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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.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
Posting Guidelines
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
How do I get a problem on the contest page?
How do I study for mathcounts?
Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
IMO 2023 P2
799786   89
N 17 minutes ago by kaede_Arcadia
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
89 replies
799786
Jul 8, 2023
kaede_Arcadia
17 minutes ago
combinatorıc
o.k.oo   0
38 minutes ago
A total of 3300 handshakes were made at a party attended by 600 people. It was observed
that the total number of handshakes among any 300 people at the party is at least N. Find
the largest possible value for N.
0 replies
o.k.oo
38 minutes ago
0 replies
Problem 2, Olympic Revenge 2013
hvaz   66
N an hour ago by MonkeyLuffy
Source: XII Olympic Revenge - 2013
Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.
66 replies
hvaz
Jan 26, 2013
MonkeyLuffy
an hour ago
Functional equation wrapped in f's
62861   35
N an hour ago by ihatemath123
Source: RMM 2019 Problem 5
Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]for all real numbers $x$ and $y$.
35 replies
62861
Feb 24, 2019
ihatemath123
an hour ago
Amc10 prep question
Shadow6885   13
N 2 hours ago by RandomMathGuy500
My question is how much of the geo and IA textbooks is relevant to AMC 10?
13 replies
Shadow6885
Yesterday at 6:20 AM
RandomMathGuy500
2 hours ago
k Is your state listed?
Chatelet1   408
N 3 hours ago by Eddie_tiger
Multiple states have announced their top students who will advance to the 2025 MATHCOUNTS National Competition in May:

• From Alabama: Henry Gladden of Mobile, Austin Lu of Birmingham, Jessie Shi of Vestavia, and Minlu Wang-He of Auburn.

• From Arkansas: Ryan Fan of Fayetteville, Vivek Kalyankar of Fayetteville, Evan Ning of Fayetteville and Charles Yao of Conway.

• From Connecticut: Hayden Hughes of Newtown, Ethan Shi of Riverside, Alex Svoronos of Greenwich and Elaine Zhou of Hamden.

• From the Department of Defense: Narmin Guliyeva of Ankara, Turkey; Taeyul Kim of Manana, Bahrain; Nathan Liang of Wiesbaden, Germany; and Lucas Sze of Okinawa, Japan.

• From Hawaii: Taehwan Jeon, Hilohak Kwak, Isaac Qian and Thien Tran, all from Honolulu.

• From Kansas: Haidan Anderson & Jayden Xue of Overland Park, Christopher Spencer of Manhattan, and Ruby Jiang of Lawrence.

• From Maine: Ana Kanitkar & Connor Kirkham of Falmouth, Anna McClary of Hermon and Poppy Sandin of Bar Harbor.

• From Massachusetts: Eric Huang of Acton, Shlok Mukund & Brandon Ni of Lexington, and Soham Samanta of Medford.

• From Missouri: Lucas Lai of Columbia, Kevin Shi of St. Louis, Charles Yong & Jay Zhou of Chesterfield.

• From Montana: Titus Gilder of Missoula, Otis Heggem of Billings, Kaleb Houtz of Great Falls and Evan Newcomer of Missoula.

• From Nevada: Solomon Dumont of Las Vegas, Aaron Lei of Reno, Leeoz Nebat of Henderson and Maxwell Tsai of Las Vegas.

• From New Mexico: Mark Goldman, Daniel He, Iris Huang and Patrick McArdle, all from Albuquerque.

• From New York: Derrick Chen of Great Neck, Victor Yang of Great Neck, Hanru Zhang of Jericho and Ryan Zhang of Jericho.

• From Rhode Island: Kahlan Anderson of the Wheeler School, Julian Bernhoft & Colin Hegstrom of Providence, and Theodora Watson of Barrington.

• From South Carolina: Yukai Hu of Elgin, Justin Peng of Clemson, Geonhoo Shim of Columbia, and Aaron Wang of Mount Pleasant.

• From South Dakota: Seth Chaplin & Maxwell Wang of Sioux Falls, Laukia Gundewar of Aberdeen, and Cohwen Heimann of Aberdeen.

• From Texas: Shaheem Samsudeen & Ayush Narayan of Plano, Nathan Liu of Richardson, and James Stewart of Southlake.

• From Vermont: Mohid Ali of South Burlington, Vivek Chadive of South Burlington, Joshua Kratze of St. Johnsbury and Albert Zhang of South Burlington.

• From Wisconsin: August Reeder & Lucy Chen of Fitchburg, Junhao Feng of Milwaukee, and Jiyan Singh of River Hills.

===
Updated on 3/15/2025:

• From Colorado: Noah Liu, Christopher Zhu, Neo Luo, and Andrew Zhao.

• From Florida: Arnav Bhatia, Gnaneswar Peddesugari, Edwin Gao, and Rananjay Parmar.

• From Indiana: Roland Li, Hrishabh Bhowmik, Sophia Chen, and Arjun Raman.

• From Kentucky: Sri Shubhaan Vulava, Joyce Liu, Victor Gong, and Brandon Tedja.

• From Maryland: Eric Xie, Angie Zhu, Roger Huang, and Leo Su.

• From Michigan: Arnav Vunnam, Eric Jin, Akshaj Malraj, and Chaithanya Budida.

• From Minnesota: Ahmed Ilyasov, Will Masanz, Anshdeep Singh, and Branden Qiao.

• From New Jersey: Ethan Imanuel, Advait Joshi, Jay Wang, and Easton Wei.

• From North Carolina: Shivank Chintalpati, Steven Wang, Lucas Li, and Leo Hong.

• From Ohio: Henry Lu, Andy Mo, Archishmen Dey, and Caleb Tan.

• From Oregon: Sophia Han, Kevin Cheng, Garud Shah, and Ryan Zhang.
408 replies
1 viewing
Chatelet1
Mar 8, 2025
Eddie_tiger
3 hours ago
probability pentagon contains center
JohnStuckey   3
N 4 hours ago by sadas123
Here's a cute problem:

Consider a regular pentagon. Choose 3 points along the perimeter of the pentagon, and form a triangle with those 3 points. What is the probability that this triangle contains the center of the pentagon.
3 replies
JohnStuckey
Yesterday at 6:27 PM
sadas123
4 hours ago
ohio mathcounts state
Owinner   38
N 4 hours ago by Andyluo
what is the cutoff for cdr for ohio? Is ohio a competitve state?
38 replies
Owinner
Mar 11, 2025
Andyluo
4 hours ago
What was your Mathcounts Chant?
ilovebender   59
N 4 hours ago by wikjay
Hi, I wanted to ask everyone, before the written competition, each state had to do their chant. What chant did you and your team make xd!

Im in WV and I think we did the least cringe chant, and it goes like this:

(After one person says something, the other person following waits ~2 seconds, and then says their line)
P1: Hi
P2: Hi
P3: Hi
P4: Bye

Lol if you were at the national competition you prob remember.
59 replies
ilovebender
May 12, 2022
wikjay
4 hours ago
Problem of the week
evt917   26
N 4 hours ago by PikaPika999
Whenever possible, I will be posting problems twice a week! They will be roughly of AMC 8 difficulty. Have fun solving! Also, these problems are all written by myself!

First problem:

$20^{16}$ has how many digits?
26 replies
evt917
Mar 5, 2025
PikaPika999
4 hours ago
Imposible
maxamc   1
N 6 hours ago by maromex
if 1+1 is 2 then what is the square root of 4 with 100 significant figures?
1 reply
maxamc
6 hours ago
maromex
6 hours ago
9 What should I do..?
Leeoz   6
N Yesterday at 2:13 PM by sadas123
So, there is a very important decision.. and I have decided to asked all of the people on AoPS :P

I am going to MathCounts nats, but there is the MathCon a day before the competition. It there anything I will miss by just going to MathCounts right on the day of the contest, or is it worth it to go to MathCon, even if I will have to leave before the awards.

Just want your ideas and opinions for this :)
also pls say if there is something that I will miss in MathCounts before the actual contest
6 replies
Leeoz
Yesterday at 6:17 AM
sadas123
Yesterday at 2:13 PM
MATHCOUNTS State Preparation
mithu542   27
N Yesterday at 1:16 PM by krish6_9
Hello!

I'm going to prepare for Mathcounts state soon. I want some advice on what to do. I am in 7th grade, and I want to make it to nationals. I know I should obviously take practice tests, but should I do something else other than that, or just grind all (or most) practice tests from previous years? Also, how much should I focus on Countdown round relative to the other tests?

(For reference, I got 43 on school, and 41 on chapter. Last year, I got 16/116 rank in state. Since then, I have done the following courses from aops:
Intro: algebra b, number theory, c&p, geometry
Intermediate: algebra, number theory, c&p)
27 replies
mithu542
Feb 14, 2025
krish6_9
Yesterday at 1:16 PM
9 Pi or Tau
jkim0656   100
N Yesterday at 5:19 AM by jkim0656
Hey Aops!
Pi = Circumfrence/Diameter
Tau = Circumfrence/Radius
I have noticed a lot of sites, including Khan Academy, in support of tau over pi...
so what do you think?
https://www.scientificamerican.com/article/let-s-use-tau-it-s-easier-than-pi/
However i am still in support of the good ol pi :)
(btw this is my first aops poll) :-D

EDIT: 50 votes!!! :play_ball:
EDIT: 100 votes!!! :jump:
EDIT: 150 votes! :trampoline:
EDIT: 200 votes! ;)
Edit: 250 votes !!!! yaya :gathering:

If u support pi pls upvote :)
100 replies
jkim0656
Mar 14, 2025
jkim0656
Yesterday at 5:19 AM
Master of Functional Equations
Bandera   13
N Sep 6, 2019 by RustyFox
Source: Own, based on a problem from Ukrainian Mathematical Olympiad
The first person to solve this problem will receive the title of Master of Functional Equations.

Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$$f(f(x)+f(y))=xyf(x+y)$$for any $x, y \in \mathbb R^+$. Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$.

Background
Acknowledgements
13 replies
Bandera
Aug 14, 2018
RustyFox
Sep 6, 2019
Master of Functional Equations
G H J
G H BBookmark kLocked kLocked NReply
Source: Own, based on a problem from Ukrainian Mathematical Olympiad
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Bandera
470 posts
#1 • 7 Y
Y by khan.academy, mathlomaniac, Kayak, integrated_JRC, don2001, centslordm, Adventure10
The first person to solve this problem will receive the title of Master of Functional Equations.

Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$$f(f(x)+f(y))=xyf(x+y)$$for any $x, y \in \mathbb R^+$. Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$.

Background
Acknowledgements
This post has been edited 1 time. Last edited by Bandera, Aug 15, 2018, 4:53 PM
Reason: Acknowledgements added
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Eray
381 posts
#2 • 12 Y
Y by Happy2020, Supercali, AmSm_9, Smita, Kayak, integrated_JRC, tenplusten, RudraRockstar, centslordm, physics-ka-ashiq, Adventure10, Mango247
Who else wants to bet for pco?
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AmSm_9
60 posts
#3 • 2 Y
Y by centslordm, Adventure10
I guessed a solution
This post has been edited 1 time. Last edited by AmSm_9, Aug 16, 2018, 1:20 AM
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Smita
514 posts
#4 • 3 Y
Y by centslordm, Adventure10, Mango247
Eray wrote:
Who else wants to bet for pco?

I would also bet for pco
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TLP.39
778 posts
#5 • 21 Y
Y by lminsl, MNJ2357, AlastorMoody, Kayak, toshihiro shimizu, NahTan123xyz, Bandera, don2001, AmSm_9, tenplusten, enthusiast101, Supercali, Aritra12, centslordm, physics-ka-ashiq, L567, Pluto1708, MathLuis, Adventure10, Mango247, Sedro
Bandera wrote:
Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$$f(f(x)+f(y))=xyf(x+y)$$for any $x, y \in \mathbb R^+$. Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$.
The only answer is $\boxed{f(x)\equiv \frac{1}{x}}.$ We will prove this through the following claims.

Let $P(x,y)$ be the assertion.

Claim 1 : $x^2f(x)$ is unbounded above.
Proof : Assume the contrary, let $a$ be the upper bound.
For any $k>\sqrt{\frac{2a}{f(2)}},$ we have $f(k)<\frac{f(2)}{2}.$
By $P(1-x,1+x),$ we can see that $(0,f(2)]\in f(\mathbb{R}).$
Thus,there exists $l$ such that $f(l)+f(k)=f(2),$ the assumed condition then imply that $l<\sqrt{\frac{2a}{f(2)}}.$
Finally, $P(k,l)\implies f(f(2))=klf(k+l)\le \frac{a}{\frac{k}{l}+2+\frac{l}{k}},$ thus by letting $k\to\infty,$ we can make the ratio $\frac{k}{l}$ as large as possible, implying that $f(f(2))\le 0,$ a contradiction. $\square$
From Claim 1, we get the following corollary :

Corollary 2 : $f(x)$ is surjective.
Proof : For any positive number $r,$ let $s$ be a positive number satisfying $s^2f(s)\ge 4r.$ Now, let $t=\sqrt{\frac{s^2}{4}-\frac{r}{f(s)}},$ then
$$P(\frac{s}{2}-t,\frac{s}{2}+t)\, ;\, f(f(\frac{s}{2}-t)+f(\frac{s}{2}+t))=r$$Hence the desired result.
Claim 3 : If $a,b,c,d$ satisfy $a<b,c<d,f(a)=f(b),$ and $f(c)=f(d),$ then $a=c$ amd $b=d.$
Proof : Assume the contrary, WLOG $a=\min\{a,b,c,d\}.$
If $a\neq c,$ then
$P(c-a,a),P(c-a,b),P(d-a,a),P(d-a,b)\implies f(c+b-a)=f(d+b-a)$
$P(b,d),P(b,c)\implies df(b+d)=cf(b+c)$
$P(a,c+b-a),P(a,d+b-a)\implies (c+b-a)f(b+c)=(d+b-a)f(b+d)\implies c(d+b-a)=d(c+b-a)\implies (c-d)(b-a)=0.$ Clearly a contradiction. Thus $a=c$ and so $b\neq d.$
Finally, by looking at the quadruple $(a',b',c',d')=(a,b,\min\{b,d\},\max\{b,d\})$ instead of $(a,b,c,d),$ the same method applies that $a=\min\{b,d\}$, still a contradiction. $\square$

Claim 3 implies that there are only one ordered pair of positive real numbers $(a,b)$ such that $a<b$ and $f(a)=f(b).$
Claim 4 : $f(x)$ is injective.
Proof : Assume the contrary, then there exists a pair of positive real numbers $(a,b)$ such that $a<b$ and $f(a)=f(b).$
$P(x,a),P(x,b)\implies af(x+a)=bf(x+b)\implies f(x+(b-a))=\frac{a}{b}f(x)\,\forall\, x>a$
Define $t=b-a>0$ and $c=\frac{a}{b}<1,$ then let $l$ be a positive real number that satisfy the system of inequalities :
$$\begin{cases}l>a \\ \frac{l^2}{4}>\frac{c^2(l+2t)^2}{4}>al\\ \frac{(l+t)^2}{4}>\frac{c(l+2t)^2}{4}>a(l+t)\end{cases}$$then there exists $x_1,x_2,x_3,x_4>a$ such that
$$\begin{cases}x_1+x_2=l \\ x_3+x_4=l+t\\ x_1x_2=\frac{c^2(l+2t)^2}{4}\\ x_3x_4=\frac{c(l+2t)^2}{4}\end{cases}$$Thus,we have $x_1x_2f(x_1+x_2)=x_3x_4f(x_3+x_4)=h^2f(2h)$ where $h=\frac{l+2t}{2}.$
Hence, by $P(x_1,x_2),P(x_3,x_4),P(h,h)$ and Claim 3, there exists different multisets $\{y_1,y_2\},\{z_1,z_2\}$ such that $y_1,y_2,z_1,z_2>a$ and $f(y_1)+f(y_2)=f(z_1)+f(z_2).$
Finally, we can see that $f(y_1+t)+f(y_2+t)=f(z_1+t)+f(z_2+t)$ and $f(y_1+2t)+f(y_2+2t)=f(z_1+2t)+f(z_2+2t).$ Thus,
$$\begin{cases}y_1y_2f(y_1+y_2)=z_1z_2f(z_1+z_2) \\ (y_1+t)(y_2+t)f(y_1+y_2)=(z_1+t)(z_2+t)f(z_1+z_2)\\ (y_1+2t)(y_2+2t)f(y_1+y_2)=(z_1+2t)(z_2+2t)f(z_1+z_2) \end{cases}$$Solving the system, we got that $y_1+y_2=z_1+z_2,y_1y_2=z_1z_2\implies \{y_1,y_2\}=\{z_1,z_2\}$, a contradiction. $\square$
Claim 5 : $x^2f(x)$ is strictly increasing.
Proof : Assume that there exists $a>b$ such that $a^2f(a)\le b^2f(b),$ then $af(a)<bf(b)$ and $f(a)<f(b).$
Let $p=\frac{bf(b)-af(a)}{f(b)-f(a)},$ then $p<b$ and thus
$P(p,a-p),P(p,b-p)\implies f(f(p)+f(a-p))=f(f(p)+f(b-p))\implies a=b,$ a contradiction. $\square$
Corollary 6 : $f(x)$ is strictly monotone.
Proof : Since $f(x)$ is bijective, it's suffice to prove that $f(x)$ is continuous.
Since $x^2f(x)=4f(2f(\frac{x}{2}))$ is surjective, and also strictly increasing from Claim 5, $x^2f(x)$ must be continuous, and thus $f(x)$ must be continuous as desired.
Claim 7 : $f(x)\equiv \frac{1}{x}$
Proof : We will split into $2$ cases :

Case 1 : $f(x)$ is strictly increasing.
In this case, we have
$$\forall y\in (0,1), \, f(f(1))<f(f(1)+f(y))=yf(1+y)<yf(2)$$This clearly implies that $f(f(1))\le 0,$ a contradiction. Hence no answer in this case.

Case 2 : $f(x)$ is strictly decreasing.
In this case, we have
$P(1,1)\implies f(1)=1$
Now, for any pair of positive numbers $a<b$,we have
$P(a-k,k),P(b-k,k)\implies (a-k)f(a)<(b-k)f(b)\,\forall\, k\in (0,a)\implies af(a)\le bf(b)$
Hence $f(x)\ge\frac{1}{x}$ for all $x\ge 1$ and $f(x)\le\frac{1}{x}$ for all $x\le 1.$
For any $t>1,$ choose $T>1$ such that $\frac{1}{t}+\frac{1}{T}<1,$ then
$P(t,T)\implies \frac{tT}{t+T}\ge f(\frac{1}{t}+\frac{1}{T})\ge f(f(t)+f(T))=tTf(t+T)\ge \frac{tT}{t+T}$
Thus, $f(\frac{1}{t}+\frac{1}{T})= f(f(t)+f(T))\implies \frac{1}{t}+\frac{1}{T}= f(t)+f(T)\implies f(t)=\frac{1}{t}.$
On the other hand, for any $t<1,$ choose $T<1$ such that $t+T<1,$ then
$P(t,T)\implies \frac{tT}{t+T}\le f(\frac{1}{t}+\frac{1}{T})\le f(f(t)+f(T))=tTf(t+T)\le \frac{tT}{t+T}$
Thus, $f(\frac{1}{t}+\frac{1}{T})= f(f(t)+f(T))\implies \frac{1}{t}+\frac{1}{T}= f(t)+f(T)\implies f(t)=\frac{1}{t}.$
Hence $f(x)\equiv \frac{1}{x}$ as desired. $\square$
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Bandera
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#6 • 3 Y
Y by centslordm, Adventure10, Mango247
TLP.39 wrote:
Define $t=b-a>0$ and $c=\frac{a}{b}<1,$ then let $l$ be a positive real number that satisfy the system of inequalities :
$$\begin{cases}l>a \\ \frac{l^2}{4}>\frac{c^2(l+2t)^2}{4}>al\\ \frac{(l+t)^2}{4}>\frac{c(l+2t)^2}{4}>a(l+t)\end{cases}$$
Why does the system have any solution?
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TLP.39
778 posts
#7 • 2 Y
Y by centslordm, Adventure10
It suffices to prove that for each inequality of this system:
$$\begin{cases}l>a \\ \frac{l^2}{4}>\frac{c^2(l+2t)^2}{4}\\ \frac{c^2(l+2t)^2}{4}>al\\ \frac{(l+t)^2}{4}>\frac{c(l+2t)^2}{4}\\ \frac{c(l+2t)^2}{4}>a(l+t)\end{cases}$$There exists $N>0$ such that for all $l>N,$ $l$ satisfy the equation.
This is clear for the first, third, and fifth as L.H.S. has higher degree than R.H.S., and has positive leading coefficient when consider as polynomials on $l$.
For the second and fourth, just see that both sides have equal degree but L.H.S. has bigger leading coefficient than R.H.S. when consider as polynomials on $l$ and we are done.
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Bandera
470 posts
#8 • 3 Y
Y by centslordm, Adventure10, Mango247
@TLP.39 Congratulations! You have won the title:
$$\boxed{\mathfrak f(x+y) = \mathfrak f(x) +\mathfrak f(y) \; \maltese \text{ Master}}$$
You may want to post another problem (here or in a new topic) for finding more FE Masters. It should be:
  1. Very hard;
  2. Solvable;
  3. New, i.e. not having published solutions (here or anywhere else on the Internet);
  4. Concerning functional equations.
This post has been edited 1 time. Last edited by Bandera, Aug 19, 2018, 10:51 AM
Reason: Uniqueness requirement added
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Bandera
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#9 • 4 Y
Y by MNJ2357, centslordm, Adventure10, Mango247
Here's my solution.
Answer The only solution is $f(x) \equiv \frac 1 x.$

Plan of proof (all variables below are assumed to be from $\mathbb R^+$; moreover, $x$ and $y$ are considered free)
0. $f(f(x)+f(y))=xyf(x+y)$
1. $y \le \tfrac {x^2} 4 f(x) \implies \exists z \colon f(z)=y$
---A0. $x^2f(x)$ is bounded, $\exists C \; \forall x \colon x^2f(x) \le C$
---A1. $\exists a \; \forall x \le a \; \exists z \colon f(z)=x$
---A2. $f$ is bounded, $\exists B \; \forall x \colon f(x) \le B$
---A3. $x \le a \implies \exists z \le 2B \colon f(z)=x$
---A4. $\exists \varepsilon_n \colon \lim_{n \to \infty} \varepsilon_n=0, \lim_{n \to \infty} f(\varepsilon_n)=0$
---A5. $\pmb \bot$ (a contradiction)
2. $x^2f(x)$ is unbounded, $\nexists C \; \forall x \colon x^2f(x) \le C$
3. $f$ is surjective, $\forall x \; \exists z \colon f(z)=x$
4. $x^2f(x)$ is surjective, $\forall x \; \exists z \colon z^2f(z)=x$
---B0. $f$ isn't injective, $\exists \, v<w \colon f(v)=f(w)$
---B1. $(f(x_1)=f(x_2)) \land (y_1\!-x_1\!=y_2\!-x_2\!>0) \implies \tfrac {f(y_1)}{f(y_2)}=\tfrac {x_2}{x_1}$
---B2. $(f(x_1\!+y_1)=f(x_2\!+y_1)) \land (f(x_1\!+y_2)=f(x_2\!+y_2)) \implies (x_1\!=x_2) \lor (y_1\!=y_2)$
---B3. $(\exists T \; \forall t>T \colon x_1f(x_1\!+t)=x_2f(x_2\!+t)) \implies f(x_1)=f(x_2)$
---B4. $\pmb \bot$
5. $f$ is injective, $f(x)=f(y) \implies x=y$
6. $xf(x)$ is non-decreasing, $x<y \implies xf(x) \le yf(y)$
7. $f$ is continuous, $\lim_{n \to \infty} x_n = x \implies \lim_{n \to \infty} f(x_n) = f(x)$
8. $\lim_{x \to +\infty} f(x) = 0$
9. $f(1)=1$
10. $f(f(x))=x$
11. $f(x)=\tfrac 1 x$

Details notation
0.> The original equation.
1.> Let $t = \tfrac y {f(x)} \in (0, \tfrac {x^2} 4]$. Then $\exists x_1>0, x_2>0 \colon x_1x_2=t, x_1+x_2=x$ (these are solutions of a quadratic equation: $x_{1,2}=\tfrac {x \pm \sqrt {x^2-4t}} 2$ ). 0$[x \gets x_1, y \gets x_2]$ : $f(f(x_1)+f(x_2))=tf(x)=y$ and we can take $z=f(x_1)+f(x_2)$.
A0.> Branching condition (assumption).
A1.> 1$[x \gets 1, y \gets x]$ : $x \le \tfrac {f(1)} 4 \implies \exists z \colon f(z)=x$, so we can take $a = \tfrac {f(1)} 4$.
A2.> If $x \le 2a$, then A1$[x \gets \tfrac x 2]$ : $\exists z \colon f(z)=\tfrac x 2$. 0$[x \gets z, y \gets z]$ : $f(2f(z))=z^2f(2z)$. A0$[x \gets 2z]$ : $4z^2f(2z) \le C$. Thus, $x \le 2a \implies f(x)=f(2f(z))=z^2f(2z) \le \tfrac C 4$. On the other hand, if $x > 2a$, then A0 gives $f(x) \le \tfrac C {x^2} < \tfrac C {4a^2}$. Hence, we can take $B = \max \{\tfrac C 4, \tfrac C {4a^2}\}$.
A3.> The same as for 1 and A1, just note that due to A2 a value of $z$ at the end of 1's proof does not exceed $2B$.
A4.> Let $x_n=n, y_n=n^2, \varepsilon_n=f(x_n)+f(y_n)$. A0$[x \gets x_n]$ : $f(x_n) \le \tfrac C {n^2}$, thus $\lim_{n \to \infty} f(x_n)=0$. Similarly, $\lim_{n \to \infty} f(y_n)=0$. Hence, $\lim_{n \to \infty} \varepsilon_n=0$. 0$[x \gets x_n, y \gets y_n]$ : $f(\varepsilon_n)=x_ny_nf(x_n+y_n)=n^3f(n+n^2)$. A0$[x \gets n+n^2]$ : $f(n+n^2) \le \tfrac C {(n+n^2)^2} < \tfrac C {n^4}$. Thus, $f(\varepsilon_n) < \tfrac {Cn^3}{n^4}$ and $\lim_{n \to \infty} f(\varepsilon_n)=0$.
A5.> Take any $d \le a$. Now use $\varepsilon_n$ from A4; let $x_n=d-f(\varepsilon_n)$, for all $n$ such that $f(\varepsilon_n)<d$ we get: $x_n \in (0,a)$, A3$[x \gets x_n]$ : $\exists z_n \le 2B \colon f(z_n)=x_n$, so $d=f(z_n)+f(\varepsilon_n)$. 0$[x \gets z_n, y \gets \varepsilon_n]$ : $f(f(z_n)+f(\varepsilon_n))=z_n \varepsilon_n f(z_n+\varepsilon_n)$. A2$[x \gets z_n+\varepsilon_n]$ : $f(z_n+\varepsilon_n) \le B$, thus $f(d)=z_n \varepsilon_n f(z_n+\varepsilon_n) \le 2B \cdot \varepsilon_n \cdot B$, a contradiction since $f(d)>0$ but $\varepsilon_n \to 0$ when $n \to \infty$.
2.> A5, dead branch defined by A0 pruned, this is the rest.
3.> From 2 it follows that $\exists x' \colon 4x \le x'^2f(x')$. 1$[x \gets x', y \gets x]$ : $\exists z \colon f(z)=x$.
4.> 0$[x \gets \tfrac z 2, y \gets \tfrac z 2]$ : $4f(2f(\tfrac z 2))=z^2f(z)$. When $z$ runs through all $\mathbb R^+$, so does $\tfrac z 2$, from 3 it follows that so does $f(\tfrac z 2)$, then $2f(\tfrac z 2)$, $f(2f(\tfrac z 2))$, and finally, $4f(2f(\tfrac z 2))=z^2f(z)$.
B0.> Branching condition (assumption).
B1.> 0$[x \gets x_1, y \gets y_1\!-x_1]$ : $f(f(x_1)+f(y_1\!-x_1))=x_1(y_1\!-x_1)f(y_1)$. 0$[x \gets x_2, y \gets y_2\!-x_2]$ : $f(f(x_2)+f(y_2\!-x_2))=x_2(y_2\!-x_2)f(y_2)$. The left sides are equal, thus $x_1f(y_1)=x_2f(y_2)$.
B2.> Assume that $y_1 \!\ne y_2$. WLOG, $y_2 \!> y_1$. B1$[x_1 \gets x_1\!+y_1, x_2 \gets x_2\!+y_1, y_1 \gets x_1\!+y_2, y_2 \gets x_2\!+y_2]$ : $\tfrac {f(x_1+y_2)}{f(x_2+y_2)}=\tfrac {x_2+y_1}{x_1+y_1}$. Since the left side is equal to $1$, the right side must be $1$ too, thus $x_1\!=x_2$.
B3.> Let $t>T$. 0$[x \gets x_1, y \gets t]$ : $f(f(x_1)+f(t))=x_1tf(x_1+t)$, 0$[x \gets x_2, y \gets t]$ : $f(f(x_2)+f(t))=x_2tf(x_2+t)$. Thus, $\forall t>T \colon f(f(x_1)+f(t))=f(f(x_2)+f(t))$. Note that $f(t)$ cannot be constant for all $t>T$, because in such case, for example, $f(T+1)=f(T+2)=f(T+3)$ and B2$[x_1 \gets T, x_2 \gets T+1, y_1 \gets 1, y_2 \gets 2]$ gives a contradiction, so we can choose $t_1>T,\,t_2>T$ such that $t_1 \ne t_2$ and $f(t_1) \ne f(t_2)$. B2$[x_1 \gets f(x_1), x_2 \gets f(x_2), y_1 \gets f(t_1), y_2 \gets f(t_2)]$ : $f(x_1)=f(x_2)$ because $f(t_1) \ne f(t_2)$.
B4.> Let $d=w-v>0, r=\tfrac w v>1, \; u=\tfrac {2v^2}{v+w}, z=\tfrac {2w^2}{v+w}$. Clearly, $z-u=2d$, $\tfrac z u = r^2$ and $0<u<v<w<z$. Now take any $y_1>v$ and let $y_2=y_1+d$. Then $y_2-w=y_1-v>0$. B1$[x_1 \gets v, x_2 \gets w]$ : $\tfrac {f(y_1)}{f(y_2)}=r$. Let $y_3=y_1+z-u=y_2+d$. Then $y_3-w=y_2-v>0$. B1$[x_1 \gets v, x_2 \gets w, y_1 \gets y_2, y_2 \gets y_3]$ : $\tfrac {f(y_2)}{f(y_3)}=r$. Multiplying, we get $\tfrac {f(y_1)}{f(y_3)}=r^2=\tfrac z u$. Denoting $t=y_1-u$, we can rewrite this as $uf(u+t)=zf(z+t)$, that is true for any $t>v-u$. B3$[T \gets v-u, x_1 \gets u, x_2 \gets z]$ : $f(u)=f(z)$. Now let $s=w+d=v+2d=v+z-u$. Then $s-z=v-u>0$. B1$[x_1 \gets u, x_2 \gets z, y_1 \gets v, y_2 \gets s]$ : $\tfrac {f(v)}{f(s)}=\tfrac z u = r^2$. On the other hand, B1$[x_1 \gets v, x_2 \gets w, y_1 \gets w, y_2 \gets s]$ : $\tfrac {f(w)}{f(s)}=\tfrac w v = r$. The left sides are equal due to $f(v)=f(w)$, but the right sides are not because $r>1$, a contradiction.
5.> B4, dead branch defined by B0 pruned, this is the rest.
6.> Let $x<y$. If $f(x) \le f(y)$, then clearly $xf(x) < yf(y)$. Assume that $f(x)>f(y)$ but $xf(x)>yf(y)$. Let $z=\tfrac {xf(x)-yf(y)}{f(x)-f(y)}$. Obviously, $0<z<x<y$. Also note that $(x-z)f(x)=(y-z)f(y)$. 0$[x \gets z, y \gets x-z]$ : $f(f(z)+f(x-z))=z(x-z)f(x)$, 0$[x \gets z, y \gets y-z]$ : $f(f(z)+f(y-z))=z(y-z)f(y)$. Hence, $f(f(z)+f(x-z))=f(f(z)+f(y-z))$. From this 5 deduces: $f(z)+f(x-z)=f(z)+f(y-z) \implies f(x-z)=f(y-z) \implies x-z=y-z \implies x=y$, a contradiction.
7.> From 6 it follows that the function $g(x) \equiv x^2f(x) \equiv x \cdot xf(x)$ is strictly increasing. 4 says that $g(x)$ is surjective. Hence, it is continuous at the whole domain and so is $f(x) \equiv \tfrac {g(x)}{x^2}$.
8.> From 5 and 7 it follows that $f$ is strictly monotonic. Moreover, from 3 it follows that only two cases are possible: $(\,\lim_{x \to 0^+} f(x) = 0\,) \land (\,\lim_{x \to +\infty} f(x) = +\infty\,)$ when $f$ is increasing and $(\,\lim_{x \to 0^+} f(x) = +\infty\,) \land (\,\lim_{x \to +\infty} f(x) = 0\,)$ when $f$ is decreasing. The first case quickly leads to a contradiction because fixing some $y$ in 0 and taking it to the limit when $x \to 0^+$ we get (employing the continuity): $f(f(y))=0 \cdot yf(y)=0$.
9.> 0$[x \gets 1, y \gets 1]$ : $f(2f(1))=f(2)$. 5$[x \gets 2f(1), y \gets 2]$ : $2f(1)=2$.
10.> Dividing the original equation by $x$ and taking it to the limit when $y \to +\infty$, we get (using 8 and 7): $\tfrac {f(f(x))} x = \lim_{y \to +\infty} yf(x+y)$. Denote the function at the left side by $h(x)$. The right side is in fact independent of $x$. Indeed, whatever $x$ is, we already know that the limit exists (it is equal to $h(x)$ ). Thus, $\lim_{y \to +\infty} yf(x+y)=$ $x \cdot 0 + \!\lim_{y \to +\infty} yf(x+y)=$ $x \!\lim_{y \to +\infty} f(x+y)+\!\lim_{y \to +\infty} yf(x+y)=$ $\lim_{y \to +\infty} xf(x+y)+\lim_{y \to +\infty} yf(x+y)=$ $\lim_{y \to +\infty} (x+y)f(x+y)=$ $\{z=x+y\}=$ $\lim_{z \to +\infty} zf(z),$ that is independent of $x$. Hence, $h(x)$ is a constant, $h(x) \equiv C$. From 9 if follows that $C=1$.
11.> 0$[y \gets f(x)]$ : $f(f(x)+f(f(x)))=xf(x)f(x+f(x))$. 10 reduces the left side to $f(f(x)+x)$. Thus, $xf(x)=1$.

Direct check shows that $f(x) \equiv \tfrac 1 x$ is indeed a solution.
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MNJ2357
644 posts
#10 • 3 Y
Y by khan.academy, centslordm, Adventure10
I'm reviving this thread since this is(or will be) one of the most difficult FE marathons. :)

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that
  • $f(xf(x+y)+yf(y-x))=f(x)^2+f(y^2)$ for all $x,y\in\mathbb{R}$
  • $f$ is continuous on $\mathbb{R}\backslash \{-1,1\}$. (Continuity at $x=1,-1$ is not provided.)

Prove that if $x\neq 1,-1$, then $f(x)=-f(-x)$.
This post has been edited 1 time. Last edited by MNJ2357, Mar 9, 2019, 4:08 PM
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MNJ2357
644 posts
#11 • 3 Y
Y by centslordm, Adventure10, Mango247
Bump!! :showoff:
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adityaguharoy
4655 posts
#12 • 3 Y
Y by centslordm, Adventure10, Mango247
Here is a problem I like a lot and at first if one doesn't know the solution it may seem difficult.
(Although pco sir solved it a while earlier -- salut)

Are there any functions $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $$f(xf(y) + yf(x)) = x^2 + y^2$$holds true $\forall (x,y) \in \mathbb{R}_{>0} ^2$.
(Here $\mathbb{R}_{>0}$ is the set of all positive real numbers only).


Source and remarks
This post has been edited 1 time. Last edited by adityaguharoy, Mar 23, 2019, 4:31 PM
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BlazingMuddy
281 posts
#13 • 4 Y
Y by MNJ2357, centslordm, Adventure10, Mango247
MNJ2357 wrote:
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that
$f(xf(x+y)+yf(y-x))=f(x)^2+f(y^2)$ for all $x,y\in\mathbb{R}$
$f$ is continuous on $\mathbb{R}\backslash \{-1,1\}$. (Continuity at $x=1,-1$ is not provided.)

Prove that if $x\neq 1,-1$, then $f(x)=-f(-x)$.

I'll demonstrate that we can actually find all such functions.

All functions

Solution
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RustyFox
491 posts
#14 • 2 Y
Y by centslordm, Adventure10
Here is not so trivial yet good FE, which may take time but try it until you get it.

Find all functions $f:\mathbb{W-{1}}\rightarrow\mathbb{W-{1}}$ such that $xf(x'y) - yx^{y}f(x) = yf(x"xy) + xyf(y'x)$, where operations $f(x'y)$ is defined as $f(f(f(f(.....y))))...$ $x$ times, i.e, $f(x'2) = f(f(x))$ and $f(x"y) = f(x)^{y}$.
This post has been edited 2 times. Last edited by RustyFox, Sep 6, 2019, 1:04 PM
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