## FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm
Anindya Biswas wrote:
Tue Mar 16, 2021 7:40 pm
Asif Hossain wrote:
Tue Mar 16, 2021 7:02 pm
2) $\forall x \in \mathbb{Q}$ , conjugate of $f(2010)$ $\times$ $f(x)=f(2010)$ $\times$ conjugate of $f(x)$ where $\times$ represent normal multiplication.
Did you mean $\overline{f(2010)}\cdot f(x)$ or $\overline{f(2010)\cdot f(x)}$? I guess the first one.
1st one
Hmm..Hammer...Treat everything as nail

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

### Solution to Problem-19

Asif Hossain wrote:
Tue Mar 16, 2021 7:02 pm
Since Nobody posting any problem here is one...
Problem 19
Determine all functions $f:\mathbb{Q}\to\mathbb{C}$ such that
1) For any rational $x_1,x_2,...,x_{2010}$, $f(x_1+x_2+\dots+x_{2010})=f(x_1)f(x_2)...f(x_{2010})$
2) $\forall x \in \mathbb{Q}$ , $\overline{f(2010)}\times f(x)=f(2010)\times\overline{f(x)}$ where $\times$ represent normal multiplication.
"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-20

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $\left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz)$ $\forall x,y,z,t\in\mathbb{R}$.
Source :
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

Someone check my soln.
However i am posting another problem.
Problem 21: Find all function such $f:R\rightarrow R$ and
$f(f(x+f(y)))(f(x)+y)=xf(x)+yf(y)+2yf(x)$ for all real x,y

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

### Re: FE Marathon!

Dustan wrote:
Sun Mar 21, 2021 8:28 am
Someone check my soln.
However i am posting another problem.
Problem 21: Find all function such $f:R\rightarrow R$ and
$f(f(x+f(y)))(f(x)+y)=xf(x)+yf(y)+2yf(x)$ for all real x,y
Solution:
Last edited by Asif Hossain on Fri Mar 26, 2021 7:19 am, edited 1 time in total.
Hmm..Hammer...Treat everything as nail

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

### Re: FE Marathon!

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:
Hmm..Hammer...Treat everything as nail

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

@above you should observe that there exists a constant function. And $f(x)=0$ is a solution.

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

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 $k≠0$

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

### Re: FE Marathon!

Dustan wrote:
Thu Mar 25, 2021 11:12 pm
@above you should observe that there exists a constant function. And $f(x)=0$ is a solution.
edited now pls confirm if its right
Last edited by Asif Hossain on Fri Mar 26, 2021 12:02 pm, edited 1 time in total.
Hmm..Hammer...Treat everything as nail