## FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
nimon
Posts:8
Joined:Sat Mar 27, 2021 7:30 am
Re: FE Marathon!
can you suggest me any book for solving this kind of function problem.

Anindya Biswas
Posts:264
Joined:Fri Oct 02, 2020 8:51 pm
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### Re: Solution to P24

~Aurn0b~ wrote:
Tue Mar 30, 2021 11:39 am
$\textbf{Problem 24}$

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
$f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.$
Warning : Don't click, graphical violence ahead...
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Anindya Biswas
Posts:264
Joined:Fri Oct 02, 2020 8:51 pm
Contact:

### Problem 25

Find functions $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)+f(f(n))+f(f(f(n)))=3n\ \ \ \forall n\in\mathbb{N}$
Where $\mathbb{N}$ is the set of all positive integers
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

### Re: Problem 25

Anindya Biswas wrote:
Sat Apr 03, 2021 9:54 pm
Find functions $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)+f(f(n))+f(f(f(n)))=3n\ \ \ \forall n\in\mathbb{N}$
Where $\mathbb{N}$ is the set of all positive integers
Again 2 days passed no solution
Hmm..Hammer...Treat everything as nail

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

### Re: Problem 25

Anindya Biswas wrote:
Sat Apr 03, 2021 9:54 pm
Find functions $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)+f(f(n))+f(f(f(n)))=3n\ \ \ \forall n\in\mathbb{N}$
Where $\mathbb{N}$ is the set of all positive integers
$\textbf{Solution 25}$

Anindya Biswas
Posts:264
Joined:Fri Oct 02, 2020 8:51 pm
Contact:

### Re: Problem 25

~Aurn0b~ wrote:
Sat Apr 10, 2021 10:31 pm
Anindya Biswas wrote:
Sat Apr 03, 2021 9:54 pm
Find functions $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)+f(f(n))+f(f(f(n)))=3n\ \ \ \forall n\in\mathbb{N}$
Where $\mathbb{N}$ is the set of all positive integers
$\textbf{Solution 25}$
Yeah, but this last statement can be nicely proven by induction.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

### Re: FE Marathon!

$\textbf{Problem 25}$

Find all functions $f:\mathbb Z\rightarrow \mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$, the following equality holds:
$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$(Here $\mathbb{Z}$ denotes the set of integers.)

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

### Problem 26

~Aurn0b~ wrote:
Sun Apr 11, 2021 10:10 am
$\textbf{Problem 25}$

Find all functions $f:\mathbb Z\rightarrow \mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$, the following equality holds:
$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$(Here $\mathbb{Z}$ denotes the set of integers.)
For the sake to continue the marathon, This is 2012 IMO P4. I am posting the next problem.
Problem 26
Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x^2-y^2)=xf(x)-yf(y)$
Hmm..Hammer...Treat everything as nail

Zeta
Posts:4
Joined:Fri Mar 18, 2022 2:26 am

### Re: FE Marathon!

Let $P(x,y)$ denote the original equation.
$P(0,0)\Rightarrow f(0)=0.$
$P(-x,y)\Rightarrow f(x)=-f(x) \Rightarrow f \text{ is odd.}$
WLOG $x,y \geq 0.$
$P(x,0)\Rightarrow f(x^2)=xf(x) \Rightarrow f(x^2-y^2)=f(x^2)-f(y^2)$
$\Rightarrow f(x^2)=f(x^2-y^2)+f(y^2).$
$\exists z\geq 0: z^2=x \Rightarrow f(x)=f(x-y)+f(y).$
$P(2z,z)\Rightarrow f(2z)=2f(z).$
$P(z+1,z)\Rightarrow f(2z+1)=(z+1)f(z+1)-zf(z).~~~~~(*)$
$P(2z+1,1)\Rightarrow f(2z+1)=f(2z)+f(1)=2f(z)+f(1).$
$P(z+1,1) \Rightarrow (z+1)f(z+1)-zf(z)=(z+1)(f(z)+f(1))-zf(z)=f(z)+(z+1)f(1).$
$(*) \Rightarrow 2f(z)+f(1)=f(z)+(z+1)f(1) \implies f(z)=zf(1) ~~~\forall z\geq 0.$
$f$ is odd give that it holds for all real $z.$
So $\boxed{f(x)=cx},$ where $c=f(1),$ which fits. $\blacksquare$