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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 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
2-var inequality
sqing   11
N 2 minutes ago by ytChen
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
11 replies
1 viewing
sqing
May 27, 2025
ytChen
2 minutes ago
Sums of products of entries in a matrix
Stear14   0
10 minutes ago
(a) $\ $Each entry of an $\ 8\times 8\ $ matrix equals either $\ 1\ $ or $\ 2.\ $ Let $\ A\ $ denote the sum of eight products of entries in each row. Also, let $\ B\ $ denote the sum of eight products of entries in each column. Find the maximum possible value of $\ A-B.\ $ In other words, find
$$ {\rm max}\ \left[ \sum_{i=1}^8\ \prod_{j=1}^8\ a_{ij} - 
\sum_{j=1}^8\ \prod_{i=1}^8\ a_{ij} \right]
$$
(b) $\ $Same question, but for a $\ 2025\times 2025\ $ matrix.
0 replies
Stear14
10 minutes ago
0 replies
a father and his son are skating around a circular skating rink
parmenides51   2
N 26 minutes ago by thespacebar1729
Source: Tournament Of Towns Spring 1999 Junior 0 Level p1
A father and his son are skating around a circular skating rink. From time to time, the father overtakes the son. After the son starts skating in the opposite direction, they begin to meet five times more often. What is the ratio of the skating speeds of the father and the son?

(Tairova)
2 replies
parmenides51
May 7, 2020
thespacebar1729
26 minutes ago
Sums of n mod k
EthanWYX2009   1
N 28 minutes ago by Martin.s
Source: 2025 May 谜之竞赛-3
Given $0<\varepsilon <1.$ Show that there exists a constant $c>0,$ such that for all positive integer $n,$
\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]Proposed by Cheng Jiang
1 reply
1 viewing
EthanWYX2009
May 26, 2025
Martin.s
28 minutes ago
Easy P4 combi game with nt flavour
Maths_VC   1
N 3 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
3 hours ago
Central sequences
EeEeRUT   14
N 4 hours ago by HamstPan38825
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
14 replies
EeEeRUT
Apr 16, 2025
HamstPan38825
4 hours ago
Elementary Problems Compilation
Saucepan_man02   32
N 5 hours ago by atdaotlohbh
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
32 replies
Saucepan_man02
May 26, 2025
atdaotlohbh
5 hours ago
Random Points = Problem
kingu   5
N 5 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
kingu
Apr 27, 2024
happypi31415
5 hours ago
Combo resources
Fly_into_the_sky   1
N 5 hours ago by Fly_into_the_sky
Ok so i never did combinatorics in my life :oops: and i am willing to be able to do P1/P4 combos (or even more)
So yeah how can i start from scratch?
Remark:i don't want compuational combo resources :noo:
1 reply
Fly_into_the_sky
5 hours ago
Fly_into_the_sky
5 hours ago
Very odd geo
Royal_mhyasd   2
N 5 hours ago by Royal_mhyasd
Source: own (i think)
nevermind
2 replies
Royal_mhyasd
Yesterday at 6:10 PM
Royal_mhyasd
5 hours ago
Polynomial Application Sequences and GCDs
pieater314159   46
N 6 hours ago by cursed_tangent1434
Source: ELMO 2019 Problem 1, 2019 ELMO Shortlist N1
Let $P(x)$ be a polynomial with integer coefficients such that $P(0)=1$, and let $c > 1$ be an integer. Define $x_0=0$ and $x_{i+1} = P(x_i)$ for all integers $i \ge 0$. Show that there are infinitely many positive integers $n$ such that $\gcd (x_n, n+c)=1$.

Proposed by Milan Haiman and Carl Schildkraut
46 replies
pieater314159
Jun 19, 2019
cursed_tangent1434
6 hours ago
c^a + a = 2^b
Havu   10
N 6 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
10 replies
Havu
May 10, 2025
Havu
6 hours ago
Own made functional equation
JARP091   0
Today at 4:10 PM
Source: Own (Maybe?)
\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]
0 replies
JARP091
Today at 4:10 PM
0 replies
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   16
N Today at 3:56 PM by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
16 replies
OgnjenTesic
May 22, 2025
JARP091
Today at 3:56 PM
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
11395 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
11395 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
11395 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
654 posts
#11
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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
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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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