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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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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
Cool Number Theory
Fermat_Fanatic108   1
N 19 minutes ago by Fermat_Fanatic108
For an integer with 5 digits $n=abcde$ (where $a, b, c, d, e$ are the digits and $a\neq 0$) we define the \textit{permutation sum} as the value $$bcdea+cdeab+deabc+eabcd$$For example the permutation sum of 20253 is $$02532+25320+53202+32025=113079$$Let $m$ and $n$ be two fivedigit integers with the same permutation sum.
Prove that $m=n$.
1 reply
Fermat_Fanatic108
20 minutes ago
Fermat_Fanatic108
19 minutes ago
ratio chasing inside a triangle, segment trisecting
parmenides51   10
N 22 minutes ago by sangsidhya
Source: CRMO 2012 Region 2 p5
Let $ABC$ be a triangle. Let $D, E$ be a points on the segment $BC$ such that $BD =DE = EC$. Let $F$ be the mid-point of $AC$. Let $BF$ intersect $AD$ in $P$ and $AE$ in $Q$ respectively. Determine $BP:PQ$.
10 replies
parmenides51
Sep 30, 2018
sangsidhya
22 minutes ago
Geo: incircle, escircle, isotomic conjugate
XAN4   0
22 minutes ago
Source: Own
For $\triangle{ABC}$, Its incircle $\odot I$ and $A-$escircle $\odot I_A$ are tangent to $BC$ at $D$ and $E$ respectively. $AI$ intersects line $BC$ at $J$. Line $AD$ intersects $\odot I$ at $F$, and line $AE$ intersects $\odot I_A$ at $G$. Line $FG$ intersects $BC$ at $H$. Prove that $BJ=CH$.
0 replies
XAN4
22 minutes ago
0 replies
f(x)+f(x+y) \leq f(xy)+f(y)
augustin_p   7
N 25 minutes ago by MuradSafarli
Source: Estonia TST 2022
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $$f(x)+f(x+y) \leq f(xy)+f(y).$$
7 replies
augustin_p
Jul 12, 2023
MuradSafarli
25 minutes ago
Integer equation in 3 variables
Kimchiks926   2
N 33 minutes ago by MuradSafarli
Source: Latvian TST for Baltic Way 2019 Problem 15
Determine all tuples of integers $(a,b,c)$ such that:
$$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$$
2 replies
Kimchiks926
May 29, 2020
MuradSafarli
33 minutes ago
Interesting inequality
sqing   2
N 36 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
2 replies
sqing
an hour ago
SunnyEvan
36 minutes ago
Interesting inequality
sqing   1
N 39 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-\frac{3}{2})(c^2-1) - \frac{9}{4}abc\geq -15$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - 2abc\geq -\frac{73}{6}$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - abc\geq -\frac{7}{2}$$$$(a^2-2)(b-\frac{3}{2})(c^2-2) - abc\geq -6$$
1 reply
sqing
an hour ago
sqing
39 minutes ago
Hard problem involving circumcenter and concurrent lines
GeoMetrix   6
N 40 minutes ago by bin_sherlo
Source: AQGO 2020 Problem 3
Let $\triangle{ABC}$ be a triangle with circumcenter $O$. Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$. Let $D$ be the projection of $A$ on $\overline{BC}$. Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$. Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$. Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$. Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix
6 replies
GeoMetrix
Jun 20, 2020
bin_sherlo
40 minutes ago
Oi! These lines concur
Rg230403   17
N an hour ago by L13832
Source: LMAO 2021 P5, LMAOSL G3(simplified)
Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$. Points $A_1, A_2$ are chosen on $\Gamma$, such that $AA_1 = AI = AA_2$, and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$. If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$.
Original Version from SL
Proposed by Mahavir Gandhi
17 replies
Rg230403
May 10, 2021
L13832
an hour ago
Find the period
Anto0110   1
N an hour ago by Anto0110
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
1 reply
Anto0110
Yesterday at 7:37 PM
Anto0110
an hour ago
inequality
senku23   2
N an hour ago by User21837561
Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).
2 replies
senku23
3 hours ago
User21837561
an hour ago
Differentiable functional
bakerbakura   2
N an hour ago by Gryphos
Find all differentiable functions $ f;\mathbb{R}\to\mathbb{R}$ such that, for all real numbers $ a,b,t$ with $ 0<t<1$, $ t^2f(a)+(1-t^2)f(b)\geq f(ta+(1-t)b)$
2 replies
bakerbakura
Jan 11, 2010
Gryphos
an hour ago
function equation
cipher703516247   2
N an hour ago by luutrongphuc

Find all the functions $f: \mathbb R^{+} \to \mathbb R^{+}$such that:
$$f(xy^n +f(y)^{2n}) )=f(x)f(y)^n +(yf(y))^n $$
2 replies
cipher703516247
Feb 14, 2025
luutrongphuc
an hour ago
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   78
N 2 hours ago by SimplisticFormulas
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
78 replies
EthanWYX2009
Jul 16, 2024
SimplisticFormulas
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
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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
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#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
11098 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
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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
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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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