Plan ahead for the next school year. Schedule your class today!

G
Topic
First Poster
Last Poster
k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Wednesday, Jul 16 - Oct 29
Sunday, Aug 17 - Dec 14
Tuesday, Aug 26 - Dec 16
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10

Prealgebra 2 Self-Paced

Prealgebra 2
Friday, Jul 25 - Nov 21
Sunday, Aug 17 - Dec 14
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Tuesday, Jul 15 - Oct 28
Sunday, Aug 17 - Dec 14
Wednesday, Aug 27 - Dec 17
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Wednesday, Jul 2 - Sep 17
Sunday, Jul 27 - Oct 19
Monday, Aug 11 - Nov 3
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8

Introduction to Number Theory
Tuesday, Jul 15 - Sep 30
Wednesday, Aug 13 - Oct 29
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Friday, Jul 18 - Nov 14
Thursday, Aug 7 - Nov 20
Monday, Aug 18 - Dec 15
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30

Introduction to Geometry
Monday, Jul 14 - Jan 19
Wednesday, Aug 13 - Feb 11
Tuesday, Aug 26 - Feb 24
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
Friday, Aug 8 - Feb 20
Tuesday, Aug 26 - Feb 24
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4

Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24

Intermediate Number Theory
Wednesday, Sep 24 - Dec 17

Precalculus
Wednesday, Aug 6 - Jan 21
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31

Advanced: Grades 9-12

Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)

MATHCOUNTS/AMC 8 Advanced
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Tuesday, Aug 26 - Nov 11
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)

AMC 10 Problem Series
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 10 - Nov 2
Thursday, Aug 14 - Oct 30
Tuesday, Aug 19 - Nov 4
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 10 Final Fives
Friday, Aug 15 - Sep 12
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30

AMC 12 Problem Series
Wednesday, Aug 6 - Oct 22
Sunday, Aug 10 - Nov 2
Monday, Aug 18 - Nov 10
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28

AIME Problem Series A
Thursday, Oct 23 - Jan 29

AIME Problem Series B
Tuesday, Sep 2 - Nov 18

F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30

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, Aug 14 - Oct 30
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3

Intermediate Programming with Python
Friday, Oct 3 - Jan 16

USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3

Physics

Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11

Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Jul 1, 2025
0 replies
IMO 2025 live scoreboards and manifold markets
wnwj   7
N 2 minutes ago by MathGuy1729
Live scoreboards:

Part 1
Part 2
Part 3
Part 4

Make predictions at Manifold Markets
7 replies
wnwj
5 hours ago
MathGuy1729
2 minutes ago
I miss Turbo
sarjinius   21
N 4 minutes ago by cocoioio7
Source: 2025 IMO P6
Consider a $2025\times2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile.

Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.

Proposed by Zhao Yu Ma and David Lin Kewei, Singapore
21 replies
sarjinius
Yesterday at 3:20 AM
cocoioio7
4 minutes ago
IMO 2025 Medal Cutoffs Prediction
GreenTea2593   53
N 16 minutes ago by AbbyWong
What are your prediction for IMO 2025 medal cutoffs?
53 replies
+1 w
GreenTea2593
Yesterday at 4:44 AM
AbbyWong
16 minutes ago
Sequence and mod 2003
Goblik   0
17 minutes ago
Define a sequence recursively by $a_0=1$, $a_1=7$, and $a_n-2n^2.a_{n-1}+n^2(n-1)^2.a_{n-2}=(n-1).n!$ for all $n \geq 2$.

Find $(\frac{a_{2002}}{2002!}-\frac{2002a_{2001}}{2001!})(mod2003)$
0 replies
Goblik
17 minutes ago
0 replies
Probably a Cool Optimization Problem
PhysicsIsChad   8
N 4 hours ago by PhysicsIsChad
What i am trying to do is finding a general solution to this earlier problem (https://artofproblemsolving.com/community/u1056796h3610029p35343126) . Here i am trying to generalise such a solution for n points . Simply stated find the min value of \[
f(x, y) = \sum_{i=1}^{n} \sqrt{(x - a_i)^2 + (y - b_i)^2}
\]. I would love to hear your insights on this problem .
My insights :-
\[
\text{Let } (x_1, y_1), (x_2, y_2), \dots, (x_n, y_n) \text{ be the points}
\]\[
\text{Minimize } f(x, y) = \sqrt{(x - x_1)^2 + (y - y_1)^2} + \sqrt{(x - x_2)^2 + (y - y_2)^2} + \cdots + \sqrt{(x - x_n)^2 + (y - y_n)^2}
\]\[
\text{Set } \frac{\partial f}{\partial x} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 0
\]\[
\frac{\partial f}{\partial x} = \frac{x - x_1}{\sqrt{(x - x_1)^2 + (y - y_1)^2}} + \frac{x - x_2}{\sqrt{(x - x_2)^2 + (y - y_2)^2}} + \cdots + \frac{x - x_n}{\sqrt{(x - x_n)^2 + (y - y_n)^2}} = 0
\]\[
\frac{\partial f}{\partial y} = \frac{y - y_1}{\sqrt{(x - x_1)^2 + (y - y_1)^2}} + \frac{y - y_2}{\sqrt{(x - x_2)^2 + (y - y_2)^2}} + \cdots + \frac{y - y_n}{\sqrt{(x - x_n)^2 + (y - y_n)^2}} = 0
\]I am not able to move ahead but probably we will find a constraint here through some inequality which will help us reach the solution .


8 replies
PhysicsIsChad
Yesterday at 3:14 PM
PhysicsIsChad
4 hours ago
Calculus-flavored sum
NamelyOrange   5
N Today at 3:54 AM by rchokler
Evaluate $\sum_{n = 0}^{\infty}\frac{1}{(4n)!}$.

(Source: AOPS user @SomeDumbChild)
5 replies
NamelyOrange
Jun 24, 2025
rchokler
Today at 3:54 AM
10 Problems
Sedro   46
N Today at 3:35 AM by Sedro
Title says most of it. I've been meaning to post a problem set on HSM since at least a few months ago, but since I proposed the most recent problems I made to the 2025 SSMO, I had to wait for that happen. (Hence, most of these problems will probably be familiar if you participated in that contest, though numbers and wording may be changed.) The problems are very roughly arranged by difficulty. Enjoy!

Problem 1: An sequence of positive integers $u_1, u_2, \dots, u_8$ has the property for every positive integer $n\le 8$, its $n^\text{th}$ term is greater than the mean of the first $n-1$ terms, and the sum of its first $n$ terms is a multiple of $n$. Let $S$ be the number of such sequences satisfying $u_1+u_2+\cdots + u_8 = 144$. Compute the remainder when $S$ is divided by $1000$.

Problem 2 (solved by fruitmonster97): Rhombus $PQRS$ has side length $3$. Point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$. Given that $QX=2$, the area of $PQRS$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 3 (solved by Math-lover1): Positive integers $a$ and $b$ satisfy $a\mid b^2$, $b\mid a^3$, and $a^3b^2 \mid 2025^{36}$. If the number of possible ordered pairs $(a,b)$ is equal to $N$, compute the remainder when $N$ is divided by $1000$.

Problem 4 (solved by CubeAlgo15): Let $ABC$ be a triangle. Point $P$ lies on side $BC$, point $Q$ lies on side $AB$, and point $R$ lies on side $AC$ such that $PQ=BQ$, $CR=PR$, and $\angle APB<90^\circ$. Let $H$ be the foot of the altitude from $A$ to $BC$. Given that $BP=3$, $CP=5$, and $[AQPR] = \tfrac{3}{7} \cdot [ABC]$, the value of $BH\cdot CH$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 5 (solved by maromex): Anna has a three-term arithmetic sequence of integers. She divides each term of her sequence by a positive integer $n>1$ and tells Bob that the three resulting remainders are $20$, $52$, and $R$, in some order. For how many values of $R$ is it possible for Bob to uniquely determine $n$?

Problem 6: There is a unique ordered triple of positive reals $(x,y,z)$ satisfying the system of equations \begin{align*} x^2 + 9 &= (y-\sqrt{192})^2 + 4 \\ y^2 + 4 &= (z-\sqrt{192})^2 + 49 \\ z^2 + 49 &= (x-\sqrt{192})^2 + 9. \end{align*}The value of $100x+10y+z$ can be expressed as $p\sqrt{q}$, where $p$ and $q$ are positive integers such that $q$ is square-free. Compute $p+q$.

Problem 7: Let $S$ be the set of all monotonically increasing six-term sequences whose terms are all integers between $0$ and $6$ inclusive. We say a sequence $s=n_1, n_2, \dots, n_6$ in $S$ is symmetric if for every integer $1\le i \le 6$, the number of terms of $s$ that are at least $i$ is $n_{7-i}$. The probability that a randomly chosen element of $S$ is symmetric is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

Problem 8: For a positive integer $n$, let $r(n)$ denote the value of the binary number obtained by reading the binary representation of $n$ from right to left. Find the smallest positive integer $k$ such that the equation $n+r(n)=2k$ has at least ten positive integer solutions $n$.

Problem 9 (solved by Math-lover1): Let $p$ be a quadratic polynomial with a positive leading coefficient. There exists a positive real number $r$ such that $r < 1 < \tfrac{5}{2r} < 5$ and $p(p(x)) = x$ for $x \in \{ r,1,  \tfrac{5}{2r} , 5\}$. Compute $p(20)$.

Problem 10 (solved by aaravdodhia): Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=995$ and $ab+bc+ca$ is a multiple of $995$.
46 replies
Sedro
Jul 10, 2025
Sedro
Today at 3:35 AM
Challenge: Make as many positive integers from 2 zeros
Biglion   45
N Today at 3:25 AM by defiw
How many positive integers can you make from at most 2 zeros, any math operation and cocatination?
New Rule: The successor function can only be used at most 3 times per number
Starting from 0, 0=0
45 replies
Biglion
Jul 2, 2025
defiw
Today at 3:25 AM
Trigonometry equation practice
ehz2701   17
N Today at 12:32 AM by ehz2701
There is a lack of trigonometric bash practice, and I want to see techniques to do these problems. So here are 10 kinds of problems that are usually out in the wild. How do you tackle these problems? (I had ChatGPT write a code for this.). Please give me some general techniques to solve these kinds of problems, especially set 2b. I’ll add more later.

Leaderboard and Solved Problems

problem set 1a (1-10)

problem set 2a (1-20)

problem set 2b (1-20)
answers 2b

General techniques so far:

Trick 1: one thing to keep in mind is

[center] $\frac{1}{2} = \cos 36 - \sin 18$. [/center]

Many of these seem to be reducible to this. The half can be written as $\cos 60 = \sin 30$, and $\cos 36 = \sin 54$, $\sin 18 = \cos 72$. This is proven in solution 1a-1. We will refer to this as Trick 1.
17 replies
ehz2701
Jul 12, 2025
ehz2701
Today at 12:32 AM
[PMO23 Qualifying I.5] Polynomial Problem
kae_3   2
N Yesterday at 10:58 PM by Siopao_Enjoyer
Find the sum of all $k$ for which $x^5+kx^4-6x^3-15x^2-8k^3x-12k+21$ leaves a remainder of $23$ when divided by $x+k$.

$\text{(a) }-1\qquad\text{(b) }-\dfrac{3}{4}\qquad\text{(c) }\dfrac{5}{8}\qquad\text{(d) }\dfrac{3}{4}$

Answer Confirmation
2 replies
kae_3
Feb 9, 2025
Siopao_Enjoyer
Yesterday at 10:58 PM
a + b + c + \fracs
SYBARUPEMULA   2
N Yesterday at 9:50 PM by SYBARUPEMULA
Given $a, b, c > 0$ such that $9a + 5b + 9c = 218$,
find the smallest value of

$$a + b + c + \frac{40}{a + 6} + \frac{72}{b + 8} + \frac{10}{c + 2}$$
2 replies
SYBARUPEMULA
Yesterday at 4:16 PM
SYBARUPEMULA
Yesterday at 9:50 PM
Expected Value of Chocolate Bar Pieces
mudkip42   1
N Yesterday at 8:55 PM by mudkip42
A $2 \times 10$ bar of chocolate has every square colored with red, green or blue uniformly at random each with $\tfrac13$ probability. All borders between squares of different colors are cut, leaving several connected pieces of squares all of which are the same color. Find the expected value of the number of pieces.
1 reply
mudkip42
Yesterday at 8:54 PM
mudkip42
Yesterday at 8:55 PM
Nt and chains
Tofa7a._.36   2
N Yesterday at 7:11 PM by Brunosalam
Let \( a_0, a_1, \ldots, a_n \) be positive divisors of the number \( 2024^{2025} \) such that:

\(\bullet\) \( a_0 < a_1 < a_2 < \cdots < a_n \)

\(\bullet\) \( a_0 \mid a_1,\, a_1 \mid a_2,\, \ldots,\, a_{n-1} \mid a_n \)

Find the largest possible value of the positive integer \( n \).
2 replies
Tofa7a._.36
Yesterday at 1:13 AM
Brunosalam
Yesterday at 7:11 PM
CMO qualifying repechange 2025
Zeroin   5
N Yesterday at 5:36 PM by Zeroin
In scalene triangle $ABC$, the circumcentre and incentre are respectively $O$ and $I$.
Let $AD$ be the altitude to line $BC$, with $D$ lying on line $BC$. Given that the radius of the
circumcircle and A-excircle are equal, prove that the points $O$, $I$, and $D$ are collinear.
5 replies
Zeroin
Yesterday at 12:24 PM
Zeroin
Yesterday at 5:36 PM
Hard Function
johnlp1234   11
N May 17, 2025 by GreekIdiot
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3
11 replies
johnlp1234
Jul 7, 2020
GreekIdiot
May 17, 2025
Hard Function
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
johnlp1234
35 posts
#1 • 1 Y
Y by alexey_phenichniy
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TuZo
19351 posts
#2
Y by
johnlp1234 wrote:
$f:R+--->R+$:
$f(x^3+f(y))=y+(f(x))^3$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aritra12
1026 posts
#3 • 3 Y
Y by abhradeep12, CatsMeow12, Wisphard
johnlp1234 wrote:
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

Can you please say what is the question,please dont post incomplete problems
please read this post's point number c Rules
This post has been edited 3 times. Last edited by Aritra12, Jul 7, 2020, 5:06 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SomeUser221104
144 posts
#4 • 1 Y
Y by abhradeep12
johnlp1234 wrote:
Find all function $f:\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that :
$$f(x^3+f(y))=y+(f(x))^3$$

FTFY
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11434 posts
#5
Y by
johnlp1234 wrote:
f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

Let $P(x,y)$ be the assertion $f\left(x^3+f(y)\right)=y+f(x)^3$.
$P(f(x),y^3+f(z))\Rightarrow f\left(f(x)^3+y+f(z)^3\right)=f(f(x))^3+y^3+f(z)$
Swapping $x,z$, we get that $f(f(x))^3=f(x)+c$ for all $x>0$ where $c\in\mathbb R$ is some constant, that is, $f(x)=\sqrt[3]{x+c}$ for all $x\in f(\mathbb R^+)$.
$P(1,x)\Rightarrow f(1+f(x))=x+f(1)^3$ so $\left(f(1)^3,\infty\right)\subseteq f(\mathbb R^+)$ and $f(x)=\sqrt[3]{x+c}$ for all $x>f(1)^3$.

Let $u>\max\left\{f(1)^3,8647\right\}$ be sufficiently large, then $u^3+f(x)>f(1)^3$ for all $x>0$ and $u>f(1)^3$, so:
$P(u,x)\Rightarrow\sqrt[3]{u^3+f(x)+c}=x+u+c$
So $f$ is a cubic polynomials, however since $f(x)=\sqrt[3]{x+c}$ for all $x>f(1)^3$ it cannot be a cubic polynomial. Hence no solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
323 posts
#6
Y by
But $f(x)=x$ clearly satisfies...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
323 posts
#7
Y by
Let $P(x,y)$ denote the assertion
Fixing $x$ and varying $y$ we see that $f$ is surjective
Let $f(a)=f(b)$ for some $a$, $b$ then
$P(x,a)-P(x,b) \implies f(x^3+f(a))-f(x^3-f(b))=a-b \implies a=b$ thus $f$ is bijective
To be continued
Ι found $f(f(x))=x \: \forall \: x \in \mathbb{R_+}$ using the substitution jasper chose above. Too lazy to writeup.
This post has been edited 1 time. Last edited by GreekIdiot, May 17, 2025, 9:49 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ektorasmiliotis
120 posts
#8
Y by
surjective in R+ or for values greater than (f(c))^3 ?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
323 posts
#9
Y by
I didnt use surjectivity anyways so it shouldnt matter...
$f$ is non constant thus picking $x$ to minimize $f(x)$ we get that $f$ is surjective on interval $(min^3\{f\}, \infty)$
This post has been edited 1 time. Last edited by GreekIdiot, May 17, 2025, 4:57 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
maromex
284 posts
#10
Y by
Actually $f$ is surjective because it's an involution.

Should I post about this FE here instead of the identical https://artofproblemsolving.com/community/c6h2179422?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
maromex
284 posts
#11
Y by
Hi. $P(f(x), y^3+ f(z)) : f(f(x)^3 + z + f(y)^3) = f(f(x))^3 + y^3 + f(z).$
And therefore by switching $x, y$ and comparing,$$f(f(x))^3 = x^3 + c.$$$P(f(x), y) : f(f(x)^3 + f(y)) = y + x^3 + c$
$P(x, f(y)) : f(x^3 + f(f(y))) = f(y) + f(x)^3$
which implies $f(x^3 + \sqrt[3]{y^3 + c}) = f(y) + f(x)^3$
Take function of both sides, and substitute into $P(f(x), y)$, get $f(f(x^3 + \sqrt[3]{y^3 + c})) = y + x^3 + c$
Cube both sides and apply some stuff, now we have$$(x^3 + \sqrt[3]{y^3 + c})^3 + c = (x^3 + y + c)^3$$for all $x, y$. This is equivalent to $(x^3 + y + c)^3 - (x^3 + \sqrt[3]{y^3 + c})^3 = c$. Fix a $y$ and let $a = y + c$ and $b = \sqrt[3]{y^3 + c}$. Both $a$ and $b$ are positive. Then we have $(x^3 + a)^3 - (x^3 + b)^3 = c$. When we factor, we get$$(a - b)((x^3 + a)^2 + (x^3 + a)(x^3 + b) + (x^3 + b)^2) = c.$$Notice that, when increasing $x$, the LHS gets arbitrarily large (and therefore not constant) if $a - b > 0$, and gets arbitrarily small (and therefore not constant) if $a - b < 0$. The only possibility is $a - b = 0$ and therefore $c = 0$. Therefore,$$f(f(x))^3 = x^3 \implies f(f(x)) = x.$$This implies $f$ is bijective. If we take $t=x^3 + f(y)$, then $f(t) > y = f(f(y))$. Now $t$ can be anything greater than $f(y)$. Because $f$ is surjective, $f(y)$ can be anything and $f$ is strictly increasing.

If, for some $x$ we have $f(x) < x$, then $x = f(f(x)) > f(x)$ contradicting the fact that $f$ is strictly increasing. If, for some $x$ we have $f(x) > x$, then $f(x) > f(f(x)) = x$ which also contradicts $f$ strictly increasing. Therefore $$f(x) = x$$for all $x$, and it works.
This post has been edited 7 times. Last edited by maromex, May 17, 2025, 8:16 PM
Reason: Done!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GreekIdiot
323 posts
#12
Y by
maromex wrote:
Actually $f$ is surjective because it's an involution.

Yeah you are right I forgot I proved that $f(f(x))=x$ :blush:
Z K Y
N Quick Reply
G
H
=
a