Random Random FE

by utkarshgupta, Feb 16, 2017, 2:09 AM

Problem :
Find all functions $f : Let$ $f:\mathbb{R}\to \mathbb{R}$ such that
$f(x^{2}+y+f(y))=2y+f(x)^2$

Okay....
Here it goes...

Solution :
Let $P(x,y)$ be the assertion
$f(x^{2}+y+f(y))=2y+f(x)^2.$
And let $k=\frac{-f(0)^2}{2}$
$P(0,k) \implies f(k)=0$
Now let if possible, there exist some $a \neq k$ such that $f(a)=0$ ,
$P(0,a) \implies f(a) = f(0)^2 + 2a \neq 0$

A contradiction !

Hence $f(t) = 0 \implies t = k$ .

Let $f(0)=c$ then $k = \frac{-c^2}{2}$ .

$P(k,0) \implies f(k^2+f(0)) = f(k)^2 = 0$ $\implies k^2 + c = k$ $\implies c = 0,2$
Let if possible $c=2 \implies k=-2$
Then $P(0,0) \implies f(2) = 4$
$P(0,k) \implies f(2) = 0$

A contradiction !

$\boxed {f(0) = 0}$ And there exists no $a \neq 0$ such that $f(a)=0$

$P(x,\frac{-f(x)^2}{2}) \implies f(x^2 + \frac{-f(x)^2}{2} + f(\frac{-f(x)^2}{2})) = 0$ $\implies x^2 + \frac{-f(x)^2}{2} + f(\frac{-f(x)^2}{2}) = 0$
Thus if $f(a) = f(b)$ then $a^2 = b^2$

Let if possible $f(a) = f(-a)$ for some $a \neq 0$
$f(a + f(a)) = 2a = -f(-a + f(-a))$
Hence $f((a+f(a))^2) = f((-a+f(a))^2)$ Since $x^2 > 0$ ...
We must have $a^2+f(a)^2+2af(a) = a^2 + f(a)^2 - 2af(a)$
That is either $a= 0$ or $f(a) = 0$
A contradiction !

Hence $f$ is injective.

$P(x,0) \implies f(x^2) = f(x)^2 \ge 0$ $P(0,y) \implies f(y+f(y)) = 2y$
Let $z^2 - y^2 + f(z^2) - f(y^2) = x^2$ (assume greater than zero or else we can assume it to be equal to $-x^2$ )
$P(x^2,y^2) \implies f(x^2+y^2 + f(y^2)) = f(x^2) + 2y^2$ $P(0,z^2) \implies f(z^2+f(z^2)) = 2z^2$ $\implies f(x^2 + f(z^2) - f(y^2)) = 2(y^2-z^2) = f(x^2 + f(x^2))$ Since $f$ is injective
$f(x^2+y^2) = f(x^2) + f(y^2)$ .
We also have $-f(x) = f(-x)$ and $f(x^2) > 0$
So by Cauchy we are done !

That is $\boxed {f(x)=x}$ is the only solution !

This post has been edited 2 times. Last edited by utkarshgupta, Feb 16, 2017, 4:31 AM

functional equation

AoPS

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Doesn't $k^2+c=k \Longrightarrow c \in \{0,-2\}$ instead?

by anantmudgal09, Feb 17, 2017, 5:15 PM

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That was a silly mistake

But a contradiction can be derived nevertheless
Will edit later

by utkarshgupta, Feb 18, 2017, 3:19 AM

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Yes. My idea for a correction was $f(f(0)+f(f(0)))=2f(0)+f(0)^2,$ and as $f(f(0))=f(0)^2,$ we obtain $f(f(0)^2+f(0))=f(f(0))^2=f(0)^4,$ so $c$ also satisfies $c^4-c^2-2c=0$ . Combined with $k^2+c=k,$ we get $c=0$ as the only possibility. Rest is as you said.

by anantmudgal09, Feb 18, 2017, 8:43 AM

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Well actually...
If $c=-2$ we will get $f(-2) = -2$ and $f(-2)=0$
So yeah...

by utkarshgupta, Feb 18, 2017, 11:13 AM

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This was pretty nice!

by TheOneYouWant, Feb 22, 2017, 12:13 PM

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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
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
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
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
Oh you certainly will..
Bu I don't think I will
by achilles04, Feb 11, 2015, 6:24 PM
Thanks
Hope you are there too ...

And hope I get selected
by utkarshgupta, Feb 11, 2015, 11:14 AM
Nice blog ..
keep it up and you might one day become a mathematician lIke me
( just joking )
Btw don't forget us after going to imotc.
by achilles04, Feb 10, 2015, 4:49 PM
nice blog!!!!!
by pradhit, Feb 7, 2015, 11:51 AM
No ideas about the cut off but should be less than 60 according to me...
So you never know
by utkarshgupta, Feb 6, 2015, 4:06 PM
Hi utkarsh,
what do u expect the cutoff to be this year??

by kgdpsdurg, Feb 6, 2015, 12:41 PM
Hi utkarsh
How many did you clear in INMO 2015?
by moldevort, Feb 1, 2015, 3:59 PM
@ Anonymous Bunny
If we both (that should have included me) are lucky enough !
by utkarshgupta, Sep 17, 2014, 2:44 AM
Nice blog. Great job!

If I am lucky enough, hopefully we shall meet at IMOTC.
by AnonymousBunny, Sep 16, 2014, 8:08 PM
Thanx all !!

If anyone wish to contribute anything just pm !

I will add u to the contributors' list !
by utkarshgupta, Sep 16, 2014, 6:35 AM
doing good progress!! btw nice blog.
by rachitgoel, Sep 15, 2014, 4:21 PM
Nice blog.
by Kezer, Aug 5, 2014, 4:04 PM
Hi Utkarsh! Wonderful blog. Keep it up.
by YESMAths, Jul 20, 2014, 1:36 PM
I need some shouts and characters

by utkarshgupta, Jul 20, 2014, 4:22 AM

48 shouts

Elementary

Random Random FE

by utkarshgupta, Feb 16, 2017, 2:09 AM

5 Comments