$f(x)=x$ doesn't work $f(x)=x$ works for the vietnam assuming that you have resolved previous oneDustan wrote: ↑Thu Mar 25, 2021 11:59 pmHowever 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 $k≠0$
FE Marathon!

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Re: FE Marathon!
Hmm..Hammer...Treat everything as nail

 Posts: 171
 Joined: Sat Jan 02, 2021 9:28 pm
Re: FE Marathon!
Next problemAsif Hossain wrote: ↑Thu Mar 25, 2021 10:32 pmFind all functions $f:$ $\mathbb{R} \to \mathbb{R}$ such that
$f(f(xy))=f(x)f(y)f(x)+f(y)xy$
source:
again posted since it washed away by posts
Hmm..Hammer...Treat everything as nail

 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
$1.$ 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):
Hmm..Hammer...Treat everything as nail
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) $ .
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) $ .
Re: FE Marathon!
$\textbf{Problem 24}$
Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
\[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]
Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
\[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]

 Posts: 171
 Joined: Sat Jan 02, 2021 9:28 pm
Re: FE Marathon!
2 days passed no solution??Violation of marathon rules.. Aurnob you should post the solution
Hmm..Hammer...Treat everything as nail
Re: FE Marathon!
Actually this is IMO 1977/P6.
Hint:
"Questions we can't answer are far better than answers we can't question"