## FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
Asif Hossain
Posts: 171
Joined: Sat Jan 02, 2021 9:28 pm

### Re: FE Marathon!

Dustan wrote:
Thu Mar 25, 2021 11:59 pm
However the soln is
$f(x)=x$ and$f(x)=0$ for all x,y.
Here, $P(x,y,)$ denotes the assertion of the following equation.

$P(0,0)$ we get,
$f(0)=0$ or $f(f(f(0)))=0$
The second motivates us to assume $f(t)=0$
$P(t,t)$ gives $f(0) \cdot t=0$
So,$f(0)=0$ from both case.
Case 1: $f(0)=0$
$P(x,0)$:
$f(f(x))\cdot (f(x))=xf(x)$
Or, $f(x)[f(f(x))-x]=0$
Hence, $f(x)=0$ or, $f(f(x))=x$
From the second one, we can switch the variable and will find
$yf(x)=f(y)x$
$y=1$ gives
$f(x)=cx$
Plugging this back, $f(x)=x$ is the solution for all x,y.

Case 2: Suppose there exist a c≠0 for which $f(c)=0$
From $P(x,0)$ we already get
$f(f(x))f(x)=xf(x)$
$P(x,c)$ gives
$f(f(x))(f(x)+c)=xf(x)+2cf(x)$

From this two,
$cf(f(x))=2cf(x)$
Or,$f(f(x))=2f(x)$
Multiplying both sides we get,
$2f(x)^2=xf(x)$ [ assume, $f(x)≠0$ in this case. We are dealing with non constant solution ]
$f(x)=\frac{x}{2}$
Let, $k≠0$ for some k $f(k)=\frac{k}{2}$
then $f(\frac{k}{2})=k$. But this impossible.
So,$f(x)=0$ for all real valued x,y.
And we are done.

I guess there is no mistakes. If you find any then tell me.
Thanks to aops for the idea of assuming $k≠0$
$f(x)=x$ doesn't work $f(x)=-x$ works for the vietnam assuming that you have resolved previous one
Hmm..Hammer...Treat everything as nail

Dustan
Posts: 65
Joined: Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

i haven't solved vietnam

Asif Hossain
Posts: 171
Joined: Sat Jan 02, 2021 9:28 pm

### Re: FE Marathon!

Asif Hossain wrote:
Thu Mar 25, 2021 10:32 pm
Find all functions $f:$ $\mathbb{R} \to \mathbb{R}$ such that
$f(f(x-y))=f(x)f(y)-f(x)+f(y)-xy$
source:
Next problem
again posted since it washed away by posts
Hmm..Hammer...Treat everything as nail

Asif Hossain
Posts: 171
Joined: Sat Jan 02, 2021 9:28 pm

### Re: FE Marathon!

You brothers are neglecting the marathon that's why adding a new rule as geo marathon
$1.$ if any problem stay more than $2$ days, then the proposer must give solution to the proposed problem
Let me be the first to follow this rule here is mine solution(it was a quick solu so it might have flaws so do confirm):
I wouldn't pose a problem rather you folks would
Hmm..Hammer...Treat everything as nail

Dustan
Posts: 65
Joined: Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

PROBLEM 23:
Find all continuouos functions $f: \mathbb{R}\to\mathbb{R}$ such that for all reals $x$ and $y$
$f(x+f(y))=y+f(x+1)$ .

~Aurn0b~
Posts: 44
Joined: Thu Dec 03, 2020 8:30 pm

### Re: FE Marathon!

Dustan wrote:
Tue Mar 30, 2021 10:37 am
PROBLEM 23:
Find all continuouos functions $f: \mathbb{R}\to\mathbb{R}$ such that for all reals $x$ and $y$
$f(x+f(y))=y+f(x+1)$ .
$\textbf{Solution 23}$

~Aurn0b~
Posts: 44
Joined: Thu Dec 03, 2020 8:30 pm

### Re: FE Marathon!

$\textbf{Problem 24}$

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
$f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.$

Asif Hossain
Posts: 171
Joined: Sat Jan 02, 2021 9:28 pm

### Re: FE Marathon!

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

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
$f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.$
2 days passed no solution??Violation of marathon rules.. Aurnob you should post the solution
Hmm..Hammer...Treat everything as nail

~Aurn0b~
Posts: 44
Joined: Thu Dec 03, 2020 8:30 pm

### Re: FE Marathon!

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

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
$f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.$
Source : IMOSL 1977

Anyone can post the next solution.

tanmoy
Posts: 305
Joined: Fri Oct 18, 2013 11:56 pm