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
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:
Vietnam 2005
Next problem
again posted since it washed away by posts
Re: FE Marathon!
Posted: Sun Mar 28, 2021 9:46 pm
by Asif Hossain
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):
It is easy to see the function can't be constant.
Let $P(x,y)$ be the assertion.
$P(x,x) \Rightarrow f(f(0))=f(x)^2 -x^2 \Rightarrow f(x)^2 -x^2 =f(0) \Rightarrow (f(x)+f(0))(f(x)-f(0))=x^2$
Now Plug $x=p$ where $p$ is an arbitrary prime.
now if $(f(p)+f(0))$ or $(f(p)-f(0))$ equals to $1$ then it would contradict that $f(p)$ isn't constant.
so $f(p)+f(0)=p=f(p)-f(0) \Rightarrow f(0)=0$
Now plugging $f(0)$ value back we get $f(x)^2 =x^2$
Not again falling into the pointwise error so now assume for $a,b$ that $f(a)=a$ and $f(b)=-b$
$P(a,b) \Rightarrow f(f(a-b))=-a-b$ a contradiction so either $f(x)=x$ which doesn't work or $f(x)=-x$ which indeed works.
So $f(x)=-x \forall x \in \mathbb{R}$
I wouldn't pose a problem rather you folks would
Re: FE Marathon!
Posted: Tue Mar 30, 2021 10:37 am
by Dustan
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) $ .
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) $ .
$\textbf{Solution 23}$
Substituting $x\rightarrow x-1$, we have,\[f(x+f(y)-1)=y+f(x)\]
Define $g:\mathbb{R}\rightarrow \mathbb{R}$ such that, $g(x)=f(x)-1$
As $f$ is continuous, $g$ is also continuous.
Substituting $g$ into the main equation we get,\[g(x+g(y))=y+g(x)\]
Let $P(x,y)$ be assertion of this equation.
$P(0,y)\Rightarrow g$ is bijective.
Let $g(x_0)=0$, $P(x,x_0)\Rightarrow x_0=0$
And so , $P(0,y)\Rightarrow g(g(y))=y$
Now, \[P(x,g(y))\Rightarrow g(x+y)=g(x)+g(y)\], which is a cauchy function, and its continuos. Therefore $g(x)=cx$ for some constant real $c$ .
Plugging in $f(x)=cx+1$ in the main equation we get $c=\pm1$
Therefore all the solution to the functional equation is $f(x)=x+1$ and $f(x)=1-x.\blacksquare$
Re: FE Marathon!
Posted: Tue Mar 30, 2021 11:39 am
by ~Aurn0b~
$\textbf{Problem 24}$
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:
\[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]
Source : IMOSL 1977
Actually this is IMO 1977/P6. Hint:
Try to prove that $f(n) \geq n$ $\forall n \in \mathbb{N}$. Then try to prove that the function is strictly increasing. That will give $f(n) \leq n$ $\forall n \in \mathbb{N}$. Combining these two will then give the solution: $f(n) = n$ $\forall n \in \mathbb{N}$.
assuming that you have resolved previous one
