FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
Asif Hossain
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Re: Asking for clearance

Unread post by Asif Hossain » Tue Mar 16, 2021 8:46 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

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Anindya Biswas
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Solution to Problem-19

Unread post by Anindya Biswas » Wed Mar 17, 2021 12:38 am

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.
First of all, let's plug in $x_1=x_2=\dots=x_{2010}=0$,
$f(0)=f(0)^{2010}\Rightarrow f(0)=0 \text{ or } f(0) \text{ is a 2009th root of unity}$

Case 1 : $f(0)=0$
For every $x\in\mathbb{Q}, f(x+\underbrace{0+0+\dots+0}_{2009})=f(x)f(0)^{2009}\Rightarrow f(x)=0$
So, we get a solution, $\forall x\in\mathbb{Q},f(x)=0$

Let's move to Case 2, which is $f(0)$ is a $2009^{\text{th}}$ root of unity.
Case 2 : $f(0)\text{ is a 2009th root of unity.}$
Note that $\forall x,y\in\mathbb{Q},f(x+y)=f(x+y+\underbrace{0+0+\dots+0}_{2008})=f(x)f(y)f(0)^{2008}=f(x)f(y)f(0)^{-1}$
$\Rightarrow f(x+y)f(0)^{-1}=f(x)f(0)^{-1}f(y)f(0)^{-1}$.

Let's define $g:\mathbb{Q}\to\mathbb{C}$ such that $g(x)=f(x)f(0)^{-1}$
Here's the things we get about $g$,
  • $g(0)=1$
  • $\forall x,y\in\mathbb{Q}, g(x+y)=g(x)g(y)$
  • By induction, we can generalize that $\forall n\in\mathbb{Z}, g(n)=g(1)^n$
  • Since $\forall q\in\mathbb{Q}, g(q)g(-q)=1$, we conclude that $\forall q\in\mathbb{Q}, g(q)\neq0$.
Finally, from $\text{Property (2)}$,
It's easy to see that,
$\forall x\in\mathbb{Q},\overline{g(2010)}\cdot g(x)=\overline{g(x)}\cdot g(2010)$
$\Rightarrow \frac{g(x)}{g(2010)}=\overline{\left(\frac{g(x)}{g(2010)}\right)}$
$\Rightarrow \frac{g(x)}{g(2010)}\in\mathbb{R}$.
By plugging in $x=2011$, we get $g(1)\in\mathbb{R}\Rightarrow g(2010)=g(1)^{2010}\in\mathbb{R}\Rightarrow\forall x\in\mathbb{Q}, g(x)\in\mathbb{R}$

Moreover, $g(x)=g\left(\frac{x}{2}\right)^2>0$
So, $\forall x\in\mathbb{Q}, g(x)\in\mathbb{R}^+$

Let $p,q$ be relatively prime integers with $q>0$.
$\therefore g\left(\frac{p}{q}\right)^q=g(p)\Rightarrow g\left(\frac{p}{q}\right)=g(1)^{\frac{p}{q}}$
So, $\forall x\in\mathbb{Q}, f(x)=f(0)g(x)=f(0)g(1)^x$ where $f(0)$ is a $2009^{\text{th}}$ root of unity.
Which is in fact satisfies our given conditions.

So, our solution is,
$\forall x\in\mathbb{Q}, f(x)=\gamma\cdot r^x$ where $r\in\mathbb{R}^+$ and $\gamma\in\mathbb{C}$ such that $\gamma^{2010}-\gamma=0$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Problem-20

Unread post by Anindya Biswas » Wed Mar 17, 2021 10:59 am

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 :
2002 IMO Shortlist A4
"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
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Re: FE Marathon!

Unread post by Dustan » Fri Mar 19, 2021 9:14 am

Let $P(x,y,z,t)$ be the assertion of
$(f(x)+f(y))(f(z)+f(t))=f(xy-zt)+f(xt+yz)$

$P(0,0,0,0)$
$f(0)=0$ or $f(0)=\frac{1}{2}$
Case 1: $f(0)=\frac{1}{2}$
$P(x,0,x,0)$:
$f(x)=\frac{1}{2}$
which fits the equation.
Case 2:Now we will use $f(0)=0$
$f(x)=0$ is a constant solution of the equation. Now we will look for non constant solution.
From $P(x,x,0,,0)$ we get
$f(x^2)=f(x)^2$
plugging $sqrt{x} $ insetead of $x$ the equation becomes
$f( sqrt{x})^2=f(x)$
Now,$P(x,y,0,0)$
$f(x)f(y)=f(xy)$ hence $f$ is multiplicative.
Since, $f(x)=f(sqrt{x})^2\ge0$
We will try to show that it's monotonically increasing on positive real.(Thanks to aops for the idea.Otherwise this soln won't complete)
$P(sqrt{x},sqrt{ y},sqrt{ y},sqrt{x}):$
$f(y-x)+f(x-y)=(f(sqrt{x})+f(sqrt {y}))^2$
From this
$f(x+y)\ge f(y-x)+f(x+y)=(f(sqrtx)+f(sqrty))^2\ge f(sqrt x)^2=f(x)$
Thus we are done.
So the general solution is $x^h$
Plugging this back we get $f(x)=x^2$

Hence, $f(x)=0$, $f(x)=1/2$, $f(x)=x^2$[\hide]

Dustan
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Re: FE Marathon!

Unread post by Dustan » 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

Asif Hossain
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Re: FE Marathon!

Unread post by Asif Hossain » Thu Mar 25, 2021 9:55 pm

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:
Plugging in $x=0$ implies either $f(0)=0$ or $f(f(f(0)))=0$
case 1.
Let $f(0)=0$ then plugging $y=0$ implies $f(f(x))f(x)=xf(x)$ so either $f$ is an involution and $f(x)=f^{-1}(x)$ or $f(x)=0 \forall x \in \mathbb{R}$
So we can rewrite the question as $ (x+f(y))(f(x)+y)=xf(x)+yf(y)+2yf(x)$
Since it is bijective, it must have an inverse function $f^{-1}$ and $f(x)=f^{-1}(x)$
Now plugging in $y=f^{-1}(y)$ implies $xf(y)+yf(x)=2f(y)f(x)$ plugging in $x=y=1$ implies
$f(1)=f(1)^2$ but since the function is bijective and $f(0)=0$ so $f(1)=1$
Now plug $y=1$ to get $f(x)=x$
Case2:
now assume $f(f(f(0)))=0$ now let $f(0)=t$ now we have that there exists $t$ ST $f(f(t))=0$
Now plugging $x=0,y=t$ implies $t(2t-f(t))=0$ so either $t=0$ or $f(t)=2t$ or $f(2t)=0$
Now plugging $x=0,y=2t$ implies $2t^2=0$ or $t=0$ and we are done so $f(x)=x,0$ $\forall x \in \mathbb{R}$ $\square$
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!

Unread post by Asif Hossain » 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:
Vietnam 2005
Hmm..Hammer...Treat everything as nail

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

Re: FE Marathon!

Unread post by Dustan » Thu Mar 25, 2021 11:12 pm

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

Dustan
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Re: FE Marathon!

Unread post by Dustan » 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 $k≠0$

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

Re: FE Marathon!

Unread post by Asif Hossain » Fri Mar 26, 2021 7:21 am

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 :D 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

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