Familiar Solution Set

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0 replies

jlacosta

Mar 2, 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

Minimum number of values in the union of sets

bnumbertheory 5

N an hour ago by Rohit-2006

Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3

For each positive integer $n$ , let $f(n)$ denote the smallest possible value of $|A_1 \cup A_2 \cup \dots \cup A_n|$ where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$ . Determine $f(n)$ for each positive integer $n$ .

5 replies

bnumbertheory

Oct 14, 2023

Rohit-2006

an hour ago

HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY

alcumusftwgrind 7

N 2 hours ago by dragoon

rant lol

Imagine a poor senior waiting for their MIT decisions just to have their hopes CRUSHED on 3/14 and they can't even celebrate pi day...

and even worse, this year's pi day is special because this year is a very special number...

7 replies

alcumusftwgrind

6 hours ago

dragoon

2 hours ago

rows are DERANGED and a SOCOURGE to usajmo .

GrantStar 26

N 2 hours ago by joshualiu315

Source: USAJMO 2024/4

Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$ , how many orderly colorings are there?

Proposed by Alec Sun

26 replies

GrantStar

Mar 21, 2024

joshualiu315

2 hours ago

Geo equals ABsurdly proBEMatic

ihatemath123 73

N 2 hours ago by joshualiu315

Source: 2024 USAMO Problem 5, JMO Problem 6

Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$ . Point $E$ is chosen on ray $BD$ so that $AE = EC$ . Let $M$ be the midpoint of $BC$ .

Show that line $AB$ is tangent to the circumcircle of triangle $BEM$ .

Proposed by Anton Trygub

73 replies

ihatemath123

Mar 21, 2024

joshualiu315

2 hours ago

two triangles have equal circumradii

littletush 5

N 3 hours ago by Taha.kh

Source: Italy TST 2009 p5

Two circles $O_1$ and $O_2$ intersect at $M,N$ . The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$ . Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.

5 replies

littletush

Mar 10, 2012

Taha.kh

3 hours ago

Equal lengths in cyclic quadrilateral

LoloChen 4

N 3 hours ago by Nari_Tom

Source: All-Russian MO 2024 9.4

In cyclic quadrilateral $ABCD$ , $\angle A+ \angle D=\frac{\pi}{2}$ . $AC$ intersects $BD$ at ${E}$ . A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$ , prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$ .

4 replies

LoloChen

Apr 22, 2024

Nari_Tom

3 hours ago

average FE

KevinYang2.71 74

N 3 hours ago by joshualiu315

Source: USAJMO 2024/5

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
$f(x^2-y)+2yf(x)=f(f(x))+f(y)$ for all $x,y\in\mathbb{R}$ .

Proposed by Carl Schildkraut

74 replies

KevinYang2.71

Mar 21, 2024

joshualiu315

3 hours ago

Two circles and many points

CHN_Lucas 5

N 3 hours ago by Captainscrubz

Source: 2022 China Second Round A2

$A,B,C,D,E$ are points on a circle $\omega$ , satisfying $AB=BD$ , $BC=CE$ . $AC$ meets $BE$ at $P$ . $Q$ is on $DE$ such that $BE//AQ$ . Suppose $\odot(APQ)$ intersects $\omega$ again at $T$ . $A'$ is the reflection of $A$ wrt $BC$ . Prove that $A'BPT$ lies on the same circle.

5 replies

CHN_Lucas

Dec 22, 2022

Captainscrubz

3 hours ago

Angle QRP = 90°

orl 12

N 4 hours ago by YaoAOPS

Source: IMO ShortList, Netherlands 1, IMO 1975, Day 1, Problem 3

In the plane of a triangle $ABC,$ in its exterior $,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$ , $\angle BCP = \angle QCA = 30^{\circ}$ , $\angle ABR = \angle RAB = 15^{\circ}$ .

Prove that

a.) $\angle QRP = 90\,^{\circ},$ and

b.) $QR = RP.$

12 replies

orl

Nov 12, 2005

YaoAOPS

4 hours ago

IMO 2014 Problem 4

ipaper 166

N 4 hours ago by hgomamogh

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$ . Let $M$ and $N$ be the points on $AP$ and $AQ$ , respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$ . Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$ .

Proposed by Giorgi Arabidze, Georgia.

166 replies

ipaper

Jul 9, 2014

hgomamogh

4 hours ago

IMO 2016 Problem 1

quangminhltv99 146

N 4 hours ago by Ilikeminecraft

Source: IMO 2016

Triangle $BCF$ has a right angle at $B$ . Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$ . Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$ . Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$ . Let $M$ be the midpoint of $CF$ . Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

146 replies

quangminhltv99

Jul 11, 2016

Ilikeminecraft

4 hours ago

A function equation

YaWNeeT 8

N 4 hours ago by HamstPan38825

Source: 2017 Taiwan TST 2nd round day 2 P4

Find all integer $c\in\{0,1,...,2016\}$ such that the number of $f:\mathbb{Z}\rightarrow\{0,1,...,2016\}$ which satisfy the following condition is minimal:
(1) $f$ has periodic $2017$
(2) $f(f(x)+f(y)+1)-f(f(x)+f(y))\equiv c\pmod{2017}$

Proposed by William Chao

8 replies

YaWNeeT

Apr 15, 2017

HamstPan38825

4 hours ago

Circumcenter lies on altitude

ABCDE 58

N 4 hours ago by cj13609517288

Source: 2016 ELMO Problem 2

Oscar is drawing diagrams with trash can lids and sticks. He draws a triangle $ABC$ and a point $D$ such that $DB$ and $DC$ are tangent to the circumcircle of $ABC$ . Let $B'$ be the reflection of $B$ over $AC$ and $C'$ be the reflection of $C$ over $AB$ . If $O$ is the circumcenter of $DB'C'$ , help Oscar prove that $AO$ is perpendicular to $BC$ .

James Lin

58 replies

ABCDE

Jun 24, 2016

cj13609517288

4 hours ago

Bananagrams

PieAreSquared 22

N 4 hours ago by youlost_thegame_1434

Source: EGMO 2023/3

Let $k$ be a positive integer. Lexi has a dictionary $\mathbb{D}$ consisting of some $k$ -letter strings containing only the letters $A$ and $B$ . Lexi would like to write either the letter $A$ or the letter $B$ in each cell of a $k \times k$ grid so that each column contains a string from $\mathbb{D}$ when read from top-to-bottom and each row contains a string from $\mathbb{D}$ when read from left-to-right.
What is the smallest integer $m$ such that if $\mathbb{D}$ contains at least $m$ different strings, then Lexi can fill her grid in this manner, no matter what strings are in $\mathbb{D}$ ?

22 replies

PieAreSquared

Apr 16, 2023

youlost_thegame_1434

4 hours ago

Familiar Solution Set

EulerMacaroni 171

N Mar 11, 2025 by Mathandski

Source: JMO/6, USAMO/4 2016

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$ ,
$(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$

171 replies

EulerMacaroni

Apr 20, 2016

Mathandski

Mar 11, 2025

Familiar Solution Set

G H J

Reveal topic tags

USAJMO

USA(J)MO

2016 USAJMO

2016 USAJMO Problem 6

function

algebra

G H BBookmark kLocked kLocked NReply

Source: JMO/6, USAMO/4 2016

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EulerMacaroni

851 posts

EulerMacaroni

#1 h Apr 20, 2016, 9:30 PM • 11 Y

Y by Davi-8191, parola, rashah76, icematrix2, son7, samrocksnature, megarnie, jhu08, GoodMorning, Adventure10, Mango247

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$ ,
$(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$

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mihajlon

374 posts

mihajlon

#2 h Apr 20, 2016, 9:30 PM • 4 Y

Y by samrocksnature, jhu08, Adventure10, Mango247

WRONG, :oops:

Let $A(x,y)$ be assertion of $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2$ .
If $A(0,0)\Longrightarrow 2(f(0))^2=(f(0))^2\Longrightarrow$ since $2\neq 1$

then $f(0)=0$ .
Let $A(0,y)\Longrightarrow (f(0))^2+f(y)\cdot f(-y)=f(y)\cdot f(-y)=(f(y))^2\Longrightarrow f(x)=0 \ \forall x\in \mathbb{R}$ or $f(y)=f(-y)... \bigstar$ .
Now let $A(x,-x)\Longrightarrow (f(x)-x^2)\cdot f(4x)+(f(-x)-x^2)\cdot f(4x)=(f(0))^2\stackrel{\bigstar}{\Longleftrightarrow} 2\cdot (f(x)-x^2)\cdot f(4x)=0$ .
So we have two possibilities:
$\textbf{1.)}$ $f(x)=x^2, \ \forall x\in \mathbb{R}$ , by checking i.e. $(x^2+xy)(x-3y)^2+(y^2+xy)(3x-y)^2=(x+y)^4$ , we get it's true.
$\textbf{2.)}$ $f(4x)=0$ . Since this hold for all $x\in \mathbb{R}$ we get $f(x)=0$ .

Thus only solutions are $\boxed{f(x)=x^2; \ f(x)=0}, \ \forall x\in \mathbb{R}$

This post has been edited 1 time. Last edited by mihajlon, Apr 20, 2016, 9:45 PM

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lucylai

238 posts

lucylai

#3 h Apr 20, 2016, 9:32 PM • 4 Y

Y by samrocksnature, jhu08, megarnie, Adventure10

mihajlon wrote:

Let $A(x,y)$ be assertion of $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2$ .
If $A(0,0)\Longrightarrow 2(f(0))^2=(f(0))^2\Longrightarrow$ since $2\neq 1$

This solution doesn't show that $f(x) = 0$ or that $f(x) = x^2$ is true for all real numbers. It just shows that this is true for each one. I made the same mistake for the 2014 USAMO.

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Wave-Particle

3690 posts

Wave-Particle

#4 h Apr 20, 2016, 9:34 PM • 11 Y

Y by El_Ectric, dantx5, ssk9208, vsathiam, Ultroid999OCPN, samrocksnature, jhu08, megarnie, Adventure10, Mango247, aidan0626

pointwise value trap :O

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adamov1

355 posts

adamov1

#5 h Apr 20, 2016, 9:35 PM • 7 Y

Y by fprosk, samrocksnature, jhu08, megarnie, Adventure10, Mango247, Nari_Tom

Hmm this problem was not nice. It took me 4 hours to come up with the correct argument to avoid pointwise value.

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mathwizard888

1635 posts

mathwizard888

#6 h Apr 20, 2016, 9:35 PM • 6 Y

Y by Wave-Particle, samrocksnature, jhu08, megarnie, Adventure10, mathfan2020

I could've sworn that this and 2014 AMO2/JMO3 were the same problem.

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Shaddoll

688 posts

Shaddoll

#7 h Apr 20, 2016, 9:36 PM • 4 Y

Y by samrocksnature, jhu08, megarnie, Adventure10

My solution outline:
1.Prove that f(0)=0
2.Prove that the function is even
3.Prove that for all x f(x)=x^2 or f(4x)=0
4.Prove that for all x f(x)=x^2 or f(x)=0
5.Using analysis to prove that limit of the ratio of LHS to RHS is 3/4 and arrive at a contradiction, if f(x)=0 and f(y)=y^2 for some x,y not equal to 0.

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aopser123

136 posts

aopser123

#8 h Apr 20, 2016, 9:39 PM • 4 Y

Y by samrocksnature, jhu08, megarnie, Adventure10

lucylai wrote:

This solution doesn't show that $f(x) = 0$ or that $f(x) = x^2$ is true for all real numbers. It just shows that this is true for each one. I made the same mistake for the 2014 USAMO.

I'm little bit confused, how does not simple checking show it's true for all reals?(I'm unexperienced...)

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zephyrcrush78

389 posts

zephyrcrush78

#9 h Apr 20, 2016, 9:41 PM • 3 Y

Y by samrocksnature, jhu08, Adventure10

I'm not sure if my solution is valid, but... I first proved that $f(0) = 0$ and I proved that $f(x) = 0$ satisfied the functional equation (I just plugged it in the functional equation), and then I proved that it is even. Then I proved that if it is not $f(x) = 0$ then it has to be $x^2 + cx$ for some $c$ and then said that it was even so it has to be $f(x) = 0$ or $f(x) = x^2$ .

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jeff10

1117 posts

jeff10

#10 h Apr 20, 2016, 9:41 PM • 3 Y

Y by samrocksnature, jhu08, Adventure10

Showed that if f(t)=0, then f(nt/2^m)=0 for all n, m positive integers. Then I said as m tends to infinity, this thing gets dense, and now I don't know if this is 7- or 0+

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WL0410

408 posts

WL0410

#11 h Apr 20, 2016, 9:41 PM • 4 Y

Y by samrocksnature, jhu08, megarnie, Adventure10

@aopser123
Those functions do work for all reals, but you didn't show that a function with $f(r)=0$ and $f(s)=s^2$ does not work.

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wwwrqnojcm

221 posts

wwwrqnojcm

#12 h Apr 20, 2016, 9:42 PM • 3 Y

Y by samrocksnature, jhu08, Adventure10

Well actually I tried to prove that the function should be continuous for an hour but got nothing so I wrote "it is easy to see that..."... Probably 2 pts deduction?

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niraekjs

1861 posts

niraekjs

#13 h Apr 20, 2016, 9:43 PM • 7 Y

Y by biomathematics, Imayormaynotknowcalculus, samrocksnature, jhu08, megarnie, Adventure10, Mango247

^ probably going to be like a 5 point deduction.

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mathtastic

3258 posts

mathtastic

#14 h Apr 20, 2016, 9:43 PM • 8 Y

Y by vsathiam, seoneo, richy, samrocksnature, jhu08, megarnie, Adventure10, Mango247

ok I will post my sketch here as well

Eventually one gets $f(x)=x^2$ or $f(x)=0$ for each $x$ . Suppose for some nonzero $j,k$ we had $f(k)=k^2$ and $f(j)=0$ . Assume they are positive because $f$ is even. Plugging in $(x,y)=(j,-j)$ implies that $f(4j)=0$ . Now $f(4^m j)=0$ , so we can assume $j$ is arbitralily large. Find suitable $x$ and $y$ so that $x-3y=k$ and $3x-y=j$ . Then we have $(f(x)+xy)k^2=f(x+y)^2$ . The LHS is clearly positive (taking $j$ large enough makes $x$ positive), so $f(x+y)^2=(x+y)^4$ . The LHS is at most $x(x+y)k^2$ . Solving for $x$ and $y$ in terms of $j$ and $k$ , and plugging in the results, it is easy to see that if $j$ is large enough a contradiction will arise.

This post has been edited 1 time. Last edited by mathtastic, Apr 21, 2016, 10:34 AM

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aopser123

136 posts

aopser123

#15 h Apr 20, 2016, 9:44 PM • 3 Y

Y by samrocksnature, jhu08, Adventure10

WL0410 wrote:

@aopser123
Those functions do work for all reals, but you didn't show that a function with $f(r)=0$ and $f(s)=s^2$ does not work.

Wgat are $r$ and $s$ ?

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IAmTheHazard

4999 posts

IAmTheHazard

#167 h Feb 23, 2021, 8:15 PM • 5 Y

Y by suvamkonar, samrocksnature, centslordm, jhu08, megarnie

Exhausting problem, but pretty cool.
The answer is $f(x)=x^2$ and $f \equiv 0$ , which we can check works.
We will denote the assertion as $P(x,y)$ .

First from $P(0,0)$ we get:
$f(0)^2+f(0)^2=f(0)^2$ from which we derive $f(0)=0$ .
Next, from $P(0,y)$ we get:
$f(y)f(-y)=f(y)^2.$ From $P(0,-y)$ we also get:
$f(y)f(-y)=f(-y)^2.$ We claim that this implies $f$ is even. Indeed, if for some $y$ we have $f(y) \neq 0$ then we can divide both sides of the first equation to get $f(y)=f(-y)$ . Otherwise, if $f(y)=0$ , the second equation gives $f(-y)=0$ as well, which also means $f(y)=f(-y)$ as desired.

Now from $P(x,-x)$ and using $f(0)=0$ with $f$ even, we get:
$f(4x)(2f(x)-2x^2)=0$ which gives $f(x)=x^2$ or $f(4x)=0$ for all $x \in \mathbb{R}$ . This means that $f(x)=0$ for some $x$ implies $f(4x)=0$ (if $x=0$ the statement is vacuous, else we are forced to have $x^2=0$ which is a contradiction).
Now, suppose that for some $a$ we have $f(4a)=0$ . We claim that this implies $f(a)=0$ as well.
From $P(x,3x)$ with $x$ arbitrary we get:
$(f(x)+3x^2)f(8x)=f(4x)^2$ Now letting $x=\tfrac{a}{2}$ gives $f(2a)=0$ , and then letting $x=\tfrac{a}{4}$ gives $f(a)=0$ . Thus for all $x$ , we either have $f(x)=x^2$ or $f(x)=0$ (which implies $f(4x)=0$ as well). We also have that:
$f(x)=0 \iff f(4x)=0$ for all $x$ . This implies $f(x) \geq 0$ for all $x$ .

Suppose now that there exists some value of $a \neq 0$ such that $f(a)=0$ . Since $f$ is even, WLOG let $a$ be positive. We will show that $f \equiv 0$ .
By $P(a,-y)$ for $y \in (0,a)$ and rearranging, we get:
$-ayf(a+3y)+(f(y)-ay)f(3a+y)-f(a-y)^2=0$ But on the other hand, we have:
$-ayf(a+3y) \leq 0$ and:
$(f(y)-ay)f(3a+y) \leq y(y-a)f(3a+y) \leq 0.$ which means that unless $f(a-y)^2=0$ , the $LHS$ will be negative. This implies that $f(a-y)=0$ for all $y \in (0,a)$ which implies $f(x)=0$ for all $x \in (0,a)$ .
We claim that this result, combined with $f(x)=0 \iff f(4x)=0$ , is enough to imply that $f(x)=0$ for all positive $x$ . Indeed, using easy induction and the previous result we get that $f(x)=0$ for all $x \in (0,4^ka)$ for some nonnegative integer $k$ . This means that for any positive $b>a$ , we can simply take a $k$ such that $4^ka>b$ (which clearly exists as $a>0$ ), and get that $f(b)=0$ . Since $b$ is arbitrary, putting this with the previous result, we get that if there exists some $a \neq 0$ such that $f(a)=0$ , then we must have $f(x)=0$ for all $x \geq 0$ . But since $f$ is even, this means that $f \equiv 0$ everywhere. This therefore eliminates all the pathological solutions, leaving us only with $f \equiv 0$ and $f(x)=x^2$ , as desired. $\blacksquare$

~~Edit: wait the pointwise step is wrong, but I have to go to bed now.~~ In the process of fixing, the pointwise solution has completely changed and made much quicker. For the sake of completeness I'll also fix the "original", which will be hidden below:

Alternate finish for pointwise

From $P(a,y)$ for $y$ positive, we get:
$ayf(a-3y)+(f(y)+ay)f(3a-y)=f(a+y)^2$ Now suppose that there exists some $y$ such that $f(a+y)=(a+y)^2$ . Then the equation becomes:
$(a+y)^4=ayf(a-3y)+(f(y)+ay)f(3a-y)$ But realize that we have:
$\begin{align*} ayf(a-3y)+(f(y)+ay)f(3a-y)&\leq ayf(a-3y)+(y^2+ay)f(3a-y)\\ &\leq ay(a-3y)^2+(y^2+ay)(3a-y)^2\\ &=10a^3y-3a^2y^2+4ay^3+y^4 \end{align*}$ which implies that $LHS-RHS$ is at least
$a^2(a-3y)^2.$ But by the trivial inequality and $a \neq 0$ , this cannot equal zero unless $y=\tfrac{a}{3}$ . But since $f(x)=0$ in $(0,a)$ , we have $f(\tfrac{a}{3})=0$ . Since we have $f(x)=0 \iff f(4x)=0$ for all $x$ , this implies that $f(\tfrac{4a}{3})=0$ , so we have $f(a+y)=0$ for all $y \in (0,\infty)$ . Combining this implies $f(x)=0$ over nonnegative reals. Then finish the same way.

This post has been edited 8 times. Last edited by IAmTheHazard, Apr 8, 2021, 7:51 PM

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PROA200

1748 posts

PROA200

#168 h Apr 2, 2021, 10:44 PM • 4 Y

Y by samrocksnature, jacoporizzo, jhu08, megarnie

Solved with jacoporizzo.

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$ ,
$(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$
The solutions are $f(x)=0$ and $f(x)=x^2$ only. Clearly, the first solution works. The second also does since
$\begin{align*} (x^2+xy)\cdot (x^2-6xy+&9y^2)+(y^2+xy)\cdot (9x^2-6xy+y^2)\\ &=x^4-6x^3y+9x^2y^2+x^3y-6x^2y^2+9xy^3\\ &\qquad +9x^2y^2-6xy^3+y^4+9x^3y-6x^2y^2+xy^3\\ &=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\ &=(x+y)^4 \end{align*}$ for all $x$ , $y$ .

Now, we show that these are the only solutions.

Plug in $(x,y)=(0,0)$ to get $2f(0)^2=f(0)^2\implies f(0)=0$ .

Plug in $(x,y)=(3i, i)$ to get $0 + (f(i)+3i^2)\cdot f(8i)=f(4i)^2$ .
Plug in $(x,y)=(i,3i)$ to get $(f(i)+3i^2)\cdot f(-8i)+0=f(4i)^2$ .
These two give that $f(8i)=f(-8i)$ for all $i\neq 0$ , so $f(x)=f(-x)$ for each $x$ .

Now, plugging $y=-x$ ,
$(f(x)-x^2)\cdot f(4x)+(f(-x)-x^2)\cdot f(4x) = 0.$ Hence, for each $x$ , either $f(x)=x^2$ or $f(4x)=0$ . Notice that this implies $f$ is nonnegative. $f(x)=x^2$ is one such solution. Now, assume that there exists some nonzero $k$ such that $f(k)=0$ .

Plug in $(x,y)=(k,k)$ . Then,
$k^2\cdot f(2k)+k^2\cdot f(2k) = f(2k)^2$ which implies $f(2k)=0$ as well, since $k$ is nonzero.

Finally, we plug in $y=k-x$ . This gives
$(f(x)+x(k-x))\cdot f(4x-3k) + (f(k-x)+x(k-x))\cdot f(4x-k) = 0$
Notice that when $x\in (0,k)$ , we have that all of the terms on the LHS are nonnegative and $x(k-x)$ is positive. Since the RHS is zero, each of the two terms on the LHS are zero. In particular, $f(4x-k)=0\implies f(i)=0$ for all $-k<i<3k$ . From here, notice that we can "double" this interval repeatedly to get that $f(x)=0$ for all positive $x$ . Then, since $f$ is even, $f(x)=0$ for all negative $x$ as well. This finishes.

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sidchukkayapally

669 posts

sidchukkayapally

#169 h Jun 1, 2021, 7:21 PM • 2 Y

Y by jhu08, megarnie

The possible are $f \equiv 0$ and $f \equiv x^2$

Let $F(x, y)$ denote the given assertation.

Firstly, we compute $f(0)$ by asserting $F(0, 0)$ . We have $2 \cdot f(0)^2 = f(0)^2$ , giving that $f(0) = 0$ .

Claim. $f$ is even.

Proof. We assert $F(0, y)$ and $F(0, -y)$ . $F(0, y) \to f(y)f(-y) = f(y)^2$ $F(0, -y) \to f(-y)(f(y) = f(-y)^2$ Hence, $f(-y)^2 = f(y)^2$ . Suppose that for a certain $y$ , we have $f(y) \ne f(-y)$ . We have $f(-y) = -f(y)$ , meaning that $f(y)f(-y) = -f(y)^2$ , contradiction. Hence, $f(y) = f(-y)$ for all $y$ as desired $\square$

Now assert $F(x, -x)$ , noting that we may substitute $f(3y - x)$ for $f(x - 3y)$ . We have $f(4x)(f(x) - x^2) = 0$ for all real $x$ , which implies that $f$ is non-negative.

To finish, assume that for some $c$ , $f(c) = 0$ . Asserting $F(c, c)$ gives that $f(c) = 0 \to f(2c) = 0$ . Notice we can scale up arbitrarily many times as needed and use the fact that $f$ is even to achieve a certain real from a real in the interval $(0, d)$ , or $(d, 0)$ . Hence, if for all reals $\in (0, d)$ or $(d, 0)$ that $f \equiv 0$ , we have that $f \equiv 0$ across all reals for any real $d$ ,. Now assert $F(a, c - a)$ , where $a \in (0, c)$ . We have $[f(a) + a(c - a)][f(4a - 3c)] + [f(c - a) + a(c - a)][f(4a - c)] = 0$ Since $a(c - a) > 0$ , we have that both terms on the left hand side are at least zero, but since the right hand side is zero, we must have that both terms are zero. Since $f$ is non-negative, $f(i) + a(c - a) > 0$ for all real $i$ , meaning that we must have $f(4a - 3c) = f(4c - a) = 0$ . Taking $f(4a - c)$ yields that for all reals in the interval $(-c, 3c)$ $f$ is zero, meaning that for all reals in the interval $(0, 3c)$ as desired.

Hence, if such a $c$ exists, $f(x) = 0$ < and otherwise we must have $f(x) - x^2 = 0$ , meaning that $f(x) = x^2$ as desired.

Remarks. It took forever to get the last part. My solution ended up looking very similar to @above's but oh well. This was very exhausting but worth it.

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puntre

159 posts

puntre

#170 h Aug 21, 2021, 9:07 AM • 2 Y

Y by jhu08, megarnie

We claim that the answers are $f(x)=0$ and $f(x)=x^2$
Set $(x,y)=(0,0)$ gives $f(0)=0$
Set $x=0$ gives $f(y)f(-y)=f(y)^2$ Which implies that $f$ is an even function
Set $y=-x$ gives $f(4x)(f(x)-x^2)=0$ .
Set $y=3x$ gives $(f(x)+3x^2)f(8x)=f(4x)^2$ which implies that $f(x)=0\Longleftrightarrow f(2x)=0$ .
Therefore, $f(x) = \begin{cases} x^2 & ?? \\ 0 & ?? \end{cases}$
Let $f(a)=0$ and $f(b)=b^2$ for some $a,b>0$ (since $f$ is odd). So, there exist $a'=2^ka>b$ , Let $a'=m+n$ and $b=m-3n$
Set $(x,y)=(m,n)$ gives $f(m)+f(n)+2mn=0$ Which is a contradiction since $m,n>0$

This post has been edited 1 time. Last edited by puntre, Aug 21, 2021, 9:18 AM

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megarnie

5531 posts

megarnie

#171 h Sep 11, 2021, 3:17 PM • 4 Y

Y by jhu08, GoodMorning, Mango247, Mango247

I claim that the only solutions are $f(x)=0$ and $f(x)=x^2$ . These clearly work.

Let $P(x,y)$ denote the given assertion.
$P(0,0): f(0)^2=2f(0)^2$ so $f(0)=0$ .
$P(0,y): f(y)f(-y)=f(y)^2$ .
So the function is even.
$P(3y,y): (f(y)+3y^2)\cdot f(8y)=f(4y)^2$ . Now if $f(8y)=0$ , then $f(4y)=0$ , so if $f(k)=0$ , then $f(\frac{k}{2})=0$ .
$P(x,-x): (f(x)-x^2)\cdot f(4x)+(f(x)-x^2)\cdot f(4x)=2((f(x)-x^2)\cdot f(4x))=0$
Either $f(x)=x^2$ or $f(4x)=0 \implies f(2x)=0 \implies f(x)=0$ .

Now it remains to prove that we can't have $f(x)=0$ for some reals and $f(x)=x^2$ for other reals in the same function, excluding $0$ .

Suppose that $f(x+y)=(x+y)^2$ where $x$ and $y$ are both positive.
We have $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(x+y)^4.$ The maximum of the LHS is $(x^2+xy)\cdot(x-3y)^2+(y^2+xy)\cdot (3x-y)^2=(x+y)^4$ (because $f(x)\ge0$ and $xy>0$ ), which only holds when $f(x)=x^2, f(y)=y^2, f(x-3y)=(x-3y)^2,$ and $f(3x-y)=(3x-y)^2$ .

Call $x$ good if $f(x)=x^2$ . Call $x$ bad if $f(x)=0$ .

We have all real values $k$ such that $0\le k \le x+y$ are good because we can write $x+y$ as $k+l$ , where both $k$ and $l$ are positive. Also, if $k=0$ or $x+y$ , $k$ is obviously good.

Let there be positive reals $m$ and $n$ such that $m+n$ is bad. If we have some good $a$ , we can multiply $a$ by a power of $2$ so that we exceed $m+n$ , which implies $m+n$ is good, a contradiction (if $a\cdot$ (the power of $2$ ) is bad, then $a$ is bad, so $a\cdot$ (the power of $2$ ) is good). So if we have a good positive number (by evenness if we have $-x$ is good, then $x$ is good), then all positive numbers are good, and because of evenness, all reals are good.

Thus, all reals are good or all reals are bad. So we are done.

This post has been edited 3 times. Last edited by megarnie, Jan 1, 2022, 2:51 AM

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GoodMorning

826 posts

GoodMorning

#172 h May 12, 2022, 7:42 PM • 2 Y

Y by megarnie, solasky

This was very fun to write-up

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$ ,
$(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$

Let $P(x, y)$ be the assertion that $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$

Then, $P(0, 0)$ yields $2f(0)^2 = f(0)^2 \rightarrow f(0)=0$ .

Claim 1: $f$ is even.

Proof: $P(0,y)$ yields $f(y)f(-y)=f(y)^2$ . Similarly, $P(0, -y)$ yields $f(-y)f(y)=f(-y)^2$ . It follows from here that $f$ is even.

Claim 2: For each $x \in \mathbb{R}$ , we have that $f(x)=0$ or $f(x) = x^2$ .

Proof: $P(x, -x) \rightarrow (f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(-4x) = (f(0))^2 \rightarrow (f(x)-x^2) \cdot f(4x) = 0$ .

So, either $f(x)=x^2$ or $f(4x) = 0$ .

$P(x, 3x) \rightarrow (f(x) + 3x^2) \cdot f(8x) = (f(4x))^2$ .

This means that $f(8x)=0 \implies f(4x)=0$ , which is equivalent to $f(2x)=0 \implies f(x)=0$ .

Therefore, $f(4x) = 0 \implies f(x)=0$ .

Thus, $f(x)=0$ or $f(x) = x^2$ .

Now, assume that the function is $0$ at some nonzero value. This implies the existence of a positive $C$ such that $f(C) = 0$ . Now, for any $0 < D \le 2C$ , the solutions to the equations
$\begin{cases} x+y = C \\ 3x-y = D \end{cases} \rightarrow \begin{cases} x = \frac{C+D}{4} \\ y = \frac{3C-D}{4}. \end{cases}$ are always strictly positive. Then $P(x, y) \rightarrow (f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$ Since the RHS is $0$ , and $(f(x)+xy)\cdot f(x-3y)$ is nonnegative and $(f(y)+xy)$ is strictly positive, we have that $f(3x-y) = f(D)$ is also $0$ , for all $0 < D \le 2C$ . A simple induction argument can show that the function is $0$ for all real values.

Therefore, the only possible functions are $\boxed{f(x)=x^2 \text{ or } f(x)=0}$ and we can indeed check that it works.

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#173 h Jun 3, 2022, 2:56 PM

Y by

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

$P(0,0)$ gives $f(0)=0.$ And $P(0,x)$ yields if $f(x)\neq 0$ then $f(x)=f(-x),$ although it works for $f(x)=0$ as well. From $P(x,-x)$ we get $f(x)=x^2$ or $f(4x)=0.$ Assume $f(4x)=0,$ then $P(3x/4,x/4)$ gives $f(2x)=0.$ Reapplication gives $f(x)=0.$

Claim: $\exists x\neq 0: f(x)=0 \implies f\equiv 0.$
Proof. WLOG $a,b >0: f(a)=0$ and $f(b)\geq 0.$ But $P(a/2,-b/2)$ forces $f(b)=0.$ $\blacksquare$

So the solutions are $f\equiv \text{Id}^2$ and $f\equiv 0.$ Both work.

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ismayilzadei1387

219 posts

ismayilzadei1387

#174 h Aug 16, 2023, 8:25 PM

Y by

$(f(x)+xy)f(x-3y)+(f(y)+xy)f(3x-y)=f(x+y)^2$
Let $P(x,y)$ be assertion of this equation.
1. $P(0,0)$
$f(0)=0$
2. $P(x,x)$
$(f(x)+x^2)(f(-2x)+f(2x))=f(2x)^2$
3. $P(x,-x)$
$f(4x)(f(x)+f(-x)-2x^2)=f(0)=0$
So,there are two cases which we should consider.
1'. $f(4x)=0$ which means $f$ is const and $f(x)=0$ and it indeed fits.
2'. $\boxed{f(x)+f(-x)=2x^2}$
4. $P(2x,-2x)$
$f(2x)+f(-2x)=8x^2$ $\implies{(f(x)+x^2).8x^2=f(2x)^2}$
5. $P(-x,-x)$
$(f(-x)+x^2)(f(2x)+f(-2x))=f(-2x)^2$
$5.-4.$ $\implies{f(-x)-f(x)=f(-2x)-f(2x)}$
it is easy to see $f(x)=f(2x)-3x^2$
Let's put this fact on 4.
then we get $f(x).2x^2=x^4+f(2x)^2$ which means $(x^2-f(x))^2=0$ thus we have $\boxed{f(x)=x^2}$
Only thing check pointwise trap.

This post has been edited 1 time. Last edited by ismayilzadei1387, Sep 30, 2023, 5:55 PM

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Infinity_Integral

306 posts

Infinity_Integral

#175 h Sep 7, 2023, 1:47 AM

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Only solution is $f(x)=0\forall x$ or $f(x)=x^2\forall x$ .
Take $x=y=0$ gives $f(0)=0$ .
Then take $x=0$ and use that $f(0)=0$ , giving $f(y)\cdot f(-y)=f(y)^2$
So $f$ is a even function or always $0$ , which is still even anyways.
Take $x=-y$ and use that $f$ is even, $f(0)=0$ , we can get that
$2(f(y)-y^2)f(4y)=0$ This gives that for any $x$ , we must have $f(x)=x^2$ or $f(x)=0$ .
The solution set follows after we deal with the pointwise trap.

Full proof here:
https://infinityintegral.substack.com/p/usajmo-2016-contest-review

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AngeloChu

471 posts

AngeloChu

#176 h Mar 8, 2024, 6:06 PM

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plug $y=0$ to get $f(0)f(3x)=0$ , so $f(0)=0$ or all $f(x)=0$
plug $x=0$ to get $f(y)f(-y)=(f(y))^2$ and $f(-y)=f(y)$
plug $y=-x$ to get $(f(x)+f(-x)-2x^2)(f(4x))=0$
either all $f(x)=0$ , or $f(x)+f(-x)=2x^2$
$f(x)=f(-x)$ , so solving $f(x)+f(-x)=2x^2$ gives $f(x)=x^2$
therefore, our only solutions are $f(x)=0$ or $f(x)=x^2$

This post has been edited 1 time. Last edited by AngeloChu, Mar 8, 2024, 6:08 PM
Reason: latex error

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peppapig_

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peppapig_

#177 h Mar 13, 2024, 2:35 AM

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First, note that plugging in $(0,0)$ gives that
$f(0)^2+f(0)^2=f(0)^2,$ which just implies that $f(0)=0$ . Using this, we plug in $(0,y)$ to get the equation
$(f(0)+0)f(-3y)+(f(y)+0)f(-y)=f(y)^2 \iff f(y)f(-y)=f(y)^2,$ and plugging in $(0,-y)$ similarly yields that $f(y)f(-y)=f(y)^2$ , meaning that either $f(y)=f(-y)\neq 0$ , or $f(y)=f(-y)=0$ . Either way, we have that $f(y)=f(-y)$ .

Now, we plug in $(x,-x)$ to get that
$(f(x)-x^2)f(4x)+(f(-x)-x^2)f(4x)=f(0)^2 \iff (f(x)-x^2)f(4x)=0,$ which implies that either $f(x)=x^2$ or $f(4x)=0$ . Replacing $x$ with $\frac{x}{4}$ yields that for all $x$ , $f(x)$ only has two possible values - $0$ and $x^2$ . Now we are simply left with proving that there is no pointwise trap.

I will prove this by proving that if $\exists a\neq 0$ such that $f(a)=0$ , then $f(x)=0$ for all real numbers $x$ . From this we can conclude that otherwise, $f(x)=x^2$ for all $x\in \mathbb{R}$ . Assume FTSOC that $\exists a\neq 0$ such that $f(a)=0$ .

C1: First, I claim that if $f(a)=0$ , then we also have that $f(2a)=f\left(\frac{a}{2}\right)=0$ . The latter can be shown by plugging in $\left(\frac{3a}{4}, \frac{a}{4}\right)$ , which gives
$\left(f\left(\frac{3a}{4}\right)+3\left(\frac{a}{4}\right)^2\right)f(0)+\left(f\left(\frac{a}{4}\right)+3\left(\frac{a}{4}\right)^2\right)f(2a)=f(a)^2,$ $\iff 0=f(a)^2=\left(f\left(\frac{a}{4}\right)+3\left(\frac{a}{4}\right)^2\right)f(2a),$ and since $f\left(\frac{a}{4}\right)$ is always $\geq 0$ (since $f(x)=0$ or $x^2$ ) and $a$ is non-zero, we have that $f\left(\frac{a}{4}\right)+3\left(\frac{a}{4}\right)^2$ must also be greater than zero. This gives that we must have $f(2a)=0$ . Reverse engineering the process, we can also get that $f\left(\frac{a}{2}\right)=0$ , similarly.

C2: Now I claim that if there exist two nonzero numbers such that $f(x)=f(y)=0$ , then $f(x-y)=0$ . This can be proved by setting $a=\frac{3y-x}{8}$ and $b=\frac{y-3x}{8}$ . Plugging in $(a,b)$ then gives that
$(f(a)+ab)f(x)+(f(b)+ab)f(y)=f(a+b)^2,$ $\iff 0=f(a+b)^2,$ $\iff f\left(\frac{x-y}{2}\right)=f\left(\frac{y-x}{2}\right)=f(a+b)=0,$ and by C1, we have that since $f\left(\frac{x-y}{2}\right)=0$ , we have that $f(x-y)=0$ , proving our claim.

C3: Finally, I claim that if $f(x)=0$ is not universal for all real numbers $x$ , then there cannot exist $a\in \mathbb{R}$ such that $a\neq 0$ and $f(a)=0$ . Assume FTSOC that there exists such an $a$ . Then let $x+y=a$ such that $f(x)\neq 0$ . By our assumption, this $x$ always exists.

First notice that since $f(x)\neq 0$ , this implies that $f(y)\neq 0$ by C2. Similarly by C2, we also get that $f(3x-y)\neq 0$ , since we have that if $f(3x-y)=0$ , we'd get
$f(a)=f(x+y)=0,$ $\iff f(a-(3x-y))=0,$ $\iff f(4x)=0 \iff f(2x)=0 \iff f(x)=0,$ using C1, a contradiction. Therefore, $f(3x-y)\neq 0$ . Similarly, we can also get that $f(x-3y)\neq 0$ . Using the fact that for all real numbers $r$ , we have that $f(r)=r^2$ , if we plug this (meaning $(x,y)$ ) into the large equation, we'd get that
$(x^2+xy)(x-3y)^2+(y^2+xy)(3x-y)^2=f(x+y)^2=f(a)^2=0,$ $\iff (x+y)(x)(x^2-6xy+9y^2)+(x+y)(y)(9x^2-6xy+y^2)=0,$ $\iff (x+y)(x^3-6x^2y+9xy^2+9x^2y-6xy^2+y^3)=0,$ $\iff (x+y)(x^3+3x^2y+3xy^2+y^3)=0,$ $\iff (x+y)(x+y)^3=a^4=0,$ a contradiction, since we assumed that $a\neq 0$ ! Therefore, our only solutions to this FE are $f(x)=0$ for all real $x$ or $f(x)=x^2$ for all real $x$ , finishing the problem.

This post has been edited 1 time. Last edited by peppapig_, Mar 13, 2024, 2:56 AM
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Bluesoul

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Bluesoul

#178 h Mar 18, 2024, 10:43 AM

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Denote the assertion as $P(x,y)$

When $x=y=0, 2f(0)^2=f(0)^2, f(0)=0$

We have $P(0,x)\implies f(x)f(-x)=f(x)^2, f(-x)=f(x)$ or $f(x)=0; P(x,x)\implies (f(x)+x^2)f(-2x)+(f(x)+x^2)f(2x)=2f(2x)(f(x)+x^2)=f(2x)^2\implies f(2x)=2(f(x)+x^2)$

Assertion $P(x,3x)\implies (f(x)+3x^2)f(8x)=f(4x)^2$

By the fact that $f(2x)=2(f(x)+x^2)$ , we can have $f(4x)=2(f(2x)+4x^2)=4f(x)+12x^2; f(8x)=2(f(4x)+16x^2)=8f(x)+56x^2$

Thus, we write $(f(x)+3x^2)f(8x)=f(4x)^2$ in terms of $f(x)$ , we attain $f(x)^2+2x^2f(x)-3x^4=0$ . We could solve $f(x)=x^2$ or $f(x)=-3x^2$ .

We have to get rid of the $f(x)=-3x^2$ since $P(1,1)$ doesn't work, $LHS=-48\neq 48=RHS$ , while it is easy to check $f(x)=x^2$ works

Thus, we have $\boxed{f(x)=0; f(x)=x^2}$ as solutions.

This post has been edited 1 time. Last edited by Bluesoul, Mar 18, 2024, 10:43 AM

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bjump

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bjump

#179 h Jun 26, 2024, 2:27 AM

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solution

Setting $x=y=0$ gives $2f(0)^2=f(0)^2$ or $f(0)=0$ . Now setting $x=0$ gives $f(y)f(-y)=f(y)^2$ which implies $f(y)=0$ or $f(y)=f(-y)$ , if $f(y)=0$ then $f(y)f(-y)=f(-y)^2$ which implies $f(-y)=0$ so $f(y)=f(-y)$ for all $y$ . Now setting $x=y$ gives $2(f(x)+x^2)f(2x)=f(2x)^2$ , which implies $f(2x)=2f(x)+2x^2$ or $f(2x)=0$ for each $x$ . Now setting $y=3x$ gives $(f(x)+3x^2)f(8x)=f(4x)^2$ , which means $f(8x)=0$ , only if $f(4x)=0$ , or $f(2x)=0$ only if $f(x)=0$ . Now if $f(4x) \neq 0$ then $f(4x)^2=(f(x)+3x^2)f(8x)=(\tfrac{f(2x)-2x^2}{2}+3x^2)(2f(4x)+32x^2)=\tfrac{f(4x)(f(4x)+16x^2)}{2}$ or $f(4x)=16x^2$ . Which implies for all $x$ , $f(x)$ is either $0$ or $x^2$ . To get rid of our pointwise trap suppose there exists a nonzero $a$ , such that $f(2a)=4a^2$ . Consider the substitution $x=a+b$ , $y=a-b$ then we obtain $(f(a+b)+a^2-b^2)f(4b-2a)+(f(a-b)+a^2-b^2)f(2(a-b))=16a^4$ . Then by casework and algebra on if $f(a+b)=0$ , or $(a+b)^2$ , $f(4b-2a)=0$ , or $(4b-2a)^2$ , $f(a-b)=0$ , or $(a-b)^2$ , we get that $f(x)=x^2$ for all $x$ . So we are finished, and I can sleep.

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ihatemath123

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ihatemath123

#180 h Sep 2, 2024, 10:39 PM

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The solutions are $f(x) = 0$ and $f(x) = x^2$ , both of which are somewhat easy to verify.

Taking $P(0,0)$ gives us $f(0) = 0$ – now, adding $P(0,x)$ and $P(0,-x)$ gives us $(f(x)-f(-x))^2 = 0$ , which implies that $f$ is even. Taking $P(x,-x)$ gives us
$f(4x)(f(x)-x^2) = 0, \qquad (\diamondsuit)$ while taking $P(x,x)$ gives us
$f(2x) (f(2x)-2f(x)-2x^2) = 0. \qquad (\heartsuit)$
Claim: We have $f(x) \in \{0, \tfrac{x^2}{2}, x^2 \}$ for all $x$ .
Proof: Assume that $x$ is a real which contradicts the claim. From $\heartsuit ( \tfrac{x}{2} )$ , we know that $f(\tfrac{x}{2}) \not \in \{0, \tfrac{x^2}{4} \}$ . Also, from $\diamondsuit( \tfrac{x}{4} )$ , we know that $f( \tfrac{x}{4} ) = \tfrac{x^2}{16}$ . But now, $\heartsuit (\tfrac{x}{4} )$ gives us
$f( \tfrac{x}{2} ) \left( f( \tfrac{x}{2} ) - \tfrac{x^2}{4} \right) = 0,$ which is a contradiction.

Claim: If $f(C)= 0$ for some nonzero $C$ , then $f(x) = 0$ for all $x$ .
Proof: Assume WLOG that $C$ is positive. Letting $x$ and $y$ be positive reals such that $x+y = C$ , $P(x,y)$ implies that both $f(x-3y)$ and $f(3x-y)$ are zero, since we know $f$ is nonnegative now. Since $x-3y$ takes values in the range $(-3C, C)$ , we can induct to show the claim.

On the other hand, if $f(x) \neq 0$ for all nonzero $x$ then $\diamondsuit (x)$ implies that $f(x) = x^2$ for all nonzero $x$ , so we're done.

If this is not a fakesolve, then this is unironically my favorite FE I've ever done. The second claim is super cool

This post has been edited 2 times. Last edited by ihatemath123, Sep 2, 2024, 10:45 PM

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Mathandski

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Mathandski

#182 h Mar 11, 2025, 9:45 PM

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Please contact westskigamer@gmail.com if there is an error with any of my solution for cash bounties by 3/18/2025.

This problem took me 3 hours during the mock...

Solution

We claim $f(x) \equiv 0$ or $f(x) \equiv x^2$ . Both can be checked to work.

1. Plug in $P(0, 0)$ to get $f(0)^2 + f(0)^2 = f(0)^2 \iff f(0) = 0$ .

2. Plug in $P(0, x)$ and $P(0, -x)$ to get $f(x) f(-x) = f(x)^2 = f(-x)^2$ . Clearly, $f(x) = f(-x)$ in all cases.

3. Plug in $P(x, -x)$ to get $(f(x) - x^2) f(4x) = 0$ so $f(x) = x^2$ or $f(4x) = 0$ for all $x$ .

4. Plug in $P(x, x)$ to get $2 (f(x) + x^2) f(2x) = f(2x)^2$ so $f(2x) = 0$ or $f(2x) = 2 f(x) + 2 x^2$ for all $x$ .

Claim: For all $x$ , we actually have $f(4x) \in \{ 0, (4x)^2 \}$ .

Proof: This is clearly true if $f(x) \neq x^2$ or $x = 0$ so assume non-zero $f(x) \neq x^2$ . Using (4), we have $f(2x) = 0$ or $4x^2$ .

If $f(2x) = 0$ , use (3) to get $f(8x) = 0$ . Now plug in $P(3x, x)$ to the equation to get,

$(f(x) + 3x^2) f(8x) = f(4x)^2$
Therefore, $f(4x) = 0$ ..

If $f(2x) = 4x^2$ , use (4) once again to get $f(4x) = 0$ or $f(4x) = 16x^2$ as desired. $\Box$

Equivalently, the claim tells us $f(x) \in \{0, x^2\}$ for all $x$ . We now resolve the point-wise trap. FTSOC, let $f(a) = 0$ and $f(b) = b^2$ for two $a, b \neq 0$ . Plug in $x = 3y + a$ to get,

$(f(y) + (3y+a) y) f(8y+3a) = f(4y+a)^2$
We take $y$ sufficiently large so that $(3y + a) y > 0$ . Note that the LHS is at most,
$(y^2 + (3y + a)y) (8y + 3a)^2$
If we expand, we see that, for all sufficiently large $y$ , this is less than $(4y+a)^4$ meaning $f(4y+a) = 0$ for all sufficiently large $y$ .

However, now plug in $x = b+ k$ and $y = -k$ . For sufficiently large $k$ , $f(x - 3y) = f(4k + b) = 0$ and $f(3x - y) = f(4k + 3b) = 0$ so the LHS is all zero. The RHS is $f(b)^2$ so $f(b) = b^2$ is impossible as desired.

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