Math Olympiads problem that I couldn't do.

[micromass]

What solution did u get for Y=0 just curious to know.

Hold on, let me guide you there step by step.

So, you have

$$f(x^2+f(y))=y-x^2$$

What happens if you take f of both sides??

 

[mtayab1994]

Hold on, let me guide you there step by step.

So, you have

$$f(x^2+f(y))=y-x^2$$

What happens if you take f of both sides??

I can't grasp what you mean by if you take the f of both sides?

 

[micromass]

I can't grasp what you mean by if you take the f of both sides?

For example, if we have a=b. If we take f of both sides, then we have f(a)=f(b).

 

[mtayab1994]

For example, if we have a=b. If we take f of both sides, then we have f(a)=f(b).

ooohh alright i got: fof(x^2+f(y))=f(y-x^2)

 

[micromass]

ooohh alright i got: fof(x^2+f(y))=f(y-x^2)

OK, and fof=... ??

 

[mtayab1994]

OK, and fof=... ??

f(x^2)=f^-1of(-x^2)

 

[micromass]

f(x^2)=f^-1of(-x^2)

No. How did you get this??

What is f(f(y))?? What is f(f(x^{^}2+f(y)))? How would you simplify f(f(x^{^}2+f(y)))=f(y-x^{^}2)??

 

[mtayab1994]

No. How did you get this??

What is f(f(y))?? What is f(f(x^{^}2+f(y)))? How would you simplify f(f(x^{^}2+f(y)))=f(y-x^{^}2)??

to simplify it you would do: f(f(x^2+f(y))=f(y-x^2) <=> f(x^2+f(y))=f^-1(f(y-x^2)

 

[micromass]

Take a look at what you did in post 20.

 

[mtayab1994]

Take a look at what you did in post 20.

Yea man I'm getting a bit tired i'll come back on and finish this tomorrow thanks for your help by the way

 

[mtayab1994]

 

[micromass]

What is k?

 

[mtayab1994]

k is a number i chose to equal x .

 

[micromass]

You write: if x>0, then

$$f(k^2+f(0))=-k^2+f(0)$$

Why is this?? Is this even true?

 

[mtayab1994]

why is it not true?

 

[micromass]

why is it not true?

Why is it true?? How did you prove it??

 

[mtayab1994]

yea alright since we chose x=0 we got f(f(y))=0 so therefore for every x in ℝ: f(f(x))=x that's what let's us say that: f(k^2+f(f(0))=-k^2+f(0)

 

[micromass]

yea alright since we chose x=0 we got f(f(y))=0 so therefore for every x in ℝ: f(f(x))=x that's what let's us say that: f(k^2+f(f(0))=-k^2+f(0)

Oh OK. But that's something different from what you wrote there!

You have now basically that

$$f(k^2)=-k^2+f(0)$$

 

[mtayab1994]

Oh OK. But that's something different from what you wrote there!

You have now basically that

$$f(k^2)=-k^2+f(0)$$

yea i had a typo on my paper i was writing the stuff fast so what's wrong with that?

 

[micromass]

yea i had a typo on my paper i was writing the stuff fast so what's wrong with that?

Nothing with that! I was just pointing out that what you wrote first was incorrect.

 

[mtayab1994]

yea and for x<0:

$$f(0)=f(k^{2}+f(f(-k^{2}))$$ which then equals

$$=f(-k^{2})-k^{2}$$

 

[micromass]

yea and for x<0:

$$f(0)=f(k^{2}f(f(-k^{2}))$$ which then equals

Typo here??

In general, you are indeed correct that f(x)=-x+c are the only possible answers. But you still need to show that they indeed satisfy the equation!

 

[mtayab1994]

Typo here??

yea i fixed it before you posted and btw the proof is as follows:

[tex]f(x)=-x+c[tex]for c+f(0)<br /> <br /> indeed: $$f(x^{2}+f(y))=x^{2}-f(y)+c$$<br /> <br /> as: $$f(y)=-y+c$$ then $$f(x^{2}+f(y))=-x^{2}+y-c+c$$<br /> <br /> and finally $$f(x^{2}+f(y))=y-x^{2}$$ for every x in ℝ[/tex][/tex]

 

[micromass]

Seems all good!

 

[mtayab1994]

yes now I'm sure i have all 4 questions correct for the olympiads !

 

[mtayab1994]

Do you know anywhere i can find olympiad like problems beside the imo-official site?

 

[mtayab1994]

The people from the olympiads gave us another test and they kept this same problem from last week !