Baltic Way 2014

by utkarshgupta, Nov 4, 2015, 1:23 PM

I hate to post this as it is too easy for a new entry after such a long time but I hate to see my blog unupdated for such a long time.


Problem (Baltic Way 2014):
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\]for all real numbers $x, y.$


Solution :

Let $P(x,y)$ be the assertion $f(f(y)) + f(x - y) = f(xf(y) - x)$

$f \equiv 1$ isn't obviously a solution.
Thus there exists some $a \in \mathbb R$ such that $f(a) \neq 1$

$P(\frac{f(a)}{f(a)-1},a) \implies$
$$ f(\frac{f(a)}{f(a)-1}-a)=0$$
$implies$ there exists a $k$ such that $f(k)=0$

$P(x,k) \implies f(0)+f(x-k)=f(-x)$
$P(0.k) \implies f(-k)=0$
$P(k,k) \implies f(0)=0$

$\implies f(0)=0$

$$P(0,y) \implies f(f(y)) + f(-y) = 0$$
$$P(\frac{-y}{f(y)-2},y) \implies f(y)=0$$for all $f(y) \neq 2$


Now let there exist some $t \in R$ such that $f(t)=2$
$f(f(t))= - f(t) = -2$
But we know that that for all $f(x) \neq 2$, we have $f(x)=0$

A contradiction.

$\implies \boxed{f \equiv 0}$ which indeed is a solution.

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Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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  • Here goes first post of 2025! Great blog.

    by math_holmes15, Jan 14, 2025, 8:53 AM

  • First post of 2024

    by Yiyj1, Feb 8, 2024, 5:40 AM

  • First post of 2023

    by HoRI_DA_GRe8, Jul 22, 2023, 7:45 AM

  • Nice blog ! Your isogonality lemma is really powerful !

    by 554183, Oct 14, 2021, 8:55 AM

  • Post plss....

    by samrocksnature, Apr 11, 2021, 10:12 PM

  • alas,this is ded

    by Hamroldt, Mar 18, 2021, 4:13 PM

  • Thanks for the nice blog.

    by Feridimo, Mar 6, 2020, 4:17 PM

  • I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it :D

    by gamerrk1004, Nov 4, 2019, 4:54 PM

  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

    by Kayak, Oct 2, 2017, 7:18 PM

  • hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!

    by kk108, Jun 17, 2017, 11:19 AM

  • Congrats on becoming a contest moderator!

    by Ankoganit, Mar 9, 2017, 5:22 AM

  • INTERSTING BLOG

    by kk108, Feb 19, 2017, 2:04 PM

  • I have no plans for this blog right now....
    No time here people !
    But lets see....
    I may try some combinatorics :P

    by utkarshgupta, Feb 15, 2017, 12:47 PM

  • Thanks for the nice blog!

    by Orkhan-Ashraf_2002, Feb 13, 2017, 6:34 PM

  • Revive it!!!
    Best blog out there, for sure!

    by rmtf1111, Jan 12, 2017, 6:02 PM

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