FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm
Re: FE Marathon!

Unread post by Asif Hossain » Sat Feb 27, 2021 8:59 am

~Aurn0b~ wrote:
Fri Feb 26, 2021 11:44 am
$\textbf{Problem 15}$
Find all functions $ f: \mathbb{N}\to \mathbb{N}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]for any two positive integers $ m$ and $ n$.
Please do mention the source (i think i have seen it in the shortlist)
Hmm..Hammer...Treat everything as nail

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: FE Marathon!

Unread post by ~Aurn0b~ » Sat Feb 27, 2021 9:44 am

Asif Hossain wrote:
Sat Feb 27, 2021 8:59 am
~Aurn0b~ wrote:
Fri Feb 26, 2021 11:44 am
$\textbf{Problem 15}$
Find all functions $ f: \mathbb{N}\to \mathbb{N}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]for any two positive integers $ m$ and $ n$.
Please do mention the source (i think i have seen it in the shortlist)
2004 IMO SL

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

Re: FE Marathon!

Unread post by Dustan » Sat Feb 27, 2021 12:49 pm

Yeah😶 sorry. But is this right?

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

Re: FE Marathon!

Unread post by Dustan » Sat Feb 27, 2021 3:56 pm

$Problem 16$
Find all function such that $f:R \rightarrow R$ and
$f(xf(y)+y)+f(xy+x)=f(x+y)+2xy$

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: FE Marathon!

Unread post by ~Aurn0b~ » Sat Feb 27, 2021 6:36 pm

Dustan wrote:
Sat Feb 27, 2021 3:56 pm
$Problem 16$
Find all function such that $f:R \rightarrow R$ and
$f(xf(y)+y)+f(xy+x)=f(x+y)+2xy$
$\textbf{Solution 16}$
Let $P(x,y)$ be the assertion.
$P(1,\frac{x}{2})\Rightarrow f(f(\frac{x}{2})+\frac{x}{2})=x \Rightarrow f$ is surjective $\Rightarrow$ there exists $a$ such that $f(a)=1$.
$P(\frac{x}{a+1},a)\Rightarrow f(x)=x\frac{2a}{a+1}$
Now, obviously $a \neq -1$, because assuming that $f(-1)=1,P(x,-1)$ gives $0=-2x$ which is not possible.
So, plugging $f(x)=cx$ in the main equation we have $f(x)=x,f(x)=-2x.\blacksquare$

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: FE Marathon!

Unread post by ~Aurn0b~ » Sat Feb 27, 2021 7:02 pm

$\textbf{Problem 17}$
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
Last edited by ~Aurn0b~ on Sat Feb 27, 2021 8:09 pm, edited 1 time in total.

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

Re: FE Marathon!

Unread post by Dustan » Sat Feb 27, 2021 7:59 pm

$f^2(x)=f(f(x)) $ or $f^2(x)=(f(x))^2$

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: FE Marathon!

Unread post by ~Aurn0b~ » Sat Feb 27, 2021 8:08 pm

Dustan wrote:
Sat Feb 27, 2021 7:59 pm
$f^2(x)=f(f(x)) $ or $f^2(x)=(f(x))^2$
OH! sorry for that, it would be $f(x)^2$

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Mehrab4226
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Location:Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Sat Feb 27, 2021 8:12 pm

THis one looks kinda okay, right?
$P(x,y)=f(x(f(x)+f(y)=f^2(x)+y$
$P(0,y)=f((y))=f^2(0)+y$
Let, $f(x_1)=f(x_2)$
Or,$f(f(x_1))=f(f(x_2))$
Or,$f^2(0)+x_1=f^2(0)+x_2$
$\therefore x_1=x_2$
Which means $f$ is one-one.
Now,
$P(x,0)=f(x(f(x)+f(0))=f^2(x)+0$
Or,$f(xf(x)+f(0))=f(f(x))$
Or,$xf(x)+f(0)=f(x)$
Implies, $f(x)=\frac{f(0)}{1-x} \forall x \in R$ and $x \neq 1$



The question changed so, this solution is no longer valid
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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

Re: FE Marathon!

Unread post by Asif Hossain » Sun Feb 28, 2021 7:44 pm

Dustan wrote:
Sat Feb 27, 2021 12:49 pm
Yeah😶 sorry. But is this right?
What??
Hmm..Hammer...Treat everything as nail

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