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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Non-homogenous Inequality
Adywastaken   5
N 14 minutes ago by ehuseyinyigit
Source: NMTC 2024/7
$a, b, c\in \mathbb{R_{+}}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
5 replies
Adywastaken
3 hours ago
ehuseyinyigit
14 minutes ago
f(m + n) >= f(m) + f(f(n)) - 1
orl   30
N 43 minutes ago by ezpotd
Source: IMO Shortlist 2007, A2, AIMO 2008, TST 2, P1, Ukrainian TST 2008 Problem 8
Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

Author: Nikolai Nikolov, Bulgaria
30 replies
orl
Jul 13, 2008
ezpotd
43 minutes ago
Classic Diophantine
Adywastaken   3
N an hour ago by Adywastaken
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
3 replies
Adywastaken
3 hours ago
Adywastaken
an hour ago
Add d or Divide by a
MarkBcc168   25
N an hour ago by Entei
Source: ISL 2022 N3
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$. Let $x_1=1$, and for $k\geq 1$, define
$$x_{k+1} = \begin{cases}
x_k + d &\text{if } a \text{ does not divide } x_k \\
x_k/a & \text{if } a \text{ divides } x_k
\end{cases}$$Find, in terms of $a$ and $d$, the greatest positive integer $n$ for which there exists an index $k$ such that $x_k$ is divisible by $a^n$.
25 replies
MarkBcc168
Jul 9, 2023
Entei
an hour ago
Alice and Bob play, 8x8 table, white red black, minimum n for victory
parmenides51   14
N an hour ago by Ilikeminecraft
Source: JBMO Shortlist 2018 C3
The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses $4$ rows and $4$ columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can win the game no matter how Bob plays.
14 replies
parmenides51
Jul 22, 2019
Ilikeminecraft
an hour ago
GEOMETRY GEOMETRY GEOMETRY
Kagebaka   71
N an hour ago by bin_sherlo
Source: IMO 2021/3
Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.
71 replies
Kagebaka
Jul 20, 2021
bin_sherlo
an hour ago
Equation of integers
jgnr   3
N 2 hours ago by KTYC
Source: Indonesia Mathematics Olympiad 2005 Day 1 Problem 2
For an arbitrary positive integer $ n$, define $ p(n)$ as the product of the digits of $ n$ (in decimal). Find all positive integers $ n$ such that $ 11p(n)=n^2-2005$.
3 replies
jgnr
Jun 2, 2008
KTYC
2 hours ago
Divisibility..
Sadigly   4
N 2 hours ago by Solar Plexsus
Source: another version of azerbaijan nmo 2025
Just ignore this
4 replies
Sadigly
Yesterday at 7:37 AM
Solar Plexsus
2 hours ago
Surjective number theoretic functional equation
snap7822   3
N 2 hours ago by internationalnick123456
Source: 2025 Taiwan TST Round 3 Independent Study 2-N
Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
[list=i]
[*] For all $m, n \in \mathbb{N}$, if $m > n$ and $f(m) > f(n)$, then $f(m-n) = f(n)$;
[*] $f$ is surjective.
[/list]
Find the maximum possible value of $f(2025)$.

Proposed by snap7822
3 replies
snap7822
May 1, 2025
internationalnick123456
2 hours ago
FE with devisibility
fadhool   0
2 hours ago
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
0 replies
fadhool
2 hours ago
0 replies
Many Equal Sides
mathisreal   3
N 2 hours ago by QueenArwen
Source: Brazil EGMO TST 2023 #1
Let $ABC$ be a triangle with $BA=BC$ and $\angle ABC=90^{\circ}$. Let $D$ and $E$ be the midpoints of $CA$ and $BA$ respectively. The point $F$ is inside of $\triangle ABC$ such that $\triangle DEF$ is equilateral. Let $X=BF\cap AC$ and $Y=AF\cap DB$. Prove that $DX=YD$.
3 replies
mathisreal
Nov 10, 2022
QueenArwen
2 hours ago
LOTS of recurrence!
SatisfiedMagma   4
N 2 hours ago by Reacheddreams
Source: Indian Statistical Institute Entrance UGB 2023/5
There is a rectangular plot of size $1 \times n$. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size $1 \times 1$, the blue tiles are of size $1 \times 1$ and the black tiles are of size $1 \times 2$. Let $t_n$ denote the number of ways this can be done. For example, clearly $t_1 = 2$ because we can have either a red or a blue tile. Also $t_2 = 5$ since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.

[list=a]
[*]Prove that $t_{2n+1} = t_n(t_{n-1} + t_{n+1})$ for all $n > 1$.

[*]Prove that $t_n = \sum_{d \ge 0} \binom{n-d}{d}2^{n-2d}$ for all $n >0$.
[/list]
Here,
\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]for integers $m,r$.
4 replies
SatisfiedMagma
May 14, 2023
Reacheddreams
2 hours ago
combi/nt
blug   1
N 2 hours ago by blug
Prove that every positive integer $n$ can be written in the form
$$n=a_1+a_2+...+a_k,$$where $a_m=2^i3^j$ for some non-negative $i, j$ such that
$$a_x\nmid a_y$$for every $x, y\leq k$.
1 reply
blug
Yesterday at 3:37 PM
blug
2 hours ago
Inequality, inequality, inequality...
Assassino9931   9
N 2 hours ago by ZeroHero
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
9 replies
Assassino9931
Today at 9:38 AM
ZeroHero
2 hours ago
Functional equation
Nima Ahmadi Pour   100
N Yesterday at 5:11 PM by jasperE3
Source: ISl 2005, A2, Iran prepration exam
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]
for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria
100 replies
Nima Ahmadi Pour
Apr 24, 2006
jasperE3
Yesterday at 5:11 PM
Functional equation
G H J
Source: ISl 2005, A2, Iran prepration exam
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cursed_tangent1434
625 posts
#101
Y by
The answer is $f(x)=2$ for all $x\in \mathbb{R}_{>0}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We first prove the following.

Claim : $f(x)\geq 2$ for all $x\in \mathbb{R}_{>0}$.
Proof : First, assume there exists $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha) <1$. Then, plugging in $x=\alpha$ and $y= \frac{\alpha}{1-f(\alpha)}$ gives
\begin{align*}
        f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\alpha + \frac{\alpha}{1-f(\alpha)}\cdot f(\alpha)\right) \\
        f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\frac{\alpha}{1-f(\alpha)}\right)
    \end{align*}from which it follows that $f(\alpha)=2$ which is a very clear contradiction. Thus, it follows that $f(x) \geq 1$ for all $x\in \mathbb{R}_{>0}$. Now, note that we have
\[f(x)^2 = 2f(x+xf(x))\geq 2\]from which it follows that $f(x) \geq \sqrt{2}$ for all $x\in \mathbb{R}_{>0}$. Similarly, a straightforward induction gives,
\[f(x) \geq 2^{\sum{\frac{1}{2^i}}}=2\]for all $x\in \mathbb{R}_{>0}$, which proves the claim.

Now, we show that there exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$. First note that swapping $x$ and $y$ gives us that
\[f(x+yf(x))=f(y+xf(y))\]for all positive real numbers $x$ and $y$. Then, say $x+yf(x)=y+xf(y)$ for all $x,y \in\mathbb{R}_{>0}$. Then, plugging in $y=1$ gives $f(x)=cx+1$ for some constant $c$ for all positive reals $x$ which is clearly false for any $c\in \mathbb{R}_{>0}$ upon substitution. This means, there exists $x_0,y_0 \in \mathbb{R}_{>0}$ such that $x_0+y_0f(x_0) \neq y_0+x_0f(y_0)$. WLOG, assume that $x_0+y_0f(x_0)>y_0+x_0f(y_0)$. Then, let $b= y_0+x_0f(y_0)$ and $a=\frac{x_0+y_0f(x_0) - (y_0+x_0f(y_0))}{f(y_0+x_0f(y_0))}>0$.
Then, plugging in $x=b$ and $y=a$ yields,
\[f(b)f(a)=2(f(b+af(b)))=2f(x_0+y_0f(x_0)) = 2f(b)\]from which it follows that $f(a)=2$. Thus, there indeed exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$.
We now have our final key claim.

Claim : $f$ is non-decreasing.
Proof : Simply note that
\[2f(x+yf(x))=f(x)f(y)\geq 2f(x)\]since $f\geq 2$. Thus, $f(x) \leq f(x+yf(x))$ for all $x,y \in\mathbb{R}_{>0}$. Thus, it follows that $f(x)\leq f(x+\epsilon)$ for any $x, \epsilon \in \mathbb{R}_{>0}$ (by varying $y$) which implies that $f$ is non-decreasing as claimed.

Now, consider $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha)=2$. Plugging in $x=y=\alpha$ gives
\[4=f(\alpha)^2=2f(\alpha+\alpha f(\alpha))=2f(3\alpha)\]which implies that $f(3\alpha)=2$. Thus, it follows by a straightforward induction that $f(3^ra)=2$ for all $r\in \mathbb{N}$, from which it follows that there exists arbitrarily large values of $x\in\mathbb{R}_{>0}$ such that $f(x)=2$. Putting this together with the non-decreasing nature of $f$ it follows that $f$ is constant 2, as desired.
Z K Y
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asdf334
7585 posts
#102
Y by
Let $P(x,y)$ be the assertion.
Claim: $f(x)\ge 1$.
Proof: Suppose otherwise. If there exists $a$ where $f(a)<1$ then we can find
\[y=a+yf(a)\]hence $P(a,y)$ yields $f(a)=2$, a contradiction.
Claim: $f(x)\ge 2$.
Proof: Suppose otherwise. Let $S$ be the range of $f$. From $P(x,x)$ with $v=f(x)<2$ we find
\[v\in S\implies \frac{v^2}{2}\in S\]hence we can generate a sequence $v_0,\dots,v_n\dots$ and note that
\[v_{i+1}\le \frac{v_0}{2}\cdot v_i\]thus values become arbitrarily small. Hence at some point values are less than $1$, violating the first claim.
Claim: $f$ is nondecreasing.
Proof: Take any $a$ and $b$ with $a<b$. Notice we can always choose $x$ and $y$ such that
\[a=x\]\[b=x+yf(x).\]Then since $f(y)\ge 2$ we find $f(a)\le f(b)$.
Claim: Either $f$ is increasing or $f\equiv 2$.
Proof: Suppose there exist $a$ and $b$ such that $f(a)=f(b)$. Then $f$ is constant in the interval $[a,b]$. At this point take $P(a,y)$ such that
\[a+yf(a)\in (a,b].\]For any such $y$ we have $f(y)=2$. Hence there exists an interval where $f\equiv 2$. Take $x$ in this interval and $P(x,x)$ implies $f(3x)=2$. Hence we can obtain arbitrarily large values outputting $2$, and so $f\equiv 2$ everywhere.
Claim: $f$ increasing is not possible.
Proof: Swap $x$ and $y$. Then
\[f(x+yf(x))=f(y+xf(y))\implies x+yf(x)=y+xf(y)\implies \frac{f(x)-1}{x}=\frac{f(y)-1}{y}\]hence $f(x)=cx+1$ for some $c$. However taking $x\to 0$ is a contradiction as we would have $f(x)\to 1$ which is less than $2$.
To conclude, we have one solution (which clearly works): $\boxed{f\equiv 2}$.
Z K Y
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WLOGQED1729
47 posts
#103
Y by
Very instructive problem! This reviews many useful techniques in FE.

The answer is $f(x)=2~~\forall x \in \mathbb{R}^+$ which is clearly satisfy the problem condition.
We’ll show that this is the only solution.
Denote $P(x,y)$ by the condition $f(x)f(y)=2f(x+yf(x))~~ \forall x,y \in \mathbb{R}^+$
Claim 1: $f(x)\geq 1~~ \forall x \in \mathbb{R}^+$
Proof: FTSOC, suppose there exist $t \in \mathbb{R}^+$ s.t. $f(t)<1$
Consider $P(t,\frac{t}{1-f(t)})$, it follows that $f(t)=2$ which leads to contradiction. $\square$

Claim 2: $f(x)\geq 2~~ \forall x \in \mathbb{R}^{+}$
Proof: Define a sequence ${a_n}$ s.t. $a_1=1$ and $a_{n+1}=\sqrt{2a_{n}}~~ \forall n \in \mathbb{N}$
It is easy to see that $lim_{n\to \infty}a_n=2$ and $a_n<2~~ \forall n \in \mathbb{N}~~(\clubsuit)$
We prove that $f(x)\geq a_n ~~\forall x \in \mathbb{R}^+ , n \in \mathbb{N}$ by induction on $n$.
The base case is proven in Claim 1
Suppose that $f(x)\geq a_m~~ \forall x \in \mathbb{R}^+$
Consider $P(x,x)$, $f(x)^2=2f(x+xf(x))\geq 2a_m~~ \forall x \in \mathbb{R}^+$
$$\implies f(x)\geq \sqrt{2a_m}=a_{m+1}~~\forall x \in \mathbb{R}^+ $$So, $f(x)\geq a_n ~~\forall x \in \mathbb{R}^+ , n \in \mathbb{N}$
Combine this with $(\clubsuit)$, $f(x)\geq 2~~ \forall x \in \mathbb{R}^{+}$, as desired. $\square$


Claim 3:$f$ is non decreasing
Proof: Let $a<b$ be any two distinct positive reals
Consider $P(a,\frac{b-a}{f(a)})$, $f(a)f(\frac{b-a}{f(a)})=2f(b)$
Since $f(x)\geq 2$ for all $x$, it follows that $f(a)\leq f(b)~~\square$

Now we divide into 2 cases
Case 1) There exist $c \in \mathbb{R}^+$ such that $f(c)=2$
Note that if we have $f(x)=2$, it implies that $f(x+xf(x))=2$ (This followed from $P(x,x)$)
We can easily prove that $f(3^nc)=2 \forall n \in \mathbb{N} $ by induction on $n$
Combine this result with Claim 2 and Claim 3, we conclude that $f(x)=2~~\forall x \in \mathbb{R}^+$

Case 2) $f(x)>2~~ \forall x \in \mathbb{R}^+$
Claim: $f$ is strictly increasing
Proof: We can easily copied the proof in Claim 3. $\square$
This implies that $f$ is injective.

The trick is to swap the variable in the problem assertion.
Consider $P(x,y)$ and $P(y,x)$, $f(x+yf(x))=f(y+xf(y)) \forall x,y \in \mathbb{R}^+$
By injectivity, $x+yf(x)=y+xf(y) \forall x,y \in \mathbb{R}^+$
$\implies \frac{f(x)-1}{x}$ is a constant $\forall x \in \mathbb{R}^+$
So, $f(x)=cx+1 \forall x \in \mathbb{R}^+$ for some constant $c \in \mathbb{R}^+$ which can be easily checked that it fails. $\blacksquare$
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joshualiu315
2534 posts
#104
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The answer is $f(x) = \boxed{2}$, which works. Denote the given assertion as $P(x,y)$. We will solve this problem through a series of claims:


Claim 1: $f(x) \ge 1$ for all $x \in \mathbb{R}^+$

Proof: FTSOC suppose that $x$ satisfies $f(x)<1$. Plugging in $P(x, \tfrac{x}{1-f(x)})$ yields

\[f(x)f\left(\frac{x}{1-f(x)}\right) = 2f \left(\frac{x}{1-f(x)}\right) \implies f(x) = 2,\]
a contradiction. $\square$


Claim 2: $f(x) \ge 2$ for all $x \in \mathbb{R}^+$

Proof: $P(x,x)$ yields

\[f(x)^2 = 2f(x+f(x)) \ge 2.\]
Hence, $f(x) \ge \sqrt{2}$. In fact, we can show that

\[f(x) \ge \lim_{k \to \infty} 2^{\tfrac{2^k-1}{2^k}} = 2\]
using an inductive process on $P(x,x)$, as desired. $\square$


Claim 3: $f$ is nondecreasing

Proof: Note that

\[2f(x+yf(x)) = f(x)f(y) \ge 2f(x),\]
which means $f(x) \le f(x+yf(x))$. Take $y = \varepsilon$ to be an arbitrarily small number, so that $f(x) \le f(x+\varepsilon)$ and $x < x+ \varepsilon$. $\square$


Claim 4: If there exists any value $x \in \mathbb{R}^+$ such that $f(x) = 2$, then $f \equiv 2$.

Proof: If $f(x) = 2$, then

\[f(3x) = f(x+xf(x)) = \frac{f(x)^2}{2} = 2.\]
Thus, the set of $x$ such that $f(x)=2$ is unbounded, and Claim 3 finishes. $\square$


Claim 5: $f$ is not injective.

Proof: Assume FTSOC that $f$ is injective. Then, comparing $P(x,y)$ and $P(y,x)$ gives

\[f(x+yf(x)) = f(y+xf(y)) \implies x+yf(x) = y+xf(y)\]\[\implies x(1-f(y)) = y(1-f(x)) = k.\]
Rearraging yields $f(y) = 1-\tfrac{k}{x}$, which implies $f(k) = 0$, contradiction. $\square$


Claim 6: $f>2$ cannot always hold.

Proof: Suppose that $x$ and $y$ satisfy $x+yf(x) < y+xf(y)$. Such values must exist due to Claim 5 preventing $x+yf(x) = y+xf(y)$. Note that all values in the interval $[x+yf(x), y+xf(y)]$ are constant.

The trick is to raise $y \to y'$ such that $x+y'f(x) = y+xf(y) \le y'+xf(y')$. Hence, we can expand the constant interval to $[x+yf(x), y'+xf(y')]$. It suffices to show that increasing $y$ infinitely many times causes the interval to be unbounded.

Let $\ell$ be the size of the current interval. Each time, we increase $y$ by $\tfrac{\ell}{f(x)}$, which increases the size of the next interval by at least $\tfrac{\ell}{f(x)}$. Hence, the new interval at least has size

\[\ell\left(1+\frac{1}{f(x)} \right),\]
which is evidently unbounded. $\square$


Claims $4$ and $6$ imply that $f$ is constant, which gives our solution set.
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ezpotd
1268 posts
#105
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The answer is the constant $2$ only, it is easy to check that this works. Let $P(x,y)$ denote the assertion.

First, observe is $f(x)$ is below $1$ at any point, we can find $y$ such that $x + yf(x) = y$, plugging in such $y$ forces $f(x) = 2$, thus $f$ can never be below $1$ (since we already said $f(x)$ was below $1$). If $f(x)$ is below $2$ at any point, we can always see there exists a value of $k$ such that $f(k) = \frac 12 f(x)^2$, which eventually goes below $1$ (it is bounded by a geometric series with common ratio less than $1$), so this returns to the case where $f(x)$ is below $1$ at some point. So we know $f \ge 2$.

Claim: $f$ is $2$ at some point.
If $f$ is always above $2$, we see $f(x) \frac{f(y)}{2} =f(x +yf(x))$, so for all $a > x$, setting $y = \frac{a - x}{f(x)}$ forces $f(a) > f(x)$. Now take $2f(x + yf(x)) = f(x)f(y) = f(y)f(x) = 2f(y +xf(y))$, since $f$ is increasing we know $f$ is injective, this gives $x + yf(x) = xf(y) + y$, for sufficiently small $y$ the left side is below $2x$ and the right side is greater, which is a contradiction, so this is impossible.

Claim: $f$ is periodic.
Proof: Take $c$ with $f(c) = 2$, then we get $P(x,c)$ forces $f(x) = f(x + 2c)$.

Claim: $f(x) = 2$ for all $x$.
Proof: We first show $f(x) = 1$ or $f(x) = 2$ for all $x$. Assume $f(x) \neq 1$, we show $f(x) = 2$. By varying $y$, we can find some value of $y$ such that there exists an integer $k$ with $x + yf(x) - y = 2kc$, this forces $f(y) = f(x + f(y)) $, so $P(x,y)$ gives $f(x) = 2$. Now take some value with $f(x) = 1$, then $P(x,x)$ gives $f(x + xf(x)) = \frac 12$, which is impossible, so such a value cannot exist, giving the desired conclusion.
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cosdealfa
27 posts
#106 • 2 Y
Y by pb_ana, Mathandski
I hope this is right :blush:

Denote by $P(x, y)$ the given assertion.
Claim 1: $f$ is monotonically increasing.
Proof: take two numbers $a < b$ and assume FTSOC $f(a) > f(b)$.
Let $r$ be the solution of the equation $a + rf(a) = b+ rf(b)$(it exists because of the way we ordered these two numbers).

$P(a, r): f(a)f(r)=2f(a + rf(a))$
$P(b, r): f(b)f(r)=2f(b + rf(b))$
So $f(a)=f(b)$, contradiction. $\square$

Now we will prove that $f$ is not injective( i.e strictly increasing)
Suppose FTSOC that $f$ is injective.
Swapping $x$ and $y$ in the original equation gives:
$x + yf(x) = y + xf(y), \forall x > 0$
$ \Rightarrow \frac{f(x)-1}{x} = k, \forall x > 0$
Plugging this in the original equation, we get a contradiction. $\square$

Since $f$ is not injective, we can find $a < b$ such that $f(a) = f(b) = c$. By monotonicity, we get that $\forall x \in [a, b], f(x) = c$.
$P(a, \frac{b-a}{c}): f(a)f(\frac{b-a}{c})=2f(b)$ so we can find $z \in \mathbb{R}$ such that $f(z)=2$
$P(z, x): f(x) = f(2x + 2), \forall x \in \mathbb{R}$
Set $x=b$ $\Rightarrow f(b) = f(2b+2) =c$. Using monotonicity again, we get that $\forall x \in [b,2b+2], f(x) = c$ Similarily, $\forall x > b, f(x) =c$
Finally, $P(a, x): cf(x) = 2(a + xc)$. Since the second argument $>a$ we get that $f(x) = 2 \forall x \in \mathbb{R} $ which is clearly the only solution $\blacksquare$
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SomeonesPenguin
128 posts
#107
Y by
storage
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OronSH
1745 posts
#108
Y by
We claim $f\equiv 2$ is the only solution, which clearly works.

First, if $f(x)<1$ then $P(x,\tfrac x{1-f(x)})$ gives $f(x)=2$. Thus $f(x)\ge 1$.

Next, we prove by induction that $f(x)\ge 2^{1-2^{-n}}$. From the above, the base case $n=0$ works, and for the inductive step $P(x,x)$ gives $f(x)^2=2f(x+xf(x))\ge 2^{2-2^{-n}}$ implying $f(x)\ge 2^{1-2^{-(n+1)}}$, completing the induction. Taking $n$ arbitrarily large implies $f(x)\ge 2$ for all $x$.

Next we show $f$ is constant or strictly increasing. From $f(y)\ge 2$ we get $f(x)\le f(x+yf(x))$. If $f$ is not strictly increasing we can choose some $x$ and some $\varepsilon$ for which $f(x)=f(x+kf(x))$ for $0<k\le\varepsilon$, and by equality case $f(k)=2$ for $0<k\le\varepsilon$. Now $P(x+\varepsilon f(x),k)$ implies $f(x)=f(x+(k+\varepsilon)f(x))$, so by equality case again $f(k)=2$ for $0<k\le 2\varepsilon$. Repeating this, $f(k)=2$ for all $k>0$.

Now if $f$ is strictly increasing, it is injective. Taking $P(x,1)$ and $P(1,x)$ implies $x+f(x)=1+xf(1)$, which fails for $x$ small enough since $f(x)\ge 2$.

Thus $f\equiv 2$ is the only possible solution.
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Mathandski
757 posts
#109 • 1 Y
Y by OronSH
Subjective Rating (MOHs)
Please contact westskigamer@gmail.com if there is an error with any of my solution for cash bounties by 3/18/2025.

Solution
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cj13609517288
1916 posts
#110 • 1 Y
Y by OronSH
The answer is $\boxed{f(x)=2}$ only, which clearly works.

Note that
\begin{align*}
f(x)f(y)f(z) &= f(y)f(z)f(x) \\
2f(x+yf(x))f(z) &= 2f(y+zf(y))f(x) \\
4f(x+yf(x)+zf(x+yf(x))) &= 4f(x+(y+zf(y))f(x)) \\
4f(x+yf(x)+\frac12\cdot zf(x)f(y)) &= 4f(x+yf(x)+zf(x)f(y)).
\end{align*}We want to prove that $f(a+\frac12 b)=f(a+b)$. To do so, choose some $x<a$, then choose $y$ such that $x+yf(x)=a$, then choose $z$ such that $zf(x)f(y)=b$.

So now we claim that $f$ is constant. But note that
\[f(x)=f(1.5x)=f(1.5^2 x)=f(1.5^3 x)=\dots=f(\alpha\cdot 1.5^n x)\]where $1\le\alpha<1.5$. This is enough to show that $f(x)=f(y)$ when $x<y$, so we are done. $\blacksquare$
This post has been edited 1 time. Last edited by cj13609517288, Mar 4, 2025, 4:03 PM
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N3bula
271 posts
#112
Y by
Let $P(x, y)$ denote the assertion.
Suppose that $f$, is injective, I will prove $f$ is not injective.
\[P(x, y)-P(y, x)\]\[f(x+yf(x))=f(y+xf(y))\]Thus from injective:
\[f(x)=\frac{y-(f(y)-1)x}{y}\]If $f(y)>1$ we can choose $x$ such that $y-(f(y)-1)x$ is negative which is a contradiction. Now suppose $f(y)< 1$. If $f(y)<1$ we can choose $x$ such that:
$x=y+xf(y)$ which means that $f(x)f(y)=2f(x)$ and thus $f(y)=2$, however we supposed that $f(y)<1$, which is a contradiction thus $f(y)\geq 1$. If $f(y)=1$ we
get that $f$ is constant and so not injective thus $f$ cannot be injective.
Thus $f(x)=\frac{2-x}{c}$, however for no $c$ is this a solution. Thus
$f$ is not injective. Now suppose that $f(a)=f(b)$ and $a\neq b$.
\[P(a, y)\]\[f(a)f(y)=f(a+yf(a))\]\[P(b, y)\]\[f(a)f(y)=f(b+yf(a))\]\[\therefore f(a+yf(a))=f(b+yf(a))\]Thus $f$ is periodic with period $a-b$. Let $\vert a-b \vert=k$. Let $u$ be a value such that for some fixed $x$ and $i$ and suppose that $f(x+i)>f(x)$, $uf(x+i)-uf(x)=nk-i$ let $n$ be a value such that $nk>i$. Note if
$f(x)>f(x+i)$ we can find a similar positive $u$.
\[P(x, u)\]\[f(x)f(u)=f(x+uf(x))\]\[P(x+i, u)\]\[f(x+i)f(u)=f(x+i+uf(x+i))\]As: $uf(x)-uf(x+i)-i=nk$, we get $f(x+uf(x))=f(x+i+uf(x+i))$, thus this implies $f(x)=f(x+i)$, which implies $f$ is constant which means $f(x)=2$ for all $x$.
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EpicBird08
1752 posts
#114 • 2 Y
Y by PikaPika999, OronSH
This is the strangest FE I've ever solved.

The only solution is $f(x) = 2,$ which clearly works. Now we will show that this is the only solution. Let $P(x,y)$ denote the assertion.

Claim: $f(x) \ge 1$ for all $x.$
Proof: Otherwise $P\left(x, \frac{x}{1-f(x)}\right)$ gives $f(x) = 2$, contradiction.

Claim: $f(x) \ge 2^{1 - \frac{1}{2^n}}$ for all nonnegative integers $n.$
Proof: We prove this inductively, with the base case $n=0$ being the claim above. Then $P(x,x)$ gives $$f(x)^2  = 2 f(x + x f(x)) \ge 2 \cdot 2^{1 - \frac{1}{2^n}} = 2^{2 - \frac{1}{2^n}}$$by the inductive hypothesis. Thus $f(x) \ge 2^{1 - \frac{1}{2^{n+1}}},$ so the induction is complete.

By taking $n \to \infty,$ we get that $\boxed{f(x) \ge 2}$ for all $x.$

Claim: $f$ is NOT injective.
Proof: Otherwise, by looking at $P(x,1)$ and $P(1,x),$ we get $f(x+f(x)) = f(1 + x f(1)),$ so $x + f(x) = 1 + x f(1),$ meaning $f(x) = 1 + x(f(1) - 1).$ But taking $x \to 0$ implies $f(x) < 2,$ contradiction.

Thus we have $f(r) = f(s)$ for some $r < s.$ Then $P\left(r, \frac{s-r}{f(r)}\right)$ gives $$f(r) f\left(\frac{s-r}{f(r)}\right) = 2 f(s) \implies f\left(\frac{s-r}{f(r)}\right) = 2.$$For simplicity, let $a = \frac{s-r}{f(r)}.$ Then $P(a,a)$ gives $4 = 2 f(a + a f(a)) = 2 f(3a),$ so $f(3a) = 2.$ Similarly, $P(a, 3a)$ gives $4 = 2 f(a + 3a f(a)) = 2 f(7a),$ so $f(7a) = 2.$ In general, this logic can be repeated to get $f((2^n - 1)a) = 2$ for all $n \in \mathbb{N}.$

Finally, if $b$ is such that $f(b) = 2$ and $x < b,$ then $P\left(x, \frac{b-x}{f(x)}\right)$ gives $f(x) f\left(\frac{b-x}{f(x)}\right) = 2 f(b) = 4.$ But $f(x), f\left(\frac{b-x}{f(x)}\right) \ge 2,$ so this forces $f(x) = 2$ for all $x < b.$ Since $b$ can be arbitrarily large, this finishes.
This post has been edited 4 times. Last edited by EpicBird08, Apr 6, 2025, 10:20 PM
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ezpotd
1268 posts
#115
Y by
We claim the answer is only the constant function $f(x) = 2$.

Assume $f(x) < 1$ for some $x$, then setting $y = \frac{x}{1 - f(x)}$ gives $f(x) = 2$, contradiction. So $f(x) \ge 1$ for all $x$. Then assume $f(a) < 2$, repeatedly taking $f(x + af(x)) = f(x) \cdot \frac a2 $ and replacing $x$ with $x + af(x)$ allows us to get $f$ with a value below one in finitely many applications of the assertion, so $f(a) \ge 2$ for all $a$. By varying $y$ over all reals, we get $f(x) \le \frac 12 f(x)f(y) \le f(x + c) $ for any $c$, so $f$ is nondecreasing.

Now assume $f$ is never equal to $2$, so $f$ is strictly increasing as the previous inequality becomes strict. Thus $f$ is injective, so consider swapping $x,y$ so we have $x + yf(x) = y + xf(y)$, taking extremely large $y$ and sufficiently small $x$ to ensure $xf(y) < y$ forces $x + yf(x) > 2y > y + xf(y) $, contradiction, so our assumption that $f$ can be strictly increasing is false.

Thus $f$ is $2$ at some input $a$, and by nondecreasing it is equal to $2$ for all of $(0,a]$. Now $f(a)f(a) = 2f(3a)$ gives $f(3a) = 2$, so $f$ is equal to $2$ over all of $(0,3a]$, repeating this infinitely allows us to acquire $f(x) = 2$ for all $x$.
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youochange
165 posts
#116
Y by
Nima Ahmadi Pour wrote:
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria

Can someone pls verify my sol??

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

$P(0,0): f(0)^2=2f(0)$

Hence, $f(0)=2$

$P(0,y): f(y)=f(2y)$
$\implies f(x)=f(2x)$

$P(2x,y): f(2x)f(y)=2f(2x+yf(2x))$
$\implies f(x)f(y)=2f(2x+yf(x))=2f(x+yf(x)$

let, $t=yf(x)$

Both $y,f(x) \ge 0$
$\implies t \ge 0$
$f(x+t)=f(2x+t)$ and $f(x)=f(2x)$

so $f(x) $is equal on shifted intervals.
So $f(x)$ must be a constant function.
and plugging $f(x)=c$ in, we get $f(x)=2$
This post has been edited 1 time. Last edited by youochange, Yesterday at 6:29 AM
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jasperE3
11320 posts
#117 • 1 Y
Y by youochange
youochange wrote:
Nima Ahmadi Pour wrote:
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria

Can someone pls verify my sol??

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

$P(0,0): f(0)^2=2f(0)$

Problem says that $P(x,y)$ is true for all positive real number $x$ and $y$, but $0$ isn't a positive real number.
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