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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.

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[*]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
f(x+f(x)+f(y))=x+f(x+y)
dangerousliri   10
N 28 minutes ago by jasperE3
Source: FEOO, Shortlist A5
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for any positive real numbers $x$ and $y$,
$$f(x+f(x)+f(y))=x+f(x+y)$$Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
10 replies
dangerousliri
May 31, 2020
jasperE3
28 minutes ago
n-variable inequality
ABCDE   66
N 30 minutes ago by ND_
Source: 2015 IMO Shortlist A1, Original 2015 IMO #5
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
66 replies
+1 w
ABCDE
Jul 7, 2016
ND_
30 minutes ago
Euler Line Madness
raxu   75
N an hour ago by lakshya2009
Source: TSTST 2015 Problem 2
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco
75 replies
raxu
Jun 26, 2015
lakshya2009
an hour ago
Own made functional equation
Primeniyazidayi   8
N an hour ago by MathsII-enjoy
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
8 replies
Primeniyazidayi
May 26, 2025
MathsII-enjoy
an hour ago
IMO ShortList 2002, geometry problem 7
orl   110
N 2 hours ago by SimplisticFormulas
Source: IMO ShortList 2002, geometry problem 7
The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.
110 replies
orl
Sep 28, 2004
SimplisticFormulas
2 hours ago
Cute NT Problem
M11100111001Y1R   6
N 2 hours ago by X.Allaberdiyev
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
6 replies
M11100111001Y1R
May 27, 2025
X.Allaberdiyev
2 hours ago
China MO 2021 P6
NTssu   23
N 2 hours ago by bin_sherlo
Source: CMO 2021 P6
Find $f: \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$, such that for any $x,y \in \mathbb{Z}_+$, $$f(f(x)+y)\mid x+f(y).$$
23 replies
NTssu
Nov 25, 2020
bin_sherlo
2 hours ago
Prove that the circumcentres of the triangles are collinear
orl   19
N 2 hours ago by Ilikeminecraft
Source: IMO Shortlist 1997, Q9
Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i = 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i+1}$ and $ A_iA_{i+2}$ (the addition of indices being mod 3). Let $ B_i, i = 1, 2, 3,$ be the second point of intersection of $ C_{i+1}$ and $ C_{i+2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.
19 replies
orl
Aug 10, 2008
Ilikeminecraft
2 hours ago
c^a + a = 2^b
Havu   9
N 2 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
9 replies
Havu
May 10, 2025
Havu
2 hours ago
An algorithm for discovering prime numbers?
Lukaluce   4
N 3 hours ago by alexanderhamilton124
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
4 replies
Lukaluce
May 18, 2025
alexanderhamilton124
3 hours ago
Orthocentroidal circle, orthotransversal, concurrent lines
kosmonauten3114   0
3 hours ago
Source: My own
Let $\triangle{ABC}$ be a scalene oblique triangle, and $P$ a point on the orthocentroidal circle of $\triangle{ABC}$ ($P \notin \text{X(4)}$).
Prove that the orthotransversal of $P$, trilinear polar of the polar conjugate ($\text{X(48)}$-isoconjugate) of $P$, Droz-Farny axis of $P$ are concurrent.

The definition of the Droz-Farny axis of $P$ with respect to $\triangle{ABC}$ is as follows:
For a point $P \neq \text{X(4)}$, there exists a pair of orthogonal lines $\ell_1$, $\ell_2$ through $P$ such that the midpoints of the 3 segments cut off by $\ell_1$, $\ell_2$ from the sidelines of $\triangle{ABC}$ are collinear. The line through these 3 midpoints is the Droz-Farny axis of $P$ wrt $\triangle{ABC}$.
0 replies
kosmonauten3114
3 hours ago
0 replies
inequality
Hoapham235   0
3 hours ago
Let $ 0 \leq a, b, c \leq 1$. Find the maximum of \[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]
0 replies
Hoapham235
3 hours ago
0 replies
3^n + 61 is a square
VideoCake   28
N 4 hours ago by Jupiterballs
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
28 replies
VideoCake
May 26, 2025
Jupiterballs
4 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   5
N 4 hours ago by AshAuktober
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
5 replies
BR1F1SZ
May 5, 2025
AshAuktober
4 hours ago
R+ Functional Equation
Mathdreams   10
N Apr 15, 2025 by TestX01
Source: Nepal TST 2025, Problem 3
Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \[f(f(x)) + xf(xy) = x + f(y)\]for all positive real numbers $x$ and $y$.

(Andrew Brahms, USA)
10 replies
Mathdreams
Apr 11, 2025
TestX01
Apr 15, 2025
R+ Functional Equation
G H J
G H BBookmark kLocked kLocked NReply
Source: Nepal TST 2025, Problem 3
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Mathdreams
1472 posts
#1 • 1 Y
Y by khan.academy
Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \[f(f(x)) + xf(xy) = x + f(y)\]for all positive real numbers $x$ and $y$.

(Andrew Brahms, USA)
This post has been edited 1 time. Last edited by Mathdreams, Apr 11, 2025, 1:28 PM
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megarnie
5611 posts
#2 • 4 Y
Y by khan.academy, KevinYang2.71, abrahms, Alex-131
The only solutions are $f(x) = \frac cx$ for some constant $c$ and $f\equiv 1$, which clearly work.

Let $P(x,y)$ be the given assertion. Clearly $1$ is the only constant solution, so assume $f$ is not constant.

Claim: $f$ is injective.
Proof: Suppose $f(a) = f(b)$ for some positive reals $a,b$.

$P(x,a)$ with $P(x,b)$ gives that $f(xa) = f(xb)$ for all $x \in \mathbb R^{+}$.

Now, $P(a,x)$ compared with $P(b,x)$ gives $af(ax) - a = b f(bx) - b$, so $a(f(ax) - 1) = b (f(bx) - 1)$. But, since $f(ax) = f(bx)$, we have \[ a(f(ax) - 1) = b(f(ax) - 1) \]Since $f$ isn't constant, we can choose $x$ where $f(ax) \ne 1$, so $a = b$. $\square$

$P(x, f(x)): f(f(x)) + xf(xf(x)) = x + f(f(x))$, so $xf(xf(x)) = x \implies f(xf(x)) = 1$.

$P(1, y): f(f(1)) = 1$.

Injectivity implies $xf(x) = f(1)$ for all $x$, so $f(x) = \frac{f(1)}{x}$, and setting $c = f(1)$ gives the desired result.
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pco
23515 posts
#3 • 4 Y
Y by khan.academy, Maksat_B, Sedro, abrahms
Mathdreams wrote:
Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \[f(f(x)) + xf(xy) = x + f(y)\]for all positive real numbers $x$ and $y$.
Let $P(x,y)$ be the assertion $f(f(x))+xf(xy)=x+f(y)$
Let $c=f(1)$ and $d=f(2)$

Subtracting $P(x,1)$ from $P(x,2)$, we get $f(2x)=f(x)+\frac{d-c}x$
Subtracting $P(2,1)$ from $P(2,x)$, we get $f(2x)=d+\frac{f(x)-c}2$

Subtracting : $f(x)=2\frac{c-d}x+2d-c$

Plugging $f(x)=\frac ax+b$ in original equation, we get $(a,b)=(\text{anything},0)$ or $(0,1)$ and solutions :
$\boxed{\text{S1 : }f(x)=1\quad\forall x>0}$, which indeed fits

$\boxed{\text{S2 : }f(x)=\frac ax\quad\forall x>0}$, which indeed fits, whatever is $a>0$
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jasperE3
11394 posts
#4 • 1 Y
Y by AlexCenteno2007
Let $P(x,y)$ be the assertion $f(f(x))+xf(xy)=x+f(y)$.
$P(x,f(x))\Rightarrow f(xf(x))=1$
$P(f(x),x)\Rightarrow f(f(f(x)))=f(x)$
$P(f(x),y)\Rightarrow f(x)f(yf(x))=f(y)$, in particular we have $f(f(x))=\frac{f(1)}{f(x)}$
$P(x,f(y))\Rightarrow xf(x)^2-(xf(y)+f(1))f(x)+f(y)f(1)=0\Rightarrow f(x)\in\left\{f(y),\frac{f(1)}x\right\}$ for each $x,y\in\mathbb R^+$ (we solved this as a quadratic in $f(x)$)
If $f(x)\ne\frac{f(1)}x$ for some $x$ then by varying $y$ over $\mathbb R^+$ we get that $f$ is constant, and testing, the only constant solution is $\boxed{f(x)=1}$ for all $x$.
Otherwise, $\boxed{f(x)=\frac cx}$ which works for any $c\in\mathbb R^+$.
This post has been edited 1 time. Last edited by jasperE3, Apr 11, 2025, 5:24 PM
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Tony_stark0094
69 posts
#6
Y by
if $f \equiv c$ then $c$ must be $1$ further assume $f$ is not constant:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:
now $P(x,f(x)): f(f(x))+xf(xf(x))=x+f(f(x)) \implies f(xf(x))=1$
from injectivity $xf(x)=f(1) \implies f(x)=\frac {f(1)}{x}$
hence $f(x)=1 \forall x \in R$ and $f(x)=\frac {f(1)}{x} \forall x \in R$ are the only solutions
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jasperE3
11394 posts
#7
Y by
Tony_stark0094 wrote:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?
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Tony_stark0094
69 posts
#8
Y by
jasperE3 wrote:
Tony_stark0094 wrote:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

$P(1,1): f(f(1))+f(1)=1+f(1) \implies f(f(1))=1$
for injectivity assume $f(a)=f(b)$
then subtract $P(a,1)$ from $P(b,1)$
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jasperE3
11394 posts
#9
Y by
Tony_stark0094 wrote:
jasperE3 wrote:
Tony_stark0094 wrote:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

$P(1,1): f(f(1))+f(1)=1+f(1) \implies f(f(1))=1$
for injectivity assume $f(a)=f(b)$
then subtract $P(a,1)$ from $P(b,1)$

and what if $f(a)=f(b)=1$ (which is indeed the particular case $f(xf(x))=f(f(1))=1$ that you use injectivity for)
This post has been edited 1 time. Last edited by jasperE3, Apr 12, 2025, 7:35 AM
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ThatApollo777
73 posts
#10
Y by
Claim : the only solutions are $f(x) = 1$ and $f(x) = \frac{c}{x}$.
Pf : Its easy to check these work, we now show these are only solutions.

Claim 1: If $f$ is not injective, its identically $1$.
Let $P(x, y)$ be the assertion. $$P(1,1) \implies f(f(1)) = 1$$Assuming $f(a) = f(b)$ for $a \neq b$. Let $r = \frac{b}{a} \neq 1$. $$P(a, y) - P(b, y) \implies a(f(ay) - 1) = b(f(by)-1)$$Putting $y = \frac{f(1)}{a}$ we can conclude: $$f(rf(1)) = 1$$$$P(\frac{x}{f(1)}, f(1)) - P(\frac{x}{f(1)}, rf(1)) \implies f(x) = f(rx)$$$$P(x, \frac{t}{x}) - P(rx, \frac{t}{x}) \implies f(t) = 1$$Since $r \neq 1$.

Now, assuming $f$ is injective consider $$P(x, \frac{f(1)}{x}) : f(f(x)) = f(\frac{f(1)}{x}) \implies f(x) = \frac{f(1)}{x}$$
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cursed_tangent1434
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The answers are $f(x) = 1$ for all $x\in \mathbb{R}^+$ and $f(x)= \frac{c}{x}$ for all $x\in \mathbb{R}^+$ for some fixed constant $c \in \mathbb{R}^+$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(f(x))+xf(xy)=x+f(y)$ for positive real numbers $x$ and $y$.

In what follows we assume that $f$ is not constant one. Say there does now exist some $x_0 \ne 1$ such that $f(x_0) \ne 1$ (i.e for all $x_0\ne 1$ we have $f(x_0)=1$). As $f$ is not constant one this indicates that $f(1) \ne 1$. Then, $P\left(x,\frac{1}{x}\right)$ yields,
\begin{align*}
f(f(x)) + xf(1) &= x+f\left(\frac{1}{x}\right)\\
f(1) + xf(1) &= x+1 \\
(x+1)f(1) &= x+1
\end{align*}which is a clear contradiction since $x+1>0$ and $f(1) \ne 1$. Thus, there indeed exists some $x_0 \ne 1$ such that $f(x_0) \ne 1$. Now, $P(1,1)$ implies that
\[f(f(1))+f(1)=1+f(1)\]from which we have $f(f(1))=1$. We now make the following observation.

Claim : The function $f$ is injective.

Proof : We first show that it is injective at all points except 1. For this, note that if there exists $t_1 \ne t_2 $ such that $f(t_1) =f(t_2) \ne 1$. Then, $P(t_1,1)$ and $P(t_2,1)$ yeild,
\[f(f(t_1))+t_1f(t_1)=t_1+f(1)\]\[f(f(t_2))+t_2f(t_2)=t_2+f(1)\]whose difference implies
\[(t_1-t_2)f(t_1)=t_1-t_2\]which since $f(t_1) \ne 1$ implies that $t_1=t_2$ which is a contradiction. Hence, $f$ is indeed injective at all points except 1.

With this observation in hand, consider $x_0 \ne 1$ such that $f(x_0) \ne 1$ and $\alpha$ such that $f(\alpha)=1$. Then, from $P\left(x_0 , \frac{\alpha}{x_0}\right)$ we have
\begin{align*}
f(f(x_0)) + x_0f(\alpha) &= x_0 + f\left(\frac{\alpha}{x_0}\right)\\
f(f(x_0)) &=  f\left(\frac{\alpha}{x_0}\right)
\end{align*}Now,
\[f(f(f(x_0))) = f\left(f\left(\frac{\alpha}{x_0}\right)\right)=f(x_0)\]which since $f(x_0) \ne 1$ implies $f(f(x_0))=x_0 \ne 1$. Thus,
\begin{align*}
f(f(x_0)) &=  f\left(\frac{\alpha}{x_0}\right)\\
f(x_0) &= \frac{\alpha}{x_0}
\end{align*}In particular, if there exists $\alpha_1,\alpha_2 \in \mathbb{R}^+$ such that $f(\alpha_1)=f(\alpha_2)=1$ we have
\[\frac{\alpha_1}{x_0} = f(x_0) = \frac{\alpha_2}{x_0}\]which implies $\alpha_1=\alpha_2$ proving the claim.

Thus, $f$ is injective and
\[f(x) = \frac{f(1)}{x}\]for all $x \ne f(1)$ and $f(f(1))=1$. This implies that indeed $f(x) = \frac{c}{x}$ for some fixed constant $c \in \mathbb{R}^+$ as desired.
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TestX01
341 posts
#12
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cutie patootie problem

We claim either $f$ is one or $\frac{c}{x}$. These clearly work. Let $P(x,y)$ denote the assertion.
Firstly, $P(1,1)$ gives $f(f(1))=1$.
$P(x,1)$ yields $f(f(x))+xf(x)=x+f(1)$. Hence $f(f(x))=xf(1)-xf(x)$. Thus, subbing back, $xf(xy)+f(1)-xf(x)=f(y)$. Now take $x\to f(1)$ so we get
\[f(1)f(f(1)y)=f(y)\]Now, we will take $y\to f(1)y$ in $xf(xy)+f(1)-xf(x)=f(y)$ to get $\frac{xf(xy)}{f(1)}+f(1)-xf(x)=\frac{f(y)}{f(1)}$. Comparing this with $xf(xy)+f(1)-xf(x)=f(y)$, we have
\[xf(xy)\left(\frac{1}{f(1)}-1\right)=f(y)\left(\frac{1}{f(1)}-1\right)\]This gives us two cases. Either $f(1)=1$ or $xf(xy)=f(y)$ for all $x,y$. In the latter case, we would have $xf(x)=f(1)$, and $f(x)=\frac{c}{x}$ which is a solution.

Now, suppose $f(1)=1$. Then, taking $P\left(x,\frac{1}{x}\right)$, we have $f(f(x))+x=x+f\left(\frac{1}{x}\right)$ hence $f(f(x))=f\left(\frac{1}{x}\right)$. Assume that $f$ is not always constant, as if it was constant then we would have $cx=x$ hence $c=1$ as desired. We shall prove that $f$ is injective, which would finish as then cancelling one $f$ we get $f(x)=\frac{1}{x}$.

Let $f(a)=f(b)$ such that WLOG $\frac{b}{a}>1$. Take $y=a,b$, so we have $f(f(x))+xf(ax)=x+f(a)$ and $f(f(x))+xf(bx)=x+f(b)$. Subtracting we have
\[x(f(ax)-f(bx))=f(a)-f(b)=0\]Hence, as $x\neq 0$, we have
\[f(ax)=f(bx)\quad f(x)=f(cx)\]where $c=\frac{b}{a}>1$ by scaling down $x$.

Now, consider $P(cx,y)$ so we get $f(f(x))+cxf(xy)=cx+f(y)$. Comparing with $P(x,y)$ we have
\[xf(xy)(c-1)=x(c-1)\]Yet $c-1>0$. Thus, we have $xf(xy)=x$ or $f(xy)=1$. Taking $y=1$ gives $f(x)=1$ for all $x$, a contradiction as we have dealt with constant $f$.

Thus, $f$ must be injective, and we are done.
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