot (1 + f(0)),\]so incrementing $x$ by $f(0)$ gives some geometric sequence. But the original equation can be rewritten as
\[f(x+f(y)) - f(x)(1+f(y)) = y(1-x),\]and if we increment $x$ by $f(0)$ here, the LHS will multiply by $1+f(0)$. However, the RHS cannot grow exponentially since it is negative for all $x>1$ and positive for all $x < 1$, something no exponential function does. This gives us a contradiction.

Now, taking $P(0,x)$ gives us $f(f(x)) = x$, implying surjectivity, and $P(1,x)$ gives us
\[f(1+f(x)) = f(1)(1+f(x)),\]which is enough to imply that $f$ is linear. Plugging this into our original equation gives the claimed solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bo18
38 posts
#18
Y by
f(x+f(y))+x.y=f(x).f(y)+f(x)+y
Let' s call this P(x,y)
P(0,0) gives us
f(f(0))=f(0)^2+f(0) this call *
P(x, 0) gives us
f(x+f(0))=f(x).f(0)+f(x)
x->-f(0) gives us
f(0)=f(-f(0)).f(0)+f(-f(0))
from this, we can get that:
f(0)/f(f(0))=f(0)+1
from this, * gives us
f(0)^2=f(-f(0)).f(f(0))=A
P(-f(0), f(0))
f(-f(0)+f(f(0)))-f(0)^2=A+f(-f(0))+f(0)
f(-f(0)+f(0)^2+f(0))=2.f(0)^2+f(0)+f(-f(0))
this give
f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0))
P(f(0), f(0)^2)
f(f(0)+f(f(0)^2))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2
f(f(0)+f(0)^2+f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2
f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2
Now, P(2f(f(0)), -f(0)), which give us
f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0)
Now P(x, f(0)) W
f(x+f(f(0)))+f(0).x=f(f(0)).f(x)+f(x)+f(0)
x->f(f(0)), I don' t know why I didn't put this here W
f(2f(f(0)))+f(0).f(f(0))=f(f(0)).f(f(f(0)))+f(f(f(0)))+f(0)
Now I want to find f(f(f(0)))
if we take this P(0, y) we will get this
f(f(y))=f(0).f(y)+f(0)+y
y->f(0), f(f(f(0)))=f(f(0)).f(0)+2.f(0)
if we call f(0) to be (a) and f(f(0)) to be (b), where b=a^2+a( from the beginning)
Then, f(2f(f(0)))+f(0).f(f(0))=f(f(0)).(f(f(0)).f(0)+2.f(0))+f(f(0)).f(0)+2.f(0)+f(0)
we will get this
f(2f(f(0)))+a.b=b.(b.a+2.a)+b.a+2.a+a, which is equivalent to f(2f(f(0)))=b(b.a+2.a)+3.a
f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0), Now, from this, we have:
f(2f(f(0))+f(-f(0)))-2.a.b=(b.(b.a+2.a)+3.a).f(-f(0))+b.(b.a+2.a)+3.a-a, call this T
From this, f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2, we have that the left side is equal to, also I want to say that f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0))=a^2+b+f(-f(0))
Now, nightmare is coming
The left side=b(a^2+b+f(-f(0)))+b+a^2-a^3
But we have that: f(0)^2=f(f(0)).f(-f(0)), which is equal to a^2=b.f(-f(0)), i.e the left side=a^2.b+b^2+a^2+b+a^2-a^3=a^2.b+b^2+2.a^2+b-a^3, call D
Now-if we multiply whole equation with b, we will get this
(The left side).b=2.b^2.a+(b^2.a+2.a.b+3.a).a^2+b^3.a+2.a.b^2+2.a.b
From D multiply with (b), we get this
a^2.b^2+b^3+2.a^2.b+b^2-a^3.b to be equal to 2.b^2.a+b^2.a^3+2.a^3.b+3.a^3+b^3.a+2.a.b^2+2.a.b
Now, if this above wasn' t nightmare, after we will replaceable b with a^2+a- this will be a real bad nightmare, let's start (teeth)
a^2.(a^2+a)^2+(a^2+a)^3+2.a^2.(a^2+a)+(a^2+a)^2-a^3.(a^2+a)=2.(a^2+a)^2.a+(a^2+a)^2.a^3+2.a^3.(a^2+a)+3.a^3+(a^2+a)^3.a+2.a.(a^2+a)^2+2.a.(a^2+a)
After we use Wolfram Alpha, oops, just a joke, I figured it all myself (sad) , we will get this
+2.a^7+3.a^6+6.a^5+5.a^4+4.a^3+a^2=0, the solution of this two in Real numbers, the one of them is approximately to some digit-negative and stranger, the other solution is 0
Then we have a=0=f(0)
Then, if we put this in this:
f(f(y)=f(0).f(y)+f(0)+y=y, i.e the function is involution
Now, let P(f(x), f(y)) gives us
f(f(x)+y)+f(x).f(y)=x.y+f(y)+x
But P(y,x) gives us
f(f(x)+y)+x.y=f(x).f(y)+f(y)+x, call G
Let sum this two, then this gives us
f(f(x)+y)=f(y)+x
Put this in G gives
x.y=f(x).f(y), y->x gives f(x)^2=x^2
f(x)=+x, f(x)=-x
This post has been edited 1 time. Last edited by bo18, Sep 18, 2024, 5:58 PM
Reason: This has a mistake
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Davud29_09
20 posts
#19
Y by
P(f(x),y) ,change x,y after this and y=0 implies that f(f(x))×(f(0)+1)=f(x)×f(f(0))+f(f(0))+x-x×f(0).
Plug x=0 we get f(f(y))=f(0)×f(y)+f(0)+y we get f is injective. Also we use this equation above and we get f(x)×(f(0)²+f(0)-f(f(0))) =-2x×f(0)-(f(0)²+f(0)-f((0))) if f(0)²+f(0)-f(f(0)) isn't equal 0 then f(x)=ax+b we check and we get f(x)=x or f(x)=-x. Other case implies that f(0)=0.In common equation we plug x=0 and we get f(f(x))=x.In next steps P(x,f(y)) and we change x,y we get x×f(y)=y×f(x) y=1 and we get f(x)=cx.It is easy to check and find f(x)=x and f(x)=-x answers.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.
lksb
177 posts
#20
Y by
$P(0,x)\implies f(f(x))=f(0)(f(x)+1)$
$P(f(x), y)\implies f(f(x)+f(y))+yf(x)=f(0)(f(x)+1)(f(y)+1)+xf(y)+x+y$
$P(f(x),y)-P(x, f(y))\implies  xf(y)=yf(x)\implies \boxed{f(x)=\pm x}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1728 posts
#21
Y by
The answer is $f(x)=x$ or $f(x)=-x$, which clearly work. Now we prove that this is the only solution. Let $P(x,y)$ denote the assertion, and let $f(0)=c$.

We have
\begin{align*}
P(0,0) &\implies f(c) = c^2+c \\
P(0,x) &\implies f(f(x)) = cf(x) + c + x \\
P(x,0) &\implies f(x+c) = (c+1)f(x)
\end{align*}
Claim 1: $f(0)=0$.
Take $P(c,1)$ which gives us on one hand,
\[f(c+f(1))+c=(c^2+c)(f(1)+1)+1\]But on the other hand, we have
\begin{align*}
f(f(1)+c)+c &= (c+1)f(f(1)) + c \\
&= (c+1)(cf(1) + c + 1) + c \\
&=(c^2+c)(f(1)+1)+2c+1
\end{align*}which proves our claim.

Now, $P(0,x)$ gives $f(f(x))=x$. Thus when we use $P(f(x),y)$ we get
\[f(f(x)+f(y))+f(x)y=xy+x+y\]Since $xy+x+y-f(f(x)+f(y))$ is symmetric, we must have $f(x)y=f(y)x$, implying that $f(x)/x$ is a constant. Since $f$ is an involution, this constant is $\pm 1$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11375 posts
#22
Y by
IAmTheHazard wrote:
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu

Let $P(x,y)$ be the assertion $f(x+f(y))+xy=f(x)f(y)+f(x)+y$.
$P(0,x)\Rightarrow f(f(x))=f(0)(f(x)+1)+x$
$P(f(x),y)\Rightarrow f(f(x)+f(y))-f(0)(f(x)+1)(f(y)+1)=xf(y)-yf(x)+x+y$
Swapping $x,y$ gives $yf(x)-xf(y)=xf(y)-yf(x)$, then setting $y=1$, we have $f(x)=xf(1)$. Testing, only $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ work.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bardia7003
22 posts
#23
Y by
Probably similar solutions are already mentioned. Anyway, I'll go on with mine.
Let $P(x, y)$ denote the given assertion.
$P(0, x): \underline{f(f(x)) = f(0)(f(x) + 1) + x}$
$P(f(x), y): f(f(x) + f(y)) + f(x)y = f(f(x))(f(y) + 1) + y = (f(0)(f(x) + 1) + x)(f(y) + 1) + y = f(0)f(x)f(y) + f(0)f(y) + xf(y) + f(0)f(x) + f(0) + x + y$
$\rightarrow f(f(x) + f(y)) = (x + y) + f(0)f(x)f(y) + f(0)(f(x)f(y) + f(y) + f(x) + 1) + xf(y) - f(x)y$
Now, if we swap $x,y$, the left side stays the same, and all the parts in the right side stay the same, except the $xf(y) - f(x)y$, so:
$$xf(y) - f(x)y = yf(x) - f(y)x \rightarrow 2xf(y) = 2yf(x) \rightarrow \frac{f(x)}{x} = \frac{f(y)}{y} \rightarrow \underline{f(x) = kx}$$And by checking this format in the equation, we find that only $k = 1, -1$ are the solutions, therefore the only answers are $\boxed{f(x) = x \quad \forall x \in \mathbb{R}}$ and $\boxed{f(x) = -x \quad \forall x \in \mathbb{R}}$. $\blacksquare$ :)
This post has been edited 1 time. Last edited by Bardia7003, Apr 1, 2025, 4:34 PM
Reason: typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shanelin-sigma
166 posts
#24
Y by
One linear:
$P(x+f(z),y)$ and swap $y,z$ you can find that $\frac{f(x)}{x}$ is a constant
Z K Y
N Quick Reply
G
H
=
a