FE Marathon!

[nimon]

can you suggest me any book for solving this kind of function problem.

[Anindya Biswas]

$\textbf{Problem 24}$

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...

We will prove by induction that $f(n)=n$ for all $n\in\mathbb{N}$

• Induction Hypothesis :
  $f(n)=n$ for all $n\in\{1,2,3,\dots,k\}$ and $1,2,3,\dots,k\not\in S_k=\{f(k+1),f(k+2),f(k+3),\dots\}$
• Base case :
  Let $S_0=\{f(1),f(2),f(3),\dots\}$ be the range of $f$.
  Let $f(m+1)$ be the least element of the set $S_1=\{f(2),f(3),f(4),\dots\}$
  Since $f(f(m))<f(m+1)$, we conclude $f(f(m))\not\in S_1$. But since $f(f(m))\in S_0$, that means $f(m)=1\Longrightarrow 1\in S_0$.
  $\therefore \min{S_0}=1$. Since $1\not\in S_1, 1\in S_0\setminus S_1\Longrightarrow f(1)=1$.
• Inductive Step :
  Let's denote the set $S_{i}=\{f(i+1),f(i+2),f(i+3),\dots\}$
  Note that $S_k=\{f(k+1),f(k+2),f(k+3)\dots,\}\subseteq\{k+1,k+2,k+3,\dots\}$ by our induction hypothesis.
  
  Let $m\in\{k+1,k+2,k+3,\dots\}$ such that $f(m+1)=\min{S_{k+1}}=\min{\{f(k+2),f(k+3),f(k+4),\dots\}}$
  Since $f(m+1)>f(f(m))$, we conclude $f(f(m))\not\in S_{k+1}$. But since $f(f(m))\in S_k$, that means $f(m)=k+1\Longrightarrow k+1\in S_k$
  $\therefore \min{S_k}=k+1$
  We can also see that $k+1\not\in S_{k+1}$. Because, if $k+1\in S_{k+1}$ then $\min{S_{k+1}}=k+1$ which contradicts the fact that there exist a smaller element in $\{k+1,k+2,k+3,\dots\}$ which does not belong to $S_{k+1}$.
  $\therefore k+1\in S_k\setminus S_{k+1}\Longrightarrow k+1=f(k+1)$

So, our claim is proved by mathematical induction. $\text{Q.E.D.}$

[Anindya Biswas]

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

[Asif Hossain]

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

[~Aurn0b~]

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

$\textbf{Solution 25}$

Its ez to see that its an injective function.

Plugging $n=1 \ \Rightarrow \ f(1)+f(f(1))+f(f((1)))=3 \ $ $\Rightarrow f(1)=f(f(1))=f(f(f(1)))=1$

Plugging $n=2 \ \Rightarrow \ f(2)+f(f(2))+f(f((2)))=6 \ $ , As the function is injective, none of the term can be $=1$ [ As an example, if $f(f(2))=1 \Rightarrow f(f(2))=1=f(f(1)) \Rightarrow 2=1$ ]

So these 3 terms have to be at least 2. Therefore,
$f(2)=f(f(2))=f(f(f(2)))=2$,

Going on like this we'll have on every step,
$f(n)=f(f(n))=f(f(f(n)))=n.\blacksquare$

[Anindya Biswas]

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

$\textbf{Solution 25}$

Its ez to see that its an injective function.

Plugging $n=1 \ \Rightarrow \ f(1)+f(f(1))+f(f((1)))=3 \ $ $\Rightarrow f(1)=f(f(1))=f(f(f(1)))=1$

Plugging $n=2 \ \Rightarrow \ f(2)+f(f(2))+f(f((2)))=6 \ $ , As the function is injective, none of the term can be $=1$ [ As an example, if $f(f(2))=1 \Rightarrow f(f(2))=1=f(f(1)) \Rightarrow 2=1$ ]

So these 3 terms have to be at least 2. Therefore,
$f(2)=f(f(2))=f(f(f(2)))=2$,

Going on like this we'll have on every step,
$f(n)=f(f(n))=f(f(f(n)))=n.\blacksquare$

Yeah, but this last statement can be nicely proven by induction.

[~Aurn0b~]

$\textbf{Problem 25}$

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]

$\textbf{Problem 25}$

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 $f(x^2-y^2)=xf(x)-yf(y)$

[Zeta]

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. $\blacksquare$