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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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
Kazakhstan 2012 ( Grade 10 problem 5)
ts0_9   6
N 15 minutes ago by Fly_into_the_sky
Function $ f:\mathbb{R}\rightarrow\mathbb{R} $ such that $f(xf(y))=yf(x)$ for any $x,y$ are real numbers. Prove that $f(-x) = -f(x)$ for all real numbers $x$.
6 replies
ts0_9
May 20, 2012
Fly_into_the_sky
15 minutes ago
Peru IMO TST 2023
diegoca1   0
16 minutes ago
Source: Peru IMO TST 2023 D1 P3
Find all positive integers $k$ such that the sequence $a_n = \binom{2n}{n}$, $n \geq 0$, is periodic modulo $k$ from some point onward; that is, there exists a positive integer $n_0 > 0$ such that the sequence $ a_n $ , for $n \geq n_0$ , is periodic modulo $k$.
0 replies
diegoca1
16 minutes ago
0 replies
Peru IMO TST 2023
diegoca1   0
29 minutes ago
Source: Peru IMO TST 2023 D1 P2
Let $n$ be a positive integer. On an $n \times n$ board, players $A$ and $B$ take turns in a game. On each turn, a player selects an edge of the board (not on the board border) and makes a cut along that edge. If after the move the board is split into more than one piece, then that player loses the game.

Player $A$ moves first. Depending on the value of $n$, determine whether one of the players has a winning strategy.
0 replies
diegoca1
29 minutes ago
0 replies
Peru IMO TST 2023
diegoca1   0
36 minutes ago
Source: Peru IMO TST 2023 D1 P1
Let $ \alpha > 0 $. Prove that there exist positive integers $m$ and $n$ such that:
\[
\left| \alpha - \frac{m}{n} \right| < \frac{1}{m}.
\]
0 replies
diegoca1
36 minutes ago
0 replies
Functional equation
avatarofakato   4
N an hour ago by Fly_into_the_sky
Source: Polish MO second round 2012
$f,g:\mathbb{R}\rightarrow\mathbb{R}$ find all $f,g$ satisfying $\forall x,y\in \mathbb{R}$:
\[g(f(x)-y)=f(g(y))+x.\]
4 replies
avatarofakato
Feb 19, 2012
Fly_into_the_sky
an hour ago
2011 Lusophon Mathematical Olympiad - Problem 6
nunoarala   10
N an hour ago by miloss.delfosontop
Source: 2011 Lusophon Mathematical Olympiad - Problem 6
Let $d$ be a positive real number. The scorpion tries to catch the flea on a $10\times 10$ chessboard. The length of the side of each small square of the chessboard is $1$. In this game, the flea and the scorpion move alternately. The flea is always on one of the $121$ vertexes of the chessboard and, in each turn, can jump from the vertex where it is to one of the adjacent vertexes. The scorpion moves on the boundary line of the chessboard, and, in each turn, it can walk along any path of length less than $d$.
At the beginning, the flea is at the center of the chessboard and the scorpion is at a point that he chooses on the boundary line. The flea is the first one to play. The flea is said to escape if it reaches a point of the boundary line, which the scorpion can't reach in the next turn. Obviously, for big values of $d$, the scorpion has a strategy to prevent the flea's escape. For what values of $d$ can the flea escape? Justify your answer.
10 replies
nunoarala
Sep 17, 2011
miloss.delfosontop
an hour ago
Peru IMO TST 2023
diegoca1   0
an hour ago
Source: Peru IMO TST 2023 pre-selection P4
Prove that, for every integer $n \geq 3$, there exist $n$ positive composite integers that form an arithmetic progression and are pairwise coprime.

Note: A positive integer is called composite if it can be expressed as the product of two integers greater than 1.
0 replies
diegoca1
an hour ago
0 replies
Peru IMO TST 2023
diegoca1   0
an hour ago
Source: Peru IMO TST 2023 pre-selection P3
Let $n$ be a positive integer and $\mathcal{A}$ an alphabet with $n$ letters. Find the largest positive integer $k$ (as a function of $n$) for which there exists a sequence of letters $A_1, A_2, \ldots, A_k$ satisfying the following conditions:
a) $A_i \in \mathcal{A}$, for all $1 \leq i \leq k$.
b) $A_j \ne A_{j+1}$, for all $1 \leq j \leq k - 1$.
c) There do not exist indices $1 \leq p < q < r < s \leq k$ such that $A_p = A_r$ and $A_q = A_s$.
0 replies
diegoca1
an hour ago
0 replies
Polynomial Squares
zacchro   28
N an hour ago by KevinYang2.71
Source: USA December TST for IMO 2017, Problem 3, by Alison Miller
Let $P, Q \in \mathbb{R}[x]$ be relatively prime nonconstant polynomials. Show that there can be at most three real numbers $\lambda$ such that $P + \lambda Q$ is the square of a polynomial.

Alison Miller
28 replies
zacchro
Dec 11, 2016
KevinYang2.71
an hour ago
Peru IMO TST 2023
diegoca1   0
an hour ago
Source: Peru IMO TST 2023 pre-selection P2
In the Cartesian plane, a point is called integer if its coordinates are integers. A triangle with integer vertices has exactly 3 integer points in its interior and exactly 8 integer points on its perimeter (including the vertices). If the incenter of the triangle is an integer point and the other 2 interior integer points lie on its incircle, find the lengths of the triangle’s sides.
0 replies
diegoca1
an hour ago
0 replies
IMO ShortList 1999, combinatorics problem 2
orl   11
N an hour ago by heheman
Source: IMO ShortList 1999, combinatorics problem 2
If a $5 \times n$ rectangle can be tiled using $n$ pieces like those shown in the diagram, prove that $n$ is even. Show that there are more than $2 \cdot 3^{k-1}$ ways to file a fixed $5 \times 2k$ rectangle $(k \geq 3)$ with $2k$ pieces. (symmetric constructions are supposed to be different.)
11 replies
orl
Nov 14, 2004
heheman
an hour ago
Peru IMO TST 2023
diegoca1   0
an hour ago
Source: Peru IMO TST 2023 pre-selection P1
Let $x, y, z$ be non-negative real numbers such that $x + y + z \leq 1$. Prove the inequality
\[
6xyz \leq x(1 - x) + y(1 - y) + z(1 - z),
\]and determine when equality holds.
0 replies
diegoca1
an hour ago
0 replies
Pair of multiples
Jalil_Huseynov   67
N 2 hours ago by SomeonecoolLovesMaths
Source: APMO 2022 P1
Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$.
67 replies
Jalil_Huseynov
May 17, 2022
SomeonecoolLovesMaths
2 hours ago
a+b+c=1
cadiTM   24
N 2 hours ago by lpieleanu
Source: Korea Final Round 2011
Find the maximal value of the following expression, if $a,b,c$ are nonnegative and $a+b+c=1$.
\[ \frac{1}{a^2 -4a+9} + \frac {1}{b^2 -4b+9} + \frac{1}{c^2 -4c+9} \]
24 replies
cadiTM
Aug 28, 2012
lpieleanu
2 hours ago
Funky function
TheUltimate123   25
N Jun 10, 2025 by CrazyInMath
Source: CJMO 2022/5 (https://aops.com/community/c594864h2791269p24548889)
Find all functions \(f:\mathbb R\to\mathbb R\) such that for all real numbers \(x\) and \(y\), \[f(f(xy)+y)=(x+1)f(y).\]
Proposed by novus677
25 replies
TheUltimate123
Mar 20, 2022
CrazyInMath
Jun 10, 2025
Funky function
G H J
Source: CJMO 2022/5 (https://aops.com/community/c594864h2791269p24548889)
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TheUltimate123
1753 posts
#1 • 1 Y
Y by kimyager
Find all functions \(f:\mathbb R\to\mathbb R\) such that for all real numbers \(x\) and \(y\), \[f(f(xy)+y)=(x+1)f(y).\]
Proposed by novus677
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Orestis_Lignos
560 posts
#2 • 1 Y
Y by StarLex1
Note that the zero and the identity functions both work. Let $f$ be a solution different from these two.

Taking $y=0$ in the given implies $f(f(0))=(x+1)f(0)$, so $f(0)=0$. Now, we present the following Claim:

Claim: $f(x)=0$ only when $x=0$.
Proof: Assume otherwise, and let $f(k)=0$ for some $k \neq 0$. Then $x=\dfrac{k}{y}$ in the given implies $f(y)=0$ for all non-zero $y$, so $f$ is identically zero, a contradiction $\blacksquare$

Now, take $x \rightarrow x/y$ in the given, so $f(f(x)+y)=(\frac{x}{y}+1)f(y)$ for all $x$ and for all nonzero $y$.

Then, $y=-f(x)$ in the above relation implies $$f(-f(x))(x-f(x))=0,$$for all $x \neq 0$.

However, if we pick a $x$ such that $f(x) \neq 0, f(x) \neq x$ (it must exist since $f$ is neither the zero nor the identity function), then using the Claim we have

$f(-f(x)) \neq 0$ and $x-f(x) \neq 0,$

a contradiction.

To sum up, the only solutions are $f \equiv 0$ and $f \equiv \rm id$.
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Pleaseletmewin
1574 posts
#3 • 1 Y
Y by StarLex1
Let $P(x,y)$ denote the assertion.
$P(-1,y)$ gives $f(f(-y)+y)=0$.
$P(f(-y)+y,1)$ gives
\begin{align*}
f(f(f(-y)+y)+1)&=(f(-y)+y+1)f(1) \\
f(1)&=(f(-y)+y+1)f(1) \\
f(1)(f(-y)+y)&=0.
\end{align*}Case 1: $f(1)\neq 0$.
Then, $f(-x)+x=0$ for all $x$, so $f(x)=x$ for all $x$, which is indeed a solution.
Case 2: $f(1)=0$.
$P\left(\tfrac{1}{x},x\right)$ for $x\neq 0$ gives
\begin{align*}
f(f(1)+x)&=\left(\frac{1}{x}+1\right)f(x) \\
f(x)&=\left(\frac{1}{x}+1\right)f(x) \\
\frac{f(x)}{x}&=0,
\end{align*}so $f$ vanishes on all nonzero reals. Now, suppose $f(0)=c\neq 0$. Then, $P(0,0)$ gives
\begin{align*}
f(f(0)+0)&=f(0) \\
f(c)&=f(0) \\
0&=f(0),
\end{align*}contradiction. So $f(0)=0$ as well, which means in this case, $f$ is the zero function.
Therefore, the only solutions are $f(x)=0$ and $f(x)=x$ for all $x$. $\blacksquare$
This post has been edited 1 time. Last edited by Pleaseletmewin, Mar 20, 2022, 7:39 AM
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MrOreoJuice
597 posts
#4
Y by
Notice that the only constant function satisfying $f$ is the zero function henceforth assume $f$ is non-constant and there exists some number such that $f(\text{that number}) \neq 0$. Let $P(x,y)$ denote the given assertion
  • $P(-1,0) \implies f(f(0))=0$.
  • $P(0,0) \implies f(0)=0$.
Fixing $y$ such that $f(y) \neq 0$ gives that $f$ is surjective.
Claim: $f$ is injective at $0$.
Proof. Assume there exists $a \neq 0$ such that $f(a)=0$ then $P(x/a , a)$ gives
$$f(f(x)+a) = 0$$but since $f$ is surjective it means that $f$ is constant which violates our assumption. Now, taking $y \neq 0$ to be a free variable choose $a$ such that $f(a)=-y$ and plugging $P(a/y , y)$ gives
$$0 = f(-y + y) = (a/y + 1)f(y) \implies a=-y$$so $f(-y)=-y$ for all nonzero $y \implies f(x)=x$ for all real $x$ (because $0$ is also satisfied by the relation). To conclude, the two functions satisfying the original equation are $f(x)=0$ and $f(x)=x$.
This post has been edited 1 time. Last edited by MrOreoJuice, Mar 20, 2022, 11:34 AM
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megarnie
5718 posts
#5
Y by
Find all functions $f\colon \mathbb{R}\to \mathbb{R}$ such that for all real numbers $x$ and $y$,\[f(f(xy)+y)=(x+1)f(y)\]The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x)=x}$.

Let $P(x,y)$ denote the given assertion.

$P(0,0): f(f(0))=f(0)$.

$P(-1,0): f(f(0))=0$.

So $f(0)=0$.


Now we take cases based on whether there exists a $k\ne 0$ with $f(k)=0$.

Case 1: There exists such a $k$.
$P\left(\frac{k}{x},x\right): f(x)=\left(\frac{k}{x}+1\right)f(x)$. Now $\frac{k}{x}+1\ne 1$, so $f(x)=0\implies f\equiv 0$ is the only solution here.

Case 2: There does not exist such a $k$.
$P(-1,-x): f(f(x)-x)=0\implies f(x)-x=0\implies f(x)=x$.
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CT17
1481 posts
#6
Y by
Misplaced?

Let $P(x,y)$ denote the assertion that $f(f(xy) + y) = (x+1)f(y)$.

First, note that $f(x)\equiv 0$ is a solution because $0 = (x+1)0$ for all $x$ and $f(x)\equiv x$ is a solution because $xy + y = (x+1)y$ for all $x$ and $y$. Hence, we can assume from now on that there exist (not necessarily distinct) $u$ and $v$ such that $f(u)\neq 0$ and $f(v)\neq v$. Let $c = f(v) - v$.

$P(x,u)\implies f(f(xu) + u) = (x+1)f(u)$. As $f(u)\neq 0$ the RHS can take any real value, so $f$ is surjective.

$P(-1,-v)\implies f(f(v) - v) = 0f(v)\implies f(c) = 0$.

$P(x,c)\implies f(f(cx) + c) = 0$. Since $c\neq 0$ and $f$ is surjective, $f(cx) + c$ can take any real value. However, this implies that $f(x) = 0$ for all $x$, contradicting the fact that $f(u)\neq 0$. Hence, the only such functions are $f(x)\equiv 0$ and $f(x)\equiv x$.
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IAmTheHazard
5007 posts
#7
Y by
The answer is $f \equiv 0$ and $f(x)=x$ only, which both work, so we show that no other solutions exist. Suppose $f \not \equiv 0$, and let $P(x,y)$ denote the assertion.
Pick some $x$ with $f(m) \neq 0$. By letting $x$ vary in $P(x,m)$, it follows that $f$ is surjective.
Now suppose $f(n)=0$ for some $n$. Then $P(x,n)$ gives $f(f(xn)+n)=0$. If $n \neq 0$, then by varying $x$ again and using surjectivity shows that $f \equiv 0$, which is impossible because then it's not surjective. Thus $f(0)=0$ and $f$ is injective about $0$.
To finish, from $P(-1,x)$ we have $f(f(-x)+x)=0 \implies f(-x)=-x \implies f(x)=x$, as desired. $\blacksquare$
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MathLuis
1602 posts
#9 • 1 Y
Y by kimyager
meh.
We claim that the only solutions to the F.E. are $f(x)=x$ and $f(x)=0$.
Let $P(x,y)$ the assertion of the functional Equation, clearly $f(x)=0$ works so here we will assume that there exists $c$ such that $f(c) \ne 0$
$P(x,0)$
$$f(f(0))=(x+1)f(0) \implies f(0)=0$$Case 1: $f(1)=0$
We use our $c$ in a smart way, we got $c \ne 0$ so now $P \left(\frac{1}{c},c \right)$
$$f(c)=\left(\frac{1}{c}+1 \right)f(c) \implies \frac{1}{c}+1=1 \implies \frac{1}{c}=0 \; \text{contradiction!!}$$Case 2: $f(1) \ne 0$
Assume that there exists a real $d \ne 0$ such that $f(d)=0$, then by $P(d,1)$
$$f(1)=(d+1)f(1) \implies d=0$$Hence $f$ is injective at $0$ and now by $P(-1,-x)$
$$f(f(x)-x)=0 \implies f(x)=x$$Hence all the functions that work are $\boxed{f(x)=0,x}$ thus we are done :D
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jasperE3
11449 posts
#10 • 1 Y
Y by megarnie
Let $P(x,y)$ be the assertion $f(f(xy)+y)=(x+1)f(y)$. If $f(0)\ne0$ then:
$P\left(\frac{f(f(0))+2021}{f(0)}-1,0\right)\Rightarrow 2021=0$
so $f(0)=0$. Suppose that $f(k)=0$ for some $k\ne0$.
$P\left(\frac kx,x\right)\Rightarrow f(x)=0$ for $x\ne0$, and since $f(0)=0$ $\boxed{f(x)=0}$ for all $x$.
Otherwise, $f(k)=0$ implies $k=0$.
$P(-1,-x)\Rightarrow f(f(x)-x)=0\Rightarrow f(x)-x=0\Rightarrow\boxed{f(x)=x}$
which are both solutions.
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Pleaseletmewin
1574 posts
#11
Y by
CT17 wrote:
Misplaced?
Agreed. Swap this w/ P1.
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math31415926535
5617 posts
#12
Y by
is it just me or kinda similar to damo p4
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megarnie
5718 posts
#13
Y by
math31415926535 wrote:
is it just me or kinda similar to damo p4

Much easier but it does hav3 the same solutions
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rama1728
801 posts
#14
Y by
math31415926535 wrote:
is it just me or kinda similar to damo p4

LOL I did not expect that :D
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Fakesolver19
106 posts
#15
Y by
Let $P(x,y)$ be our assertion then,
$P(0,0): f(f(0))=f(0)$
$P(x,0):f(f(0))=(x+1)f(0) \implies (x+1)f(0)=f(0) \implies f(0)=0$
Let $f(a)=f(b)$ for some $a,b \in \mathbb{R}$,
$P(a,1)=P(b,1) \implies (a-b)f(1)=0$
Case 1:-$a=b ; f(1) \neq 0$
Then the function is injective .
$P(-1,x):f(f(-x)+x)=0=f(0) \implies f(x)=x$
Case 2:- $a \neq b;f(1)=0$
$P(\frac{1}{x},x):f(f(1)+x)=(\frac{1}{x}+1)f(x) \implies \frac{1}{x}f(x)=0 \implies f(x)=0 \ \forall x \in \mathbb{R}-{0}$
Therefore we have two solutions,$\boxed{f(x)=0;f(x)=x}$
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WhatistheName
5 posts
#16
Y by
Let $P(x, y): f(f(xy)+y)=(x+1)f(y)$.

$P(x, 0): f(f(0))=(x+1)f(0).$
$x=-1; \ f(f(0))=0.$
$\therefore \ (x+1)f(0)=0.$
$x \neq -1; \ f(0)=0.$

$P(x, 1): f(f(x)+1)=(x+1)f(1).$

$\text{if } f(1)=0:$
$P\Big(\dfrac {1} {x}, x\Big): f(f(1)+x)=\Big(\dfrac {1}{x}+1\Big)f(x).$
$\therefore \ \dfrac {1}{x} f(x)=0.$
$x \neq 0; \ \color{blue}{f \equiv 0.}$

$\text{if } f(1) \neq 0:$
$\text{let } f(a)=f(b). $
$\Rightarrow P(a, 1) \text{ vs } P(b, 1):$
$(LHS): \text{same.}$
$(RHS): (a+1)f(1)=(b+1)f(1).$
$f(1) \neq 0; \ a=b, f: \text{injective}.$

$f(y) \neq 0; $
$P\Big(\dfrac{a}{f(y)}-1, y\Big): f(\text{Some complicated expressions})=a.$
$\Rightarrow f: \text{ BIJECTIVE.}$

$\text{let } f^{-1}: \text{ inverse function of } f.$
$P\Big(\dfrac{f^{-1}(-y)}{y}, y\Big): f(f(f^{-1}(-y))+y)=\Big(\dfrac{f^{-1}(-y)}{y}+1\Big)f(y). $
It seems a bit complicated, but after some cleanups, we can say that:
$\Big(\dfrac{f^{-1}(-y)}{y}+1\Big)f(y)=0. (\because f(f^{-1}(-y))=-y.)$
$f(y) \neq 0.$
$\therefore \dfrac{f^{-1}(-y)}{y}+1=0.$
$\Rightarrow f^{-1}(-y)=-y, -y=f(-y).$

$\therefore f(y)=y \text{ for } \forall y \neq 0.$
$f(0)=0.$
$\therefore \color{blue} {f(x)=x.}$

$\color{blue} f \equiv 0, f(x)=x._{\small \blacksquare}$
This post has been edited 1 time. Last edited by WhatistheName, Apr 7, 2022, 2:03 PM
Reason: Typos
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827681
163 posts
#17 • 1 Y
Y by Mango247
100 posts
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ZETA_in_olympiad
2211 posts
#18
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Let $P(x,y)$ denote the given assertion.
We have $f(0)=0$ from $P(0,0)$ and $P(x,0).$

A. $\exists a\neq 0: f(a)=0:$
Then $P(ab^{-1},b)$ yields that $f$ is identically vanishing, fits.
B. $f(a)=0\iff a=0:$
Then $P(-1,-b)$ yields that $f$ is the identity, fits.

We are done after exhausting cases.
This post has been edited 1 time. Last edited by ZETA_in_olympiad, May 28, 2022, 2:56 PM
Reason: Fits
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Taco12
1757 posts
#19 • 2 Y
Y by RedFireTruck, Mango247
We claim the only solutions are $f(x)=0$ and $f(x)=x$.

$P(0,0) \rightarrow f(f(0))=f(0)$
$P(-1,0) \rightarrow f(f(0))=0$

Thus, we have $f(0)=0$.

We have 2 cases on whether or not there is a $n\ne 0$ with $f(n)=0$.

Case 1: There does exist $n$.
$P\left(\frac{n}{x},x\right) \rightarrow f(x)=\left(\frac{n}{x}+1\right)f(x)$. We have $f(x)=0$.

Case 2: There does not exist $n$
$P(-1,-x)\rightarrow f(x)-x=0\rightarrow f(x)=x$.

Both of these solutions work, so we are done. $\blacksquare$
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CyclicISLscelesTrapezoid
376 posts
#20 • 1 Y
Y by Mango247
We claim that the only solutions are $f(x) \equiv 0$ and $f(x) \equiv x$, which work.

Let $P(x,y)$ denote the assertion that $f(f(xy)+y)=(x+1)f(y)$. By $P(x,0)$, we have $f(f(0))=(x+1)f(0)$, so $f(0)=0$. If there exists $a \ne 0$ such that $f(a)=0$, then for nonzero $x$, $P(\tfrac{a}{x},x)$ gives $f(f(a)+x)=(\tfrac{a}{x}+1)f(x)=f(x)$, and since $\tfrac{a}{x}+1 \ne 1$, we have $f(x)=0$. This gives the solution $f(x) \equiv 0$. Otherwise, $f(a)=0$ implies that $a=0$. Then, $P(-1,-x)$ gives $f(f(x)-x)=0$, so $f(x)=x$.
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math_comb01
665 posts
#21
Y by
We claim that the only solutions are $f(x) \equiv 0$ and $f(x) \equiv x$, which obviously work.
Let $P(x,y)$ denote the assertion,
$P(x,1) \rightarrow$ $f(1)=0$ or $f$ is injective
CASE $1$ : f is injective
then $P(0,0) \rightarrow$ $f(0)=0$
then $P(-1,y) \rightarrow$ combined with injectivity gives $f(y)=y$
$$CASE-2: f(1)=0$$$P(x,1) \rightarrow$ $f(f(x)+1)=0$
then $0=f(f(f(xy)+y)+1) = f((x+1)f(y)+1)$
However the thing wrapped in function in RHS is clearly surjective,
$\rightarrow f(x)=0$
$\blacksquare$
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ihatemath123
3495 posts
#22 • 2 Y
Y by megarnie, OronSH
Solved with GrantStar.

The answer is $f(x) = x$ and $f(x) = 0$, obviously working.

Letting $x=-1$ shows that there exists a root of $f$.

Letting $x = \frac{r}{y}$, where $r$ is some root of $f$, gives us
\[0 = r \cdot f(y) \]for all nonzero $y$.

So, we have two cases:
  • If the only root is $0$, take $x=-1$, giving us
    \[f(f(-y)+y) = 0,\]so $f(-y)+y = 0$, forcing $f(y) = y$.
  • If there exists a nonzero root $r$, then $f(y) = 0$ for all nonzero $y$. Plugging in $y=0$ gives us
    \[ f(f(0)) = (x+1) f(0),\]so $f(0) = 0$, giving us the solution $f(x) = 0$.
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pi271828
3398 posts
#23 • 2 Y
Y by peace09, imagien_bad
The answer is $f(x) \equiv x$ or $f(x) \equiv 0$. These both clearly work, so henceforth assume $f$ is nonzero.

Claim: $f(0) = 0$

Proof. Observe \begin{align*} P(x, 0) \implies f(f(0)) = (x+1)f(0)\end{align*}so $f(0) = 0$. $\square$

Claim: $f$ is injective

Proof. For contradiction, assume not. Take $f(a) = f(b)$ where $a \neq b$. From \begin{align*} P(a, 1) \implies f(f(a)+1) = (a+1)f(1) ~~~ (\clubsuit) 
\\ P(b, 1) \implies f(f(b)+1) = (b+1)f(1) ~~~ (\spadesuit) \end{align*}Equating $(\clubsuit)$ and $(\spadesuit)$, we require either $f(1) = 0$ or $a = b$. If $f(1) = 0$, then we have \begin{align*} P \left( x, \tfrac{1}{x} \right) \implies f \left( f(1) + \tfrac{1}{x} \right) = f(\tfrac{1}{x}) = (x+1)f(\tfrac{1}{x})\end{align*}clearly implying $f$ must be zero. $\square$

To finish, observe that \begin{align*} P(-1, -y) \implies f(f(y)-y) = f(0) = 0\end{align*}This implies that $f(y) - y = 0$. We can check for the pointwise trap, and we are done. $\blacksquare$
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jasperE3
11449 posts
#24
Y by
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that for all positive real numbers $x$ and $y$:
$$f(f(xy)+y)=(x+1)f(y).$$
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Giant_PT
65 posts
#25
Y by
The solutions to this functional equation are $\boxed{f(x)\equiv 0, f(x)\equiv x}$, which obviously works. Let $P(x,y)$ be the usual assertion.
$P(-1,0)\implies f(f(0))=0 ,$
$P(0,0)\implies f(f(0))=f(0)\implies f(0)=0$
Now we separate the problem into two cases.

Case 1: There exists a $c\in \mathbb{R}$ not equal to $0$ such that $f(c)=0$.
$P(\frac{c}{y},y)\implies f(f(c)+y)=(\frac{c}{y}+1)f(y)\implies f(y)=0$ or $\frac{c}{y} =0$.
Clearly, $\frac{c}{y}=0$ is impossible since $c\neq 0.$ Therefore in this case, the only possible solution for $f$ is $\boxed{f(x)\equiv0}.$

Case 2: There doesn't exist a $c\in \mathbb{R}$ not equal to $0$ such that $f(c)=0$.
$P(x,1)\implies f(f(x)+1)=(x+1)f(1)$
This clearly shows that $f$ is bijective since $f(1)\neq 0$.
$P(-1,y)\implies f(f(-y)+y)=0=f(0)\implies f(-y)+y=0\implies f(y)=y$.
Therefore in this case, the only possible solution for $f$ is $\boxed{f(x)\equiv x}.$
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ezpotd
1360 posts
#26
Y by
The solutions are $f(x) = x, f(x) = 0$, both of which clearly work.

Assume $f$ is nonzero for at least one input. Taking $y$ such that $f(y)$ is nonzero, $f$ is bijective. Set $x = 0$, then $f(f(0) + y)= f(y) \rightarrow y = y + f(0)$, so $f(0) = 0$. Then set $x = -1$, so $f(f(-y) + y) = 0$, so $-y = f(-y)$, so $f$ is the identity as desired.
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CrazyInMath
477 posts
#27
Y by
solving FE because I got nothing left (btw I hate FE)

solution
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