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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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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
Find the number of integral solutions
Mathlover08092002   1
N 6 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 2 :-
What is the number of integral solutions of the equation $a^{b^2}=b^{2a}$, where a > 0 and $|b|>|a|$
[list=1]
[*] 3
[*] 4
[*] 6
[*] 8
[/list]
1 reply
Mathlover08092002
Apr 9, 2020
quasar_lord
6 minutes ago
What can you say about f(x)
Mathlover08092002   3
N 10 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 1 :-
Let $f : (0, \infty) \to \mathbb{R}$ is differentiable such that $\lim \limits_{x \to \infty} f(x)=2019$ Then which of the following is correct?
[list=1]
[*] $\lim \limits_{x \to \infty} f'(x)$ always exists but not necessarily zero.
[*] $\lim \limits_{x \to \infty} f'(x)$ always exists and is equal to zero.
[*] $\lim \limits_{x \to \infty} f'(x)$ may not exist.
[*] $\lim \limits_{x \to \infty} f'(x)$ exists if $f$ is twice differentiable.
[/list]
3 replies
Mathlover08092002
Apr 9, 2020
quasar_lord
10 minutes ago
FE on Stems
mathscrazy   4
N 18 minutes ago by SatisfiedMagma
Source: STEMS 2025 Category B4, C3
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in \mathbb{R}$, \[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]Proposed by Aritra Mondal
4 replies
mathscrazy
Dec 29, 2024
SatisfiedMagma
18 minutes ago
xyz(x+y+z)=4
RnstTrjyn   5
N 19 minutes ago by sqing
Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$. Prove that
$(x+y)^2+3(z+y)^2+(x+z)^2 \geq 8\sqrt7$
5 replies
1 viewing
RnstTrjyn
Feb 3, 2019
sqing
19 minutes ago
One inequality 1
prof.   2
N 20 minutes ago by invisibleman
If $x>0$ prove inequality $$\sqrt{x}\cdot (x+1)+x\cdot (x-4)+1\ge0.$$
2 replies
prof.
3 hours ago
invisibleman
20 minutes ago
Inequality
JK1603JK   1
N 21 minutes ago by CHESSR1DER
Source: unknown
Let a,b,c>=0 and ab+bc+ca>0 then prove \sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\ge 2\sqrt[4]{ab+bc+ca}+\sqrt{\frac{a(b-c)^2+b(c-a)^2+c(a-b)^2}{ab+bc+ca}}
1 reply
1 viewing
JK1603JK
Today at 2:58 AM
CHESSR1DER
21 minutes ago
geometry coordinates
CHESSR1DER   0
29 minutes ago
Source: simplified version of Belarus TST
Points $A, B, C$ with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points $D, E, F$ with integer coordinates such that $AB = DE$, $AC = DF$, $BC = EF$
0 replies
CHESSR1DER
29 minutes ago
0 replies
Cutting a big square into smaller squares
nAalniaOMliO   5
N 32 minutes ago by RagvaloD
Source: Belarusian National Olympiad 2020
A $20 \times 20$ checkered board is cut into several squares with integer side length. The size of a square is it's side length.
What is the maximum amount of different sizes this squares can have?
5 replies
nAalniaOMliO
Jan 29, 2025
RagvaloD
32 minutes ago
Ah yes, very interesting
Quidditch   23
N 40 minutes ago by quantam13
Source: EGMO 2024 P4
For a sequence $a_1<a_2<\cdots<a_n$ of integers, a pair $(a_i,a_j)$ with $1\leq i<j\leq n$ is called interesting if there exists a pair $(a_k,a_l)$ of integers with $1\leq k<l\leq n$ such that $$\frac{a_l-a_k}{a_j-a_i}=2.$$For each $n\geq 3$, find the largest possible number of interesting pairs in a sequence of length $n$.
23 replies
Quidditch
Apr 14, 2024
quantam13
40 minutes ago
Nice FE as the First Day Finale
swynca   3
N 43 minutes ago by PerfectPlayer
Source: 2025 Turkey TST P3
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$,
$$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
3 replies
swynca
Mar 18, 2025
PerfectPlayer
43 minutes ago
keep one card and discard the other
Scilyse   1
N an hour ago by g0USinsane777
Source: CGMO 2024 P2
There are $8$ cards on which the numbers $1$, $2$, $\dots$, $8$ are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let $S$ be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer $N$ for which Bob can guarantee that $S$ is at least $N$.
1 reply
Scilyse
Jan 28, 2025
g0USinsane777
an hour ago
Simultaneous eqs. with matrix
RenheMiResembleRice   7
N an hour ago by RenheMiResembleRice
Source: Ningyi Hou
Solve the attached with steps
7 replies
RenheMiResembleRice
4 hours ago
RenheMiResembleRice
an hour ago
Find the minimum
sqing   6
N an hour ago by sqing
Source: 2019 China Mathematical Olympiad Hope League Summer Camp
Let $x,y,z $ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+2(y+z)^2+3(z+x)^2.$
6 replies
1 viewing
sqing
Aug 10, 2019
sqing
an hour ago
Interesting inequality
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 0  . $ Prove that
$$ a^4+b^4 +kab\geq\left(\sqrt{k(k+2)}-k\right)ab(a+b+k)$$Where $ k>0 . $
$$ a^4+b^4 +ab\geq (\sqrt 3-1)ab(a+b+1)$$$$ a^4+b^4 +2ab\geq 2(\sqrt 2-1)ab(a+b+2)$$
2 replies
sqing
5 hours ago
sqing
2 hours ago
Functional Equation
anantmudgal09   19
N Nov 9, 2023 by IAmTheHazard
Source: India TST 2018 D1 P3
Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $$f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$$for all $x,y \in \mathbb{R}$.
19 replies
anantmudgal09
Jul 18, 2018
IAmTheHazard
Nov 9, 2023
Functional Equation
G H J
G H BBookmark kLocked kLocked NReply
Source: India TST 2018 D1 P3
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anantmudgal09
1979 posts
#1 • 3 Y
Y by Amir Hossein, megarnie, Adventure10
Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $$f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$$for all $x,y \in \mathbb{R}$.
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anantmudgal09
1979 posts
#2 • 7 Y
Y by Amir Hossein, sa2001, AopsUser101, YC1math, ashrith9sagar_1, Commander_Anta78, Adventure10
The only functions which work are the identity and the zero-function. It is clear that both satisfy the equation. Now we show any valid function that is not zero is the identity. Suppose $f(x_0) \ne 0$ for some $x_0 \in \mathbb{R}$.

If $f(0) \ne 0$ then $f(0)f(yf(0)-1)=-f(0)$ but $f \equiv -1$ is not a solution. So $f(0)=0$.

Put $x=x_0, y=0$ so $f(-1)=-1$. Put $y=x$ so $f(xf(x)-1)=x^2-1$ for all $x$; hence $f$ is surjective over $(-1, \infty)$. Suppose $f(a)=0$ then $x=a, y=x_0$ implies $a=0$. So $f$ is injective at $0$.

Now $f(f(1)-1)=0$ so $f(1)=1$. Put $x=1$ to get $f(y-1)=f(y)-1$ for all $y$ so $f(y-N)=f(y)-N$ for all integers $N \ge 0$. Now $f(y)=f(y+N)-N$ so $f(y+N)=f(y)+N$ for all integers $N \ge 0$. Consequently, $f$ is surjective over all of $\mathbb{R}$.

Plug $f(yf(x)-1)=f(yf(x))-1$ to conclude $f(x)f(yf(x))=x^2f(y)$. Put $y=1$ so $x^2=f(x)f(f(x))$ hence $f(yf(x))=f(y)f(f(x))$ for all $x \ne 0$. For $x=0$, last claim is obvious. Now surjectivity shows $f(ty)=f(t)f(y)$ for all $t,y \in \mathbb{R}$ so $f$ is multiplicative.

Thus, $f(z^2)=f(z)^2 \ge 0$ and $f(-z^2)=f(-1)f(z)^2=-f(z)^2 \le 0$ hence $f$ preserves the sign of the argument. Now pick $x>1$ and so $1 \ge x-\lfloor x \rfloor \ge 0$ hence $f(x-\lfloor x \rfloor) \ge 0$ and $f(x-\lfloor x \rfloor-1) \le 0$ hence $$\lfloor x \rfloor \le f(x) \le \lfloor x \rfloor+1$$so $$x-1 \le f(x) \le x+1$$for all $x>1$. Now suppose $f(t_0) \ne t_0$ for some $t_0>1$.

1. If $f(t_0)>t_0$.

Now for $y>1$ sufficiently large, $$yf(t_0)^2-2f(t_0) \le f(t_0)f(yf(t_0)-1)=t_0^2f(y)-f(t_0) \le yt_0^2+t_0^2-f(t_0)$$hence $$y \le \frac{t_0^2+f(t_0)}{f(t_0)^2-t_0^2}$$which fails as $y \rightarrow \infty$.

2. If $f(t_0)<t_0$.

Again for $y>1$ sufficiently large, $$yf(t_0)^2 \ge f(t_0)f(yf(t_0)-1)=t_0^2f(y)-f(t_0) \ge yt_0^2-t_0^2-f(t_0)$$hence $$y \le \frac{t_0^2+f(t_0)}{-f(t_0)^2+t_0^2}$$which fails as $y \rightarrow \infty$.

Finally, shifting down by large integers $N$, we obtain $f(x)=x$ for all $x \in \mathbb{R}$ as desired. $\blacksquare$
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falantrng
249 posts
#3 • 3 Y
Y by Amir Hossein, ImbecileMathImbaTation, Adventure10
Easy for P3 (if problems order in difficult)and simple solution.
Let $P(x,y)$ be the assertion of $f(x)f(yf(x)-1)=x^2f(y)-f(x)$
$P(0,0)\to f(0)\cdot f(-1)=-f(0),$ then $1) f(0)=0,$ or $f(-1)=-1.$
$1)$ $P(x,0)\to f(x)\cdot f(-1)=-f(x).$
$1.1)$ For all $x,$ $f(x)\equiv 0,$ Indeed this solution work.
$1.2)$ $\exists a\in\mathbb{R} ,$ such that $f(a)\not= 0,$ them from $P(a,0)\to f(-1)=-1.$
$2)$ $P(-1,0)\to f(0)=0.$
Then from both condition we can get $f(0)=0,f(-1)=-1.$
Lemma: $f(a)\equiv 0 \iff a\equiv 0.$
Proof: we know $f(0)=0,$ let show $f(a)=0\to a=0.$
$P(a,-1)\to -a^2=0\to a=0.$ As desired.
From $P(1,1)\to f(1)\cdot f(f(1)-1)=0.$
$2.1)$ $f(1)=0.$
From $P(1,x)\to $ for all $x,$ $ f(x)=0,$ but $f(-1)=-1.$ contradiction.
$2.2)$ $f(1)\not= 0\to f(f(1)-1)=0.$ From lemma we get $f(1)=1.$
Then $P(1,x+1)\to f(x+1)=f(x)+1,$ (or $f(x-1)=f(x)-1.$)
Then our equation equivalent to $Q(x,y):f(x)\cdot f(yf(x))=x^2f(y).$
From $Q(x,1)\to f(x)\cdot f(f(x))=x^2.$
Also from $Q(x+1,1)\to (f(x)+1)\cdot f(f(f(x))+1)=f(x)\cdot f(f(x))+f(x)+f(f(x))+1=x^2+2x+1,$ then we get $f(f(x))=2x-f(x).$
From $f(x)\cdot f(f(x))=2x,$ use $f(f(x))=2x-f(x),$ we get $(x-f(x))^2=0.$
Then $f(x)\equiv x,$ for all $x.$ Indeed this solution work.
This post has been edited 2 times. Last edited by falantrng, Dec 28, 2018, 8:50 AM
Reason: Typo
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math_pi_rate
1218 posts
#4 • 4 Y
Y by rocketscience, mijail, Adventure10, Mango247
Nice problem. Here's my solution: Let $P(x,y)$ denote the given assertion. Note that the only constant function which satisfies the given equation is the zero function. So from now on we assume that $f$ is non-constant.

CLAIM For $c \in \mathbb{R}$, we have $f(c)=0 \Leftrightarrow c=0$.

Proof: Note that we have $$P(-1,0) \Rightarrow f(-1)^2=f(0)-f(-1) \Rightarrow f(-1)(f(-1)+1)=f(0)$$Also, $$P(0,0) \Rightarrow f(0)f(-1)=-f(0) \Rightarrow f(0)=0 \text{ OR } f(-1)=-1$$If $f(0)=0$, then using the first relation, we get that either $f(-1)=0$ or $f(-1)=-1$. But, when $f(-1)=0$, we have $P(x,0) \Rightarrow f(x)=0$, which contradicts the fact that $f$ is non-constant. That means $f(0)=0$ implies $f(-1)=-1$. And, when $f(-1)=-1$, then (again from the first relation), we get $f(0)=0$. Summarizing the above, we can say that $f(0)=0$ and $f(-1)=-1$ are both true simultaneously.

Now, suppose that $f(c)=0$. Then $P(c,-1) \Rightarrow c^2f(-1)=0 \Rightarrow c=0$. $\Box$

Return to the problem at hand. Then we get $P(x,-1) \Rightarrow f(x) \cdot f(-f(x)-1)=-x^2-f(x)$

$$P(-1,f(x)) \Rightarrow -f(-f(x)-1)=f(f(x))+1 \Rightarrow f(x) \cdot f(-f(x)-1)=-f(x) \cdot (f(f(x))+1)$$$$\Rightarrow x^2+f(x)=f(x) \cdot (f(f(x))+1) \Rightarrow f(f(x)) \cdot f(x)=x^2 \text{ } (*)$$
Now, $P(1,1) \Rightarrow f(1) \cdot f(f(1)-1)=0$. Using our Claim, we get that $f(1)-1=0 \Rightarrow f(1)=1$ $($as $f(1) \neq 0)$. Then $$P(1,y) \Rightarrow f(y-1)=f(y)-1 \Rightarrow \text{ By an easy induction, }f(y+n)=f(y)+n \text{ } \forall n \in \mathbb{N}$$
Finally, Putting $x \Longrightarrow x+n$ in $(*)$, and using the above relation, we get that $$(f(f(x))+n)(f(x)+n)=x^2+2nx+n^2=f(f(x)) \cdot f(x)+2nx+n^2$$$$\Rightarrow n(f(f(x))+f(x))=2nx \Rightarrow f(f(x))+f(x)=2x \text{ } (**)$$
Using $(*)$ and $(**)$, we can easily find that $f(f(x))=f(x)=x$. Thus, our final answers are $$\boxed{f \equiv 0 \text{ AND } f(x)=x \text{ } \forall x \in \mathbb{R}}$$Now one can easily verify that these solutions actually work. Hence, done. $\blacksquare$
This post has been edited 2 times. Last edited by math_pi_rate, Dec 20, 2018, 11:10 AM
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bruckner
106 posts
#5 • 2 Y
Y by Adventure10, Mango247
I think this solution is more elementary than the others, although many parts are similar to the given above.
Solution
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math_pi_rate
1218 posts
#6 • 2 Y
Y by Adventure10, Mango247
Sorry for double posting, but I have a different ending after proving the claim in post #4. Here's my solution: As proved in post #4, we have $f(0)=0,f(-1)=-1$ and that $f$ is non-constant. Now, $$P(-1,y) \Rightarrow -f(-y-1)=f(y)+1 \overset{y \rightarrow -y}{\Longrightarrow} f(y-1)=-f(-y)-1$$Again, as shown in post #4, one easily gets that $f(y-1)=f(y)-1$. Thus, we have $f(-y)=-f(y)$, i.e. $f$ is odd. We will show that $f$ is injective also. Suppose we have $f(a)=f(b)$ for some $a,b \in \mathbb{R}$. Then $$P(a,-1)-P(b,-1) \Rightarrow a^2=b^2 \Rightarrow b=a \text{ OR } b=-a$$However, we cannot have $f(-a)=f(a)$, as $f$ is odd. Hence, $f$ must be injective.

Now, as $f(y-1)=f(y)-1$, so we can rewrite the problem condition as $$P(x,y):= f(x)(f(yf(x))-1)=x^2f(y)-f(x) \Rightarrow f(x)f(yf(x))=x^2f(y)$$Then, $P(x,x)$ gives that $f(xf(x))=x^2$. This means that $f$ is surjective for $x>0$. But, as $f(-x)=-f(x)$, so we get that $f$ is in fact always surjective. Now, using injectivity, one can easily prove that $f(1)=1$, and so we have $$P(x,1) \Rightarrow x^2=f(x)f(f(x)) \Longrightarrow f(x)f(yf(x))=x^2f(y)=f(x)f(f(x))f(y) \Rightarrow f(yf(x))=f(f(x))f(y)$$However, as $f$ is surjective, so we can take $f(x)=z$, giving that $f$ is multiplicative also. Let $w=\frac{1}{f(x)}$ (As $f$ is surjective, so $\frac{1}{f}$ is also surjective). Then, using multiplicity, we have $$P \left(y+\frac{1}{f(x)},x \right) \Rightarrow x^2f(y+w)=f(x)f(yf(x)+1)=f(x)(f(yf(x))+1)= x^2f(y)+f(x)$$$$\Rightarrow \text{ As } x^2=f(x)f(f(x)) \text{, we get that } f(f(x))(f(y+w)-f(y))=1$$But, $f(x)=\frac{1}{w}$, and so we get $$f(y+w)-f(y)=\frac{1}{f \left(\frac{1}{w} \right)}=f(w) \Rightarrow f(y+w)=f(y)+f(w)$$where we use that $f \left(\frac{1}{w} \right) f(w)=f(1)=1$. This means that $f$ is both additive and multiplicative, in which case it is well known that $f$ must be the identity function. Hence, done. $\blacksquare$
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Mathotsav
1508 posts
#8 • 2 Y
Y by Adventure10, Mango247
Nice problem. Here is my solution:
Answer
Solution
Edit: Just realised that my solution is similar to @2above.
This post has been edited 7 times. Last edited by Mathotsav, May 22, 2019, 11:45 AM
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gnoka
245 posts
#9 • 1 Y
Y by Adventure10
Good problem. Nice solutions. Learned much.
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Physicsknight
635 posts
#13 • 1 Y
Y by Adventure10
Nice problem :)
Solution
$f (x)\equiv 0$ if $f (x)\neq 0$ for some $x $.

If $f (x)=f (y) $ then $x=y $ or $x=-y $
$P (0,y)\implies f (0)=0$ or $f (yf (0)-1)=-1\implies f (0)=0$
$P (x,0)\implies f (x)f (-1)=-f (x)\implies f (-1)=-1$
$P (-1,y)\implies -f(-y-1)=f (y)+1$

$\text {Restate the original equation as} $

$f (x)f (-yf (x))=x^2f (y)\implies f (x)f (f (x))=x^2$
$P (x,x): f (-xf (x))=x^2=-f (xf (-x))f (x)+f (-x)=0$
$f (y+1)=f (y)+1 f (x+1)+f (f (x+1))=(x+1)^2$
$f (x)+f (f (x))=2x $, and $f (x)=x $
Hence, $f(x)=0$ $\forall x\in\mathbb R $ and $f(x)=x $. Easy to verify.
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TinTin028
19 posts
#14 • 2 Y
Y by AlastorMoody, Adventure10
Bonus for a TST P3!
anantmudgal09 wrote:
Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $$f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$$for all $x,y \in \mathbb{R}$.

Answers: $f\equiv 0 $ and $f \equiv x \ \forall x \in \mathbb{R}$.
Clearly both these functions satisfy, so we continue by proving them to be the only ones. Denote by $(x,y)$ the assertion $f(x)\cdot f(yf(x)-1) = x^2f(y)-f(x)$.
$(x,x) \implies f(x)f(xf(x)-1) = f(x)(x^2-1) ... (1)$. Now, $(0,y) \implies f \equiv 0$(which is a solution) or $f(0)=0$. Assuming $f$ to be non-constant,
take $f(0)=0$. Now suppose there exists $x_0 \neq 0 \in \mathbb{R}$ such that $f(x_0) =0$. Then, $(x_0,y) \implies x_0^2 \cdot f(y) = 0 \iff x_0 = 0$, contradiction and thus $f$ is injective at $0 \implies f(xf(x) -1) = x^2-1 \implies f$ is surjective over $[-1,\infty)$.
Now, $(x,1) \implies f(x) f(f(x)-1) = x^2 - f(x)$. We now show that $f$ is injective. Indeed, if $f(a) =f(b)$, then $a^2 = b^2$. So suppose $a+b=0$. Then we get that $$ f(af(x)-1) = f(-af(x)-1) \forall x \in \mathbb{R}$$. Using surjectivity, pick $\alpha$ such that $f(\alpha) = \frac{1}{a}$. Then we have $f(0)=f(2)$, contradiction $\implies f$ is injective. Now using similar arguments get that $f(x) = -f(-x)$ which implies $f$ is a bijection on $\mathbb{R}$. Then $(x,f(y)) \implies x^2 \cdot f(f(y)) \cdot f(y) = y^2 \cdot f(x) \cdot f(f(x))$ and using $f(1)=1$ (obtainable from $(1,1)$) we establish $f(x) f(f(x)) = x^2$ which in turn implies $f(x+1) = f(x) + 1 \forall x$ and moreover, $f(x) f(yf(x)) = x^2 f(y) = f(x)f(f(x)) f(y) \implies f$ is multiplicative $\implies f(y)f(x+1) = f(y)f(x) + f(y)f(1) \iff f(xy+y) = f(xy) + f(y) \implies f$ is Cauchy and multiplicative and thus $f$ is identity.
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aops29
452 posts
#15 • 1 Y
Y by Adventure10
Not that hard for P3
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Keith50
464 posts
#16 • 1 Y
Y by megarnie
$\clubsuit \color{magenta}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}$.
$\blacklozenge \color{blue}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the given FE. Let $P(x,y)$ be the given assertion, we have \[P(x,0): f(x)(f(-1)+1)=x^2f(0)\]if $f(-1)\ne -1$, we will have $f(x)=cx^2$ for some real number $c$, plugging it into our FE, \[c^4x^6y^2+\ldots =c^2x^2(cx^2y-1)^2=cx^2y^2-cx^2\]which means $c=0$ and $f(x)=0$ for all real $x$. If $f(-1)=-1$, then $f(0)=0$. If there exists a real number $u$, such that $f(u)=0$, then \[P(u,x): u^2f(x)=0 \quad \forall x\in \mathbb{R}\]which if $f(x)\ne 0 $ for all real $x$, then $u=0$. So, $f$ is injective at $0$ and $f$ is not the zero function. Then, \[P(x,x): f(xf(x)-1)=x^2-1 \quad \forall x\ne 0\]but since $f(0)=0$, $ f(xf(x)-1)=x^2-1 \quad \forall x\in \mathbb{R}$. Now, by comparing $P(x, 1)$ and $P(1,f(x))$, we have \[\frac{x^2-f(x)}{f(x)}=f(f(x))-1 \implies f(x)f(f(x))=x^2 \quad \forall x\ne 0\]but again $f(0)=0$, so this also holds for all real $x$. Plugging this back to our original FE, we have \[P(x,y):f(yf(x)-1)=f(f(x))f(y)-1\]and \[P(1,x+1): f(x+1)=f(x)+1.\]Most importantly, $f$ is odd since \[P(-1,x): f(-x)-1=f(-x-1)=-f(x)-1 \implies f(-x)=-f(x) \quad \forall x\in \mathbb{R}.\]Therefore, since $f$ is odd and $f(xf(x)-1)=x^2-1$, $f$ is surjective over $\mathbb{R}$. This means \[P(x,f(y)): f(f(x)f(y))=f(f(x))f((y)) \implies f(xy)=f(x)f(y)\]$f$ is multiplicative and in particular, $f(x^2)=f(x)^2$. Finally, as \[\begin{cases} f(x+1)=f(x)+1 \\ f(x^2)=f(x)^2, \end{cases}\]this implies $f(x)=x$ (it's a well-known FE). $\quad \blacksquare$
This post has been edited 1 time. Last edited by Keith50, Mar 21, 2021, 5:25 AM
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Wizard0001
336 posts
#17
Y by
anantmudgal09 wrote:
Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $$f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$$for all $x,y \in \mathbb{R}$.
Easy, but nice.
Let $P(x,y)$ denote the given assertion. $f \equiv 0$ is obviously a solution. Henceforth, assume that $f \not \equiv 0$. Now $$P(1,1): f(1)f(f(1)-1)=0 \implies \exists c \quad \text{such that} f(c)=0$$$$P(c,y): c^2f(y)=0$$since $f \not \equiv 0$, we conclude that $f(t)=0 \iff t=0$. So $f(1)-1=0 \implies f(1)=1$. Observe that $$P(x,1): f(f(x)-1)= \frac{x^2}{f(x)}-1 \quad (i)$$and $$P(1,f(x)): f(f(x)-1)=f(f(x))-1 \quad (ii)$$On comparing $(i)$ and $(ii)$ we have that $f(x)f(f(x))=x^2 \forall x \in \mathbb{R} \quad (iii)$. Now note that for all $x\not = 1,-1$ $$P(x,x): f(xf(x)-1)=x^2-1 \implies f(x^2-1)=f(f(xf(x)-1))=\frac{(xf(x)-1)^2}{f(xf(x)-1)}=\frac{(xf(x)-1)^2}{x^2-1} (iv)$$$$P(f(x),f(x)): f(f(x)f(f(x))-1)= f(x)^2-1 \overset{\text{using} (iii)}{\implies} f(x)^2-1=f(x^2-1)= \frac{(xf(x)-1)^2}{x^2-1}$$Upon simplification, the above result is equivalent to $(f(x)-x)^2=0 \implies f(x)=x$. Also $P(x,0): f(-1)=-1$. Hence we have two solutions, i.e.
$$f(x)=x \quad \forall x \in \mathbb{R}$$$$f(x)=0 \quad \forall x \in \mathbb{R}$$Hence, we are done.
Note
This post has been edited 4 times. Last edited by Wizard0001, Apr 25, 2021, 5:04 PM
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jasperE3
11100 posts
#18 • 2 Y
Y by megarnie, kido2006
$\boxed{f(x)=0}$ works. Assume now that $\exists j:f(j)\ne0$.

$P(1,1)\Rightarrow f(1)f(f(1)-1)=0\Rightarrow\exists k:f(k)=0$
$P(k,j)\Rightarrow k^2f(j)=0\Rightarrow k=0$ (injectivity at $0$)

Either $f(1)=0$ or $f(f(1)-1)=0$. If $f(1)=0$ then $1=0$, absurd, thus $f(1)-1=0\Rightarrow f(1)=1$.

$P(1,x+1)\Rightarrow f(x+1)=f(x)+1$
The assertion becomes $Q(x,y):f(x)f(yf(x))=x^2f(y)$.
$Q(x,1)\Rightarrow f(x)f(f(x))=x^2$
$Q(x+1,1)\Rightarrow f(x)+f(f(x))=2x$
So $f(f(x))=2x-f(x)$, thus $f(x)(2x-f(x))=x^2$, which factors as $(x-f(x))^2=0$, hence $\boxed{f(x)=x}$, which is the only remaining solution.
This post has been edited 3 times. Last edited by jasperE3, Apr 25, 2021, 5:57 PM
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Project_Donkey_into_M4
136 posts
#19 • 1 Y
Y by HoRI_DA_GRe8
Solution
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douglasmorales
21 posts
#20
Y by
Hello, the equation in hand is f (x) f (y f(x) - 1) = x ^ 2 f (y) - f (x)
Assume there is some other function since f (x) =0
x = y = 0 => f (0) f (-1) = -f (0) => f(0) = 0 /f(-1) = -1
Taking f (0) = 0 and substitute y = 0
=> f (-1) = -1
Taking f (-1) = -1 and substitute y = 0
=> f (0) = 0
This proves that both statements provided before are equivalent
Considering that there exists an a which is not equal to 0 such that f (a) = 0,
Substituting x = a and y = -1. This will eliminate the case.
Substituting P (x, y)
P (-1, y) => -f (-y -1) = f(y) + 1 => f (y-1) = -f (-y) -1


Now substitute this in the original equation available to get
F (x) f (f (x)) = x^2
P(1, 1)
f(1) f (f(1) -1) = 0
f(1) = 1
Then, P (1,y) = f; This is an odd function

Use f (x +1) = f(x) +1 to get rid of the 1


Put x = x + 1, y = 1 to get
F (f(x)) = 2 x - f (x)

Now substitute f (x) f (f(x)) = x ^2 with f (f (x)) = 2x - f (x)
(f(x) - x) ^ 2 = 0
f(x) = x


Working functions: f (x) = 0 and f (x) =x
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mathscrazy
113 posts
#21
Y by
We will prove that $f\equiv0$ and $f(x)=x$ are the only solutions.
Let $P(x,y)$ be the assertion in $f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x)$.
It can be easily seen that the only constant function that works is $\boxed{f \equiv 0}$, which is our first solution.
Let $f$ be non-constant.
Claim 1 : $f(0)=0$.
Proof :
$P(0,y) : f(0)f(yf(0)-1)=-f(0)$
Assume possible $f(0)\neq0$.
Then, $f(yf(0)-1)=-1$.
As $f(0)\neq0$, as we vary $y$ over $\mathbb{R}$, $yf(0)-1$ also varies over $\mathbb{R}$.
Hence, $f(x)=-1 \forall x $.But $f\equiv-1$ doesn't satisfy given equation.
Contradiction!
Hence proved claim 1!
Claim 2 : $f$ injective.
Proof :
Let $f(a)=f(b)$.
Let $c$ be such that $f(c)\neq0$.
$P(a,c)-P(b,c) : a^2=b^2 \implies a = \pm b$.
Hence, $f(a)=f(b) \implies a= \pm b ...(1)$.
Assume that $a=-b$ ($a,b \neq 0$) is possible.
$P(c,a)-P(c,b) : f(af(c)-1)=f(bf(c)-1) \overset{(1)}{\implies} af(c)-1= \pm (bf(c)-1) \overset{a=-b}{\implies} af(c)-1= \pm (-af(c)-1)$.
If $af(c)-1=-af(c)-1$, we get $a=0$. Contradiction.
If $af(c)-1=-(-af(c)-1)$, we get $-1=1$. Contradiction!
Hence, if $f(a)=f(b)$, $a=-b$ is not possible.
Combining with $(1)$, we get $f(a)=f(b)\implies a=b$.
Hence proved claim 2!

Hence note that , $f(x) \neq 0 \forall x\neq 0$.
Claim 3 : $f(1)=1$.
Proof :
$P(1,1) : f(1)\cdot f(f(1)-1)=0 \implies f(f(1)-1)=0 \implies f(1)-1=0 \implies f(1)=1$.
(Here we used claim 1 and claim 2)
Hence Proved claim 3!
Claim 4 : $f(-1)=-1$.
Proof :
$P(x,0) : f(x)f(-1)=-f(x) \implies f(-1)=-1$. (Here we used claim 1 and claim 2)
Hence proved claim 4!
Claim 5 : The given equation reduces to $f(x)f(yf(x))=x^2f(y)$.
Proof :
$P(1,y) : f(y-1)=f(y)-1...(2)$. (Here we used claim 3)
$y\rightarrow yf(x) : f(yf(x)-1)=f(yf(x))-1$.
Substituting this in given equation, we get $$f(x)f(yf(x))=x^2f(y)$$.
Hence proved claim 4!

Let $R(x,y)$ be the assertion in $f(x)f(yf(x))=x^2f(y)$.
Claim 6 : $f(x+1)=f(x)+1$ and $f(x^2)=f(x)^2$.
Proof :
$(2) : f(y)=f(y-1)+1 \implies f(y+1)=f(y)+1$.

$R(f(x),f(x)) : f(f(x)f(f(x)))=f(x)^2$.
$R(x,1) : f(x)f(f(x))=x^2 \implies f(f(x)f(f(x))=f(x^2)$.
Hence, $f(x)^2=f(x^2)$ (From last two equations).
Hence proved claim 6!
From claim 6, its well-known that ${f(x)=x}$ is the solution.
$\boxed{f(x)=x}$ indeed works and is our second solution!
Hence, we are done :D
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megarnie
5532 posts
#22
Y by
Let $P(x,y)$ denote the given assertion.

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

So either $f(0)=0$ or $f(xf(0)-1)=-1$ for any $x$. If $f(0)\ne 0$, then $xf(0)-1$ can take on any real value, which implies $f\equiv -1$, which is not a solution. Thus, we have $f(0)=0$.

Now, noting $\boxed{f\equiv 0}$ works, we can assume $f$ is non-constant.

Claim: $f$ is injective.
Proof: Suppose $f(a)=f(b)$ with $a\ne b$.

$P(a,x): f(a)f(xf(a)-1)=a^2f(x)-f(a)$.

$P(b,x): f(b)f(xf(b)-1)=b^2f(x)-f(b)$.

This implies $a^2f(x)=b^2f(x)$. If we set $x$ such that $f(x)\ne 0$, then $a^2=b^2\implies a=\pm b\implies a=-b$, since $a\ne b$. Then $f(a)=f(-a)$. In fact, this implies $f$ is injective at $0$.

$P(a,-a): f(a)f(-af(a)-1)=a^2f(a)-f(a)$.

$P(-a,a): f(a)f(af(a)-1)=a^2f(a)-f(a)$.

If $f(a)\ne 0$, then we have $f(-af(a)-1)=f(af(a)-1)$. However, this implies either $af(a)+1=af(a)-1$, or $-af(a)-1=af(a)-1$, both are absurd. So $f(a)\ne f(-a)$.

If $f(a)=0$, then $a=0$.

Since $f$ is injective at $0$, $f$ is injective. $\blacksquare$

$P(x,0): f(x)f(-1)=-f(x)$. If we set $x\ne 0$, then we get $f(-1)=-1$.

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

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

Now we rearrange the FE.

We have $f(x)(f(yf(x))-1)=x^2f(y)-f(x)$, so \[f(x)f(yf(x))=x^2f(y)\]
Let $Q(x,y)$ be the assertion here.

$Q(x,1): f(x)f(f(x))=x^2\implies f(f(x))=\frac{x^2}{f(x)}$.

$P(x+1,1): (f(x)+1)f(f(x))=(x+1)^2-f(x)-1=x^2+2x-f(x)\implies f(f(x))=\frac{x^2+2x-f(x)}{f(x)+1}$.

This implies\begin{align*}
\frac{x^2}{f(x)}=\frac{x^2+2x-f(x)}{f(x)+1} \\
\implies x^2(f(x)+1)=f(x)(x^2+2x-f(x)) \\
\implies x^2f(x)+x^2=x^2f(x)+2xf(x)-f(x)^2 \\
\implies x^2=2xf(x)-f(x)^2 \\
\implies f(x)^2-2xf(x)+x^2=0 \\
\implies (f(x)-x)^2=0 \\
\implies \boxed{f(x)=x} \\
\end{align*}which clearly works.
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ZETA_in_olympiad
2211 posts
#23
Y by
Also Singapore 2015 aops.com/community/c6h1618668p25462682
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IAmTheHazard
5000 posts
#24
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The answer is $f(x)=x$ and $f \equiv 0$ only, which both work. Hence suppose $f \not \equiv 0$. Let $P(x,y)$ denote the assertion.

From $P(x,0)$ we have $f(x)(f(-1)+1)=x^2f(0)$. If $f(-1)+1 \neq 0$, then $f(x)=cx^2$ for some nonzero constant $c \in \mathbb{R}$. But this clearly doesn't work, so $f(-1)=-1$ and thus $f(0)=0$. Then from $P(-1,y)$ we obtain $f(-y-1)=-f(y)-1$.

If $f(x)=0$ for some $x$, by picking $y$ such that $f(y) \neq 0$ we find that $x=0$. Then for $x \neq 0$, $P(x,x)$ gives $f(xf(x)-1)=x^2-1$, hence the range of $f$ contains $(-1,\infty)$. Since $f(-y-1)=-f(y)-1$, the range of $f$ also contains $(-\infty,1)-1$, so $f$ is surjective. Additionally, $P(-1,-1)$ implies $f(-2)=-2$, hence $-f(1)-1=f(-2) \implies f(1)=1$.

Using $f(-y-1)=-f(y)-1$, from $P(x,-1)$ we obtain $f(x)(f(-f(x)-1)+1)=-x^2 \implies f(x)f(f(x))=x^2$. Thus we can rewrite the assertion as $f(yf(x)-1)=f(f(x))f(y)$. From surjectivity, this becomes $f(xy-1)=f(x)f(y)-1$. Comparing $(x,y)$ with $(xy,1)$ and using $f(1)=1$ implies that $f$ is multiplicative. On the other hand, by plugging in $y=1$ we also get $f(x-1)=f(x)-1$.

Since $f$ is multiplicative, we have $f(x^2)=f(x)^2>0$ for all $x>0$, hence $f$ sends positive reals to positive reals. Thus let $g(x)=\log f(e^x)$ sending real numbers to real numbers, so $g$ is additive (by only considering the multiplicativity of $f$ over $\mathbb{R}^+$). On the other hand, $f(x-1)=f(x)-1$ implies that $f(x)>1$ for $x>1$, hence $g(x)>0$ for $x>0$. Therefore $g$ is linear, so $f(x)=x^k$ for some $k \in \mathbb{R}$. Then we have $2^k-1=1$, hence $k=1$ and $f(x)=x$ as desired. $\blacksquare$
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