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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.

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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.
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0 replies
jlacosta
Mar 2, 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
Sharygin 2025 CR P17
Gengar_in_Galar   5
N 4 minutes ago by HVP
Source: Sharygin 2025
Let $O$, $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$; $D$, $E$, $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$, $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$, then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$.
Proposed by: P.Puchkov,E.Utkin
5 replies
Gengar_in_Galar
Mar 10, 2025
HVP
4 minutes ago
Not actually combo
MathSaiyan   0
4 minutes ago
Source: PErA 2025/5
We have an $n \times n$ board, filled with $n$ rectangles aligned to the grid. The $n$ rectangles cover all the board and are never superposed. Find, in terms of $n$, the smallest value the sum of the $n$ diagonals of the rectangles can take.
0 replies
MathSaiyan
4 minutes ago
0 replies
Sharygin 2025 CR P13
Gengar_in_Galar   5
N 6 minutes ago by HVP
Source: Sharygin 2025
Each two opposite sides of a convex $2n$-gon are parallel. (Two sides are opposite if one passes $n-1$ other sides moving from one side to another along the borderline of the $2n$-gon.) The pair of opposite sides is called regular if there exists a common perpendicular to them such that its endpoints lie on the sides and not on their extensions. Which is the minimal possible number of regular pairs?
Proposed by: B.Frenkin
5 replies
Gengar_in_Galar
Mar 10, 2025
HVP
6 minutes ago
Nice, simple geo
MathSaiyan   0
7 minutes ago
Source: PErA 2025/4
Let \( ABC \) be an acute-angled scalene triangle. Let \( B_1 \) and \( B_2 \) be points on the rays \( BC \) and \( BA \), respectively, such that \( BB_1 = BB_2 = AC \). Similarly, let \( C_1 \) and \( C_2 \) be points on the rays \( CB \) and \( CA \), respectively, such that \( CC_1 = CC_2 = AB \). Prove that if \( B_1B_2 \) and \( C_1C_2 \) intersect at \( K \), then \( AK \) is parallel to \( BC \).
0 replies
MathSaiyan
7 minutes ago
0 replies
The last nonzero digit of factorials
Tintarn   1
N 8 minutes ago by AshAuktober
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
1 reply
Tintarn
2 hours ago
AshAuktober
8 minutes ago
Stronger than Iran 96, not completely symmetric
MeoMayBe   2
N 9 minutes ago by CHESSR1DER
Source: Own
Let a, b, c\geq 0. Prove that
(bc+ca+ab)\left[\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}+\frac{1}{(a+b)^{2}}\right]\geq\frac{9}{4}+\frac{2abc(b-c)^{2}}{(a+b)^{2}(a+c)^{2}(b+c)}.
PS. 2 is also the best constant.
PS (2). Sorry again, I cannot type LaTeX since new members cannot share image.
2 replies
MeoMayBe
May 14, 2023
CHESSR1DER
9 minutes ago
Equilateral triangle geo
MathSaiyan   0
10 minutes ago
Source: PErA 2025/3
Let \( ABC \) be an equilateral triangle with circumcenter \( O \). Let \( X \) and \( Y \) be two points on segments \( AB \) and \( AC \), respectively, such that \( \angle XOY = 60^\circ \). If \( T \) is the reflection of \( O \) with respect to line \( XY \), prove that lines \( BT \) and \( OY \) are parallel.
0 replies
MathSaiyan
10 minutes ago
0 replies
Inspired by youthdoo
sqing   1
N 11 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $      \frac{3}{a^2+6}+\frac{1}{b^2+2}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $   \frac{2}{a^2+4}+\frac{3}{b^2+6}+\frac{2}{c^2+4}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $    \frac{3}{a^2+6}+\frac{2}{b^2+4}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 8$$Let $ a,b,c $ be real numbers such that $  \frac{3}{a^2+6}+\frac{4}{b^2+8}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 10$$
1 reply
sqing
25 minutes ago
sqing
11 minutes ago
Clean number theory
MathSaiyan   0
13 minutes ago
Source: PErA 2025/2
Let $m$ be a positive integer. We say that a positive integer $x$ is $m$-good if $a^m$ divides $x$ for some integer $a > 1$. We say a positive integer $x$ is $m$-bad if it is not $m$-good.
(a) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-bad positive integers?
(b) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-good positive integers?
0 replies
MathSaiyan
13 minutes ago
0 replies
Many-solutions combogeo
MathSaiyan   1
N 14 minutes ago by Speedysolver1
Source: PErA 2025/1
Let $S$ be a set of at least three points of the plane in general position. Prove that there exists a non-intersecting polygon whose vertices are exactly the points of $S$.
1 reply
MathSaiyan
17 minutes ago
Speedysolver1
14 minutes ago
2 var inquality
sqing   6
N 18 minutes ago by ionbursuc
Source: Own
Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{3}{2}. $ Show that$$ a+b+ab\geq1$$Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{5}{6}. $ Show that$$ a+b+ab\geq2$$
6 replies
sqing
Today at 4:06 AM
ionbursuc
18 minutes ago
Sharygin 2025 CR P2
Gengar_in_Galar   4
N 27 minutes ago by FKcosX
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
4 replies
Gengar_in_Galar
Mar 10, 2025
FKcosX
27 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 31 minutes ago by John_Mgr
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
32 replies
Asjmaj
Dec 31, 2024
John_Mgr
31 minutes ago
Where is the equality?
AndreiVila   2
N an hour ago by MihaiT
Source: Romanian District Olympiad 2025 9.3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2 replies
AndreiVila
Mar 8, 2025
MihaiT
an hour ago
Functional equation on (0,infinity)
mathwizard888   55
N Today at 3:43 AM by Ilikeminecraft
Source: 2016 IMO Shortlist A4
Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$, $$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$$
55 replies
mathwizard888
Jul 19, 2017
Ilikeminecraft
Today at 3:43 AM
Functional equation on (0,infinity)
G H J
Source: 2016 IMO Shortlist A4
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mathwizard888
1635 posts
#1 • 11 Y
Y by Davi-8191, CanVQ, itslumi, A-Thought-Of-God, megarnie, rama1728, ImSh95, Adventure10, Mango247, deplasmanyollari, Funcshun840
Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$, $$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$$
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FEcreater
74 posts
#2 • 11 Y
Y by khan.academy, Aryan-23, Modesti, guptaamitu1, megarnie, myh2910, Inconsistent, ImSh95, nibaba, Adventure10, Mango247
Post before! Anyway, my solution
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Snakes
10620 posts
#3 • 8 Y
Y by rmtf1111, bobcats4thewin, microsoft_office_word, sans.., ImSh95, channing421, Adventure10, vrondoS
$x=1, y=1: f(1)f(f(1))+f(f1))=f(1)f(f(1))+f(1)f(f(1))\implies \\
\implies f(1)f(f(1))=f(f(1))$, but as $f(f(1))\in \mathbb{R}^{+}\implies f(1)=1$
$x=1, y=x: f(f(x))+f(xf(1))=f(x)+f(x)f(f(x^2))\implies$
$$f(f(x))=f(x)f(f(x^2))$$$y=1: xf(x^2)=f(x)\Longleftrightarrow \boxed{x^2f(x^2)=xf(x)}$
$x,x: xf(x^2)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2))=2\frac{f(x)}{x}f(f(x^2))=2\frac{f(f(x))}{x}$ or
$xf(x)f(f(x))+xf(xf(x))=2f(f(x))$
Put $x\rightarrow x^2$ in the above and get
$x^2f(x^2)f(f(x^2))+x^2f(x^2f(x^2))=2f(f(x^2))\Longleftrightarrow$ $\Longleftrightarrow xf(xf(x))=f(f(x))(2-xf(x))\\ \text{and} \\
x^2f(x^2f(x^2))=f(f(x^2))(2-xf(x))\implies \\
\frac{x}{f(f(x))}=\frac{x^2}{f(f(x^2))}\implies f(x)=\frac{1}{x}$ for any positive real number $x$
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mathwizard888
1635 posts
#4 • 3 Y
Y by ImSh95, Adventure10, Mango247
As usual, let $P(x,y)$ denote the problem condition when $(x,y)$ is plugged in.

$P(1,1)$ is $f(1)f(f(1))+f(f(1))=2f(1)f(f(1))$, so $f(1)=1$. $P(1,x)$ and $P(x,1)$ are
\begin{align*}f(f(x))+f(x) &= f(x)\left(1+f(f(x^2))\right),\\ xf(x^2)+f(f(x)) &= f(x)\left(f(f(x^2))+1\right), \end{align*}which reduce to
\begin{align} f(f(x))&=f(x)f(f(x^2)),\\ xf(x^2)&=f(x), \end{align}where we used $(1)$ to get $(2)$. $P(x,1/x)$ is $$xf(x^2)f\left(f\left(\frac{1}{x}\right)\right) + f\left(\frac{f(x)}{x}\right) = f(f(x^2)) + f\left(f\left(\frac{1}{x^2}\right)\right),$$so $$f(x)f\left(f\left(\frac{1}{x}\right)\right) = f\left(f\left(\frac{1}{x^2}\right)\right) = \frac{f\left(f\left(\frac{1}{x}\right)\right)}{f\left(\frac{1}{x}\right)}.$$Therefore, $f(x)f(1/x)=1$ for all $x$.

Using $(2)$, rewrite the problem condition as $$f(x)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2))+f(f(y^2))\right).$$$P(x^2,x)$ gives
\begin{align*} f(x^2)f(f(x)) + f(f(x)) &= f(x^3) \left(f(f(x^4))+f(f(x^2))\right)\\ &= f(x^3) \left(\frac{f(f(x^2))}{f(x^2)}+\frac{f(f(x))}{f(x)}\right)\\ &= f(f(x))f(x^3) \left(\frac{1}{f(x^2)f(x)}+\frac{1}{f(x)}\right)\\ &= \frac{f(f(x))f(x^3)(f(x^2)+1)}{f(x^2)f(x)}. \end{align*}Therefore, $f(x^2)f(x)=f(x^3)$. $P(x,x)$ is $$f(x)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2)).$$Note that
\begin{align*} 0 < f(xf(x)) &= 2f(x^2)f(f(x^2)) - f(x)f(f(x))\\ &= 2f(x^2)f(f(x^2)) - f(x)^2f(f(x^2))\\ &= 2f(x^2)f(f(x^2)) - xf(x^2)f(x)f(f(x^2))\\ &= f(x^2)f(f(x^2))(2-xf(x)), \end{align*}so $xf(x)<2$ for all $x$. Let $g(x)=xf(x)$, and note that $f(x)f(1/x)=1$ implies $g(x)g(1/x)=1$.

I claim that $g(x^3)=g(x)^2$. Note that $f(x)=\frac{g(x)}{x}$, and by $(2)$, $f(x^2)=\frac{g(x)}{x^2}$. Therefore, $f(x^3)=f(x)f(x^2)=\frac{g(x)^2}{x^3}$, proving the claim.

Suppose $g(x)>1$ for some $x$. Then $g(x^{3^n})=g(x)^{2^n}>2$ for sufficiently large integers $n$, a contradiction. If $g(x)<1$, then $g(1/x)>1$, which gives a contradiction as before. Therefore, $g(x)=1$ for all $x$, so $\boxed{f(x)=\frac{1}{x}}$ is the only possible solution. We can easily check that this works.
This post has been edited 1 time. Last edited by mathwizard888, Jul 29, 2017, 1:34 AM
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uraharakisuke_hsgs
365 posts
#5 • 4 Y
Y by uvuvwevwevwe, ImSh95, Adventure10, Mango247
My solution If $f$ is constant then $f \equiv 0$ which is a contradiction
If $f$ is nonconstant , let $P(x_0,y_0)$ be the assertion
$P(1,1) : f(f(1)) = f(1).f(f(1)) \implies f(1) = 1$
$P(x,1) : xf(x^2)+f(f(x)) = f(x) ( f(f(x^2))+1)$
$P(1,y) : f(f(y))+f(y) = f(y)(f(f(y^2))+1) \implies f(x)+f(f(x)) = f(f(x))+xf(x^2) \implies f(x^2) = \frac{f(x)}{x}$
Once again , $P(x,1)$ gives $f(x)+f(f(x)) = f(x)+f(x)f(f(x^2)) \implies f(f(x^2)) = \frac{f(f(x))}{f(x)} = f(\frac{f(x)}{x}) = f(f^2(x))$
Now we prove $f$ is injective . Suppose that there exist $f(a) = f(b) = c$
From the conditions we have $f(x)f(f(y)) + f(yf(x)) =xf(x^2)f(f(y)+f(yf(x)) =  f(xy)(f(f(x^2))+f(f(y^2))) = f(xy).(f(f^2(x)+f(f^2(y)))$
Now we set $x = a , x = b$ respectively ; then , we receive $f(ay) = f(by) \forall $ $y \in \mathbb R^{+}$
Then $f(a^2) = f(ab) = f(b^2) = \frac{f(a)}{a} = \frac{f(b)}{b} = \frac{c}{a} = \frac{c}{b} \implies a = b$, which is a contradiction
Then $f$ is injective . From above we almost have $f(f(x^2)) = f(f^2(x)) \implies f^2(x) = f(x^2) = \frac{f(x)}{x} \implies f(x) = \frac{1}{x}$
Then , the solution is $f(x) = \frac{1}{x} $ $\forall$ $x $ $\in $ $\mathbb R^{+}$
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anantmudgal09
1979 posts
#6 • 5 Y
Y by vsathiam, itslumi, ImSh95, Adventure10, Mango247
mathwizard888 wrote:
Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$, $$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$$

Answer: The only functions which work are $f(x)=\tfrac{1}{x}$ for all $x>0$.

Let $P(x, y)$ denote the given functional equation: $$xf(x^2)f(f(y))+f(yf(x))=f(xy)\cdot \left(f(f(x^2))+f(f(y^2))\right),$$then it is easy to see that $f(x) \equiv \tfrac{1}{x}$ is indeed a solution. We now show that this is the only one.

Note that $$P(1, 1) \implies f(1)=1$$and $$P(1, y) \implies f(f(y))=f(y)f(f(y^2)),$$for all $y>0$. Also, $$P(x, 1) \implies xf(x^2)=f(x),$$for all $x>0$. Hence, we can rewrite $P$ as $$f(x)f(f(y))+f(yf(x))=f(xy) \cdot \left(\frac{f(f(x))}{f(x)}+f(f(y^2))\right).$$
Claim: $f$ is injective.

(Proof) Suppose $f(a)=f(b)$ for some $a \ne b$. Together, $P(a, y)$ and $P(b, y)$ yield that $f(ay)=f(by)$ for all $y>0$; hence, for $t=\tfrac{1}{a}$ we have $f(t)=f(1)=1$. However, this yields $f(t^2)=f(t \cdot t)=f(t)=1,$ so $$f(t)=tf(t^2)=tf(t) \implies t=1,$$hence $\boxed{a=b}$. $\blacksquare$

Finally, we observe that $$f\left(f(x)^2\right)=\frac{f(f(x))}{f(x)}=f(f(x^2))=f\left(\frac{f(x)}{x}\right) \implies f(x)^2=\frac{f(x)}{x} \implies f(x) \equiv \frac{1}{x},$$due to injectivity, hence we conclude that the initial assertion is valid. $\blacksquare$
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juckter
322 posts
#7 • 4 Y
Y by Delray, Mobashereh, ImSh95, Adventure10
The only such function is $f(x) = \frac{1}{x}$ for all $x \in \mathbb{R}^{+}$, which is easily seen to satisfy the given functional equation. Let $P(x, y)$ denote the usual.

Using $P(1, 1)$, we get $f(1)f(f(1)) + f(f(1)) = 2f(1)f(f(1))$, and hence $f(1) = 1$. Now, using $P(1, x)$, we get

$$f(f(x)) + f(x) = f(x)f(f(x^2)) + f(x)$$$$\implies$$
$$\boxed{f(f(x)) = f(x)f(f(x^2))} \quad (1)$$
Thus, using $P(x, 1)$ we obtain:

$$xf(x^2) + f(f(x)) = f(x)f(f(x^2)) + f(x) = f(x) + f(f(x))$$$$\implies$$
$$\boxed{xf(x^2) = f(x)} \quad (2)$$
We now show that $f(t) = 1$ implies $t = 1$. In this case then using $(2)$ and $(1)$ with $x = t$ we get $f(t^2) = \frac{1}{t}$ and $f\left(\frac{1}{t}\right) = 1$. Then $P(t, t)$ gives

$$t \cdot \frac{1}{t} + 1 = \frac{2}{t}$$$$\implies$$$$t = 1$$
As desired. We now proceed to show that $f$ is injective. Assume $f(a) = f(b)$, then $(1)$ and $(2)$ give $af(a^2) = bf(b^2)$ and $f(f(a^2)) = f(f(b^2))$, so comparing $P(a, x)$ and $P(b, x)$ gives $f(ax) = f(bx)$ for all positive real $x$. Take $x = \frac{1}{a}$ to get $f\left(\frac{b}{a}\right) = f(1) = 1$, and hence $a = b$ by the previously shown statement. Thus $f$ is injective. Finally, plug in $f(x)$ in $(2)$ and compare with $(1)$ to obtain

$$f(x)f(f(x)^2) = f(f(x)) = f(x)f(f(x^2))$$$$\implies$$$$f(f(x)^2) = f(f(x^2))$$$$\implies$$$$f(x)^2 = f(x^2)$$
Finally, plug this relation into $(2)$ to obtain $f(x) = \frac{1}{x}$ for all positive real $x$.
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math_pi_rate
1218 posts
#8 • 3 Y
Y by ImSh95, Adventure10, Mango247
My solution: Let $P(x,y)$ denote the given assertion.
$$P(1,1) \Rightarrow f(1)f(f(1))+f(f(1))=2f(1)f(f(1) \Rightarrow f(1)=1$$$$P(1,x) \Rightarrow f(f(x))+f(x)=f(x)(1+f(f(x^2))) \Rightarrow f(f(x))=f(x)f(f(x^2)) \text{ } \text{ } \text{ } -(A)$$$$P(x,1) \Rightarrow xf(x^2)+f(f(x))=f(x)f(f(x^2))+f(x)=f(f(x))+f(x) \Rightarrow f(x)=xf(x^2) \text{ } \text{ } \text{ } -(B) \text{ } \text{ } \text{ } \text{[Using }A \text{]}$$On putting $x \rightarrow f(x)$ in $B$, we get that $f(f(x))=f(x)f(f(x)^2)$. Then, using $A$, we have $f(f(x)^2)=f(f(x^2))$ $-(C)$.

$$P(x,x) \Rightarrow xf(x^2)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2)) \Rightarrow f(x)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2)) \text{ } \text{ } \text{ } -(D) \text{ } \text{ } \text{ } \text{[Using }B \text{]}$$
Now, Let $f(a)=f(b)$ for some $a,b \in \mathbb{R^+}$. Then, from $C$, we have $f(f(a^2))=f(f(b^2))$.

Also, From $P(a,y)$ and $P(b,y)$, along with $B$, we get that $f(ay)=f(by) \text{ } \forall y \in \mathbb{R^+}$.

Putting $y=f(a)=f(b)$ above, we have $f(af(a))=f(bf(a))=f(bf(b))$. Thus, On putting $x=a$ and $x=b$ in $D$, we get that $f(a^2)=f(b^2)$. But then, from $B$, we have $a=b$. This means that $f$ is injective.

Finally, Using injectivity on $C$, we have $f(x^2)=f(x)^2$. And with the help of $B$, we get $f(x)=xf(x^2)=xf(x)^2 \Rightarrow f(x)=\frac{1}{x} \text{ } \forall x \in \mathbb{R^+}$
This post has been edited 2 times. Last edited by math_pi_rate, Sep 23, 2018, 7:18 PM
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MNJ2357
644 posts
#9 • 2 Y
Y by ImSh95, Adventure10
Here's a solution that doesn't use injectivity.
Let $P(x,y)$ be the given assertion.
$P(1,1) : f(1)f(f(1)) + f(f(1)) = 2f(1)f(f(1))$, so we have $f(1)=1$
Combining $P(1,x)$ and $P(x,1)$ gives $xf(x^2)=f(x)...(1)$.
We rewrite the original equation as
$$f(x)f(f(y))+f(yf(x))=f(xy)\left(\frac{f(f(x))}{f(x)}+\frac{f(f(y))}{f(y)}\right)$$$P(x,\frac{1}{x}) : f(x)f(f(\frac{1}{x}))+f(\frac{f(x)}{x})=f(f(x^2))+\frac{f(f(\frac{1}{x}))}{f(\frac{1}{x})}=f(\frac{f(x)}{x})+\frac{f(f(\frac{1}{x}))}{f(\frac{1}{x})}\Longrightarrow f(x)f(\frac{1}{x})=1$
$P(x,x) : f(x)f(f(x))+f(xf(x))=\frac{2f(x^2)}{f(x)}f(f(x))=\frac{2}{x}f(f(x))...(2)$
$P(x,\frac{1}{xf(x)}) : f(x)f(f(\frac{1}{xf(x)}))+f(\frac{1}{x})=f(\frac{1}{f(x)})\left(\frac{f(f(x))}{f(x)}+\frac{f(f(\frac{1}{xf(x)}))}{f(\frac{1}{xf(x)})}\right)=f(\frac{1}{x})+f(\frac{1}{f(x)})\frac{f(f(\frac{1}{xf(x)}))}{f(\frac{1}{xf(x)})}$
$\Longrightarrow f(x)f(f(x))=f(xf(x))=(\frac{2}{x}-f(x))f(f(x)) (\because (2))\Longrightarrow f(x)=\frac{1}{x}$.
Hence $\boxed{f(x)=\frac{1}{x}}$ is the only solution.
This post has been edited 1 time. Last edited by MNJ2357, Oct 6, 2018, 3:30 AM
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yayups
1614 posts
#10 • 3 Y
Y by ImSh95, Adventure10, Mango247
Run of the mill FE. Not easy, but standard.

The only solution is $f(x)=1/x$. It is easy to see that this works. Now, suppose $f$ is the function satisfying the FE, and let $P(x,y)$ be the statement of the FE. We see that
\[P(1,1)\implies f(1)f(f(1))+f(f(1))=f(1)\cdot 2f(f(1))\implies f(1)+1=2f(1)\implies f(1)=1.\]Note that we can cancel stuff since the domain and codomain exclude $0$. We now see that
\begin{align*}
P(x,1) &\implies xf(x^2)+f(f(x)) = f(x)f(f(x^2))+f(x) \\
P(1,x) &\implies f(f(x))+f(x)=f(x)f(f(x^2))+f(x).
\end{align*}Thus, $f(x^2)=f(x)/x$, and $f(f(x^2))=f(f(x))/f(x)$. Using this, we simplify the FE to
\[f(x)f(f(y))+f(yf(x))=f(xy)\left(\frac{f(f(x))}{f(x)}+\frac{f(f(y))}{f(y)}\right),\]and call the above statement $Q(x,y)$. Note that
\[Q(x,x)\implies f(x)f(f(x))+f(xf(x))=f(x^2)\cdot\frac{2f(f(x))}{f(x)}=\frac{2f(f(x))}{x}.\]Thus,
\[f(xf(x))=f(f(x))\left(\frac{2}{x}-f(x)\right).\]
Claim: $f$ is injective.

Proof of Claim: Suppose $f(a)=f(b)=k$. Then, $Q(a,b)$ implies
\[kf(k)+f(bk)=f(ab)\cdot 2\frac{f(k)}{k}.\]But
\[f(bk)=f(bf(b))=f(k)(2/b-k),\]so
\[\frac{2f(k)}{b}=\frac{2f(k)}{k}f(ab).\]Thus, $b=k/f(ab)$. Similarly, we get $a=k/f(ab)$, so $a=b$. Thus, $f$ is injective. $\blacksquare$

Now, we have
\[f(f(x^2))=f(f(x))/f(x)=f(f(x)^2),\]so $f(x^2)=f(x)^2$. Thus, $f(x)^2=f(x)/x$, which means that $f(x)=1/x$, as desired. $\blacksquare$
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william122
1576 posts
#11 • 2 Y
Y by ImSh95, Adventure10
Denote the assertion as $P(x,y)$. $P(1,1)$ yields $f(1)f(f(1))+f(f(1))=2f(1)f(f(1))\implies f(1)+1=2f(1)$, so $f(1)=1$.
$P(1,x)$ gives $f(f(x))=f(f(x^2))f(x)$ and $P(x,1)$ yields $xf(x^2)=f(x)\,\,\,(*)$.
Now, consider the two equations: $P(x,y)$ and $P(y,x)$. As the RHS is symmetric, these two have the same LHS. In particular, $$f(x)f(f(y))+f(yf(x))=f(y)f(f(x))+f(xf(y))$$Suppose that $f(a)=f(b)$. Substituting $(a,b)$ into the equation above, we get that $f(af(b))=f(bf(a))$. Now, considering $P(a,b)$ and $P(a,a)$, it is not hard to see that the LHS is the same for both, as well as the second factor of the RHS, so we have $f(ab)=f(a^2)$. Similarly, $f(ab)=f(b^2)$, so $f(a^2)=f(b^2)$. Now, compare $f(a,b)$ and $f(b,a)$. The RHS is the same for both, as well as the latter term on the LHS, so we have that $$af(a^2)f(f(b))=bf(b^2)f(f(a))$$and therefore $a=b$. So, $f$ is injective.

To finish, note that if we plug in $f(x)$ into $(*)$, we get $f(x)f(f(x)^2)=f(f(x))$, so $f(f(x)^2)=f(f(x^2))$, and injuectivity gives $f(x^2)=f(x)^2$. Using this in $(*)$ will give $xf(x)=1$, or $f(x)=\frac{1}{x}$, so this is the only solution.
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sa2001
281 posts
#12 • 3 Y
Y by ImSh95, Adventure10, Mango247
We claim that $f(x) \equiv 1/x$ is the only solution. By substitution, we see that it works.
Let $P(x, y)$ be the assertion in the problem statement.
$P(1, 1)$ gives $f(1) = 1$.
$P(x, y)$ combined with $P(y, x)$ gives:
$f(x^2) = f(x)/x$ - $(1)$
$f(f(x)) = f(x)f(f(x^2))$ - $(2)$
Now, suppose $f(a) = f(b)$.
Then, $P(a, y)$ combined with $P(b, y)$ combined with $(1)$ and $(2)$ gives $f(ay) = f(by)$ which gives $f(x) = f(b/a * x)$.
Then, $(1)$ gives $f(b/a) = f(b/a*b/a)*b/a = f(b/a)*b/a$ which gives $a = b$. Thus, $f$ is injective.
Considering $(2)$, $f(f(x^2)) = f(f(x))/f(x) = f(f(x)^2)$ by $(1)$, and applying injectivity gives $f(x^2) = f(x)^2$. Putting this in $(1)$ gives $f(x) = 1/x$, as desired.
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IndoMathXdZ
691 posts
#13 • 3 Y
Y by ImSh95, Adventure10, Mango247
We claim that $f(x) = \frac{1}{x}$ is the only solution. It is easy to check that this function indeed works.
Now, we will prove that it is the only one.
Let $P(x,y)$ denote the assertion of $x$ and $y$ to the given functional equation.
Now, notice that $P(1,1)$ gives us $f(1) = 1$.
Furthermore, $P(x,1)$ and $P(1,x)$ gives us:
\[ xf(x^2) + f(f(x)) = f(x) (f(f(x^2)) + 1) \]\[ f(f(x)) + f(x) = f(x) ( f(f(x^2)) + 1) \]From here, we get that $f(f(x)) = f(x) f(f(x^2)) $ and $f(x) = xf(x^2)$.

Now, notice that the given functional equation is now equivalent to
\[ f(x)f(f(y)) + f(yf(x)) = f(xy) \left( \frac{f(f(x))}{f(x)} + \frac{f(f(y))}{f(y)} \right) \]Take $P(x,x)$ now :
\[ f(x)f(f(x)) + f(xf(x)) = f(x^2) \cdot \frac{2f(f(x))}{f(x)} = \frac{2f(f(x))}{x} \]Change $x$ to $x^2$ in the equation above, and by simplifying:
\[  \frac{f(f(x))}{x} + f(xf(x)) = \frac{2f(f(x)))}{x^2 f(x)} \]Comparing last two equations we have,
\[ f(f(x)) \left( \frac{2}{x} - f(x) \right) = f(f(x)) \left( \frac{2}{x^2f(x)} - \frac{1}{x} \right) \]\[ \frac{3}{x} = f(x) + \frac{2}{x^2 f(x)} \]Solve this give us $f(x) = \frac{1}{x}$ or $f(x) = \frac{2}{x}$.
Checking back to the equation, the one satisfy is only $f(x) = \frac{1}{x}$.
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khina
993 posts
#14 • 2 Y
Y by ImSh95, Adventure10
This proof is basically the proof above, but i'm an arrogant kid that wants to flex his math skills so this i'm leaving this for storage.

First of all we claim that $f(x) = \dfrac{1}{x}$ is the only solution. This clearly works. Also, denote the assertion as $P(x,y)$

$P(1, 1)$ gives us that $f(1) = 1$. Plugging $P(1,x)$ then gives us our first major observation:
\begin{align*}
f(f(x^2)) &= \dfrac{f(f(x))}{f(x)}
\end{align*}In addition, plugging in $P(x,1)$ gives us that:
\begin{align*}
xf(x^2) + f(f(x)) &= f(x) + f(f(x)) \\
f(x^2) &= \dfrac{f(x)}{x}
\end{align*}Note that this simplifies the function equation into:
\begin{align*}
f(x)f(f(y)) + f(yf(x)) &= f(xy) \sum_{cyc} \dfrac{f(f(x))}{f(x)}
\end{align*}Also note that we have:
\begin{align*}
f( \frac{f(x)}{x} ) &= f(f(x^2)) \\
&= \dfrac{f(f(x))}{f(x)}
\end{align*}This finally gives us that plugging in $P(x, \dfrac{1}{x})$ and simplifying, we have:
\begin{align*}
f(\frac{1}{x}) &= \dfrac{1}{f(x)}
\end{align*}Our final step now is to plug in $P(x,x)$. We have that after some algebra we find:
\begin{align*}
f(xf(x)) &= f(f(x)) \left (\dfrac{2}{x} - f(x) \right )
\end{align*}Now, since $f(x^2) = \dfrac{f(x)}{x}$, we have that $xf(x) = x^2f(x^2)$. This gives us that by plugging in $x^2$ into the equation above and equating, we find
\begin{align*}
f(x) &= \dfrac{1}{x} \thickspace \text{or} \thickspace \dfrac{2}{x}
\end{align*}for all real $x$.

Finally using the fact that $f(\frac{1}{x}) = \dfrac{1}{f(x)}$ gives us that $\boxed{f(x) = \dfrac{1}{x}}$ for all positive reals $x$.
This post has been edited 1 time. Last edited by khina, Sep 13, 2019, 3:07 AM
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Physicsknight
635 posts
#15 • 2 Y
Y by ImSh95, Adventure10
My Solution
Let $P (x,y) $ be the assertion of $xf (x^2)f (f (y))+f (y f (x))=f (xy)(f (f (x^2))+f (f (y^2)) $
$P (1,1) $ gives $f (1)=1$ hence, $P (1,x) $ shows $f (f (x))=f (x)f (f (x^2))\cdots (1) $
From $P (x,1) $
we get $xf (x^2)=f (x) $ or $f (x^2)=\frac {f (x)}{x}\cdots  (2) $
Now, $P\left (x,\frac {1}{x}\right)=f (x)f (f \left(\frac {1}{x}\right)=f (f \left(\frac {1}{x^2})\right) $
Finally, $f (x)f (\frac {1}{x})=1$
Applying $P \left(\frac {1}{x},\frac {1}{f (x)}\right)=1$
$\frac {1}{xf (x^2)f (f (f (x)))}+\frac {1}{f (f (x)^2)}=\frac {1}{f (xf (x))}\left (\frac {1}{f (f (x^2))}+\frac {1}{f (f (f (x^2)))}\right) $
Simplifying that,
$f (xf (x))^2 f (f (x))=f (x)^2 f (f (x))f (f (x)^2) $
Simplifying $(1) $ and $(2) $
$f (x f (x))^2=f (x)f (f (x)) $
$P\left (\frac {1}{x},\frac {1}{x}\right)\implies f (f (x))=x^4f (x)^3$
Note that, $x^4f (x)^3=f (f (x))=f (x)f (f (x^2))=f (x)x^8f (x^2)^3=x^5f (x)^4$
Therefore, $\boxed {f (x)=\frac {1}{x}}$ $ \forall x\in\mathbb {R^+} $
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