Math Olympiads problem that I couldn't do.

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1. Dec 3, 2011

mtayab1994

1. The problem statement, all variables and given/known data
for every (x,y) in ℝ^2:

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

2. Relevant equations

Find all functions.

3. The attempt at a solution
I was wondering if rewriting it as x^2+f(y)=f^-1(y-x^2)?

2. Dec 3, 2011

micromass

So, what is the question?? And what is f??

3. Dec 3, 2011

deluks917

I believe the idea is to find all functions satisfying that condition.

4. Dec 3, 2011

mtayab1994

Yes that's what it is. Find all functions satisfying that condition.

5. Dec 3, 2011

micromass

Note that

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

So if we define

$$g_x(y):=x^2+f(y)$$

Then what can you say about $g_x\circ g_x$?

6. Dec 3, 2011

mtayab1994

I don't get what you're trying to tell me with gx∘gx.

I understand that its a category from gx to gx, so u can say: gx(g(x)). right?

7. Dec 3, 2011

Quinzio

I'm sorry for my intrusion, but after having spent some time on this I would like to add my comment.
From intuition and some trials I can say the $f(x) = -x$ works.
So that:
$f(y) = -y$
$f(y) + x^2= -y+x^2$
$f(f(y) + x^2)= y-x^2$

But is there an analytical or mechanical way to solve these problems ?

I can't get this too. Maybe one should apply the derivative ???

8. Dec 3, 2011

Dick

I'd try just differentiating both sides with respect to x.

9. Dec 3, 2011

micromass

Yes, but what if f is not differentiable??

10. Dec 3, 2011

Quinzio

But it's not a function of 2 variables. There's only one "input".

11. Dec 3, 2011

mtayab1994

f(f(y)+x^x)=y−x^2 can also be written as fof(y)+f(x^2)=y-x^2

12. Dec 3, 2011

Quinzio

But the problem says "find all the solutions". So there are more than 1 ? What the others look like then ?

13. Dec 3, 2011

Dick

I'd say find all the differentiable ones first, then try thinking about if there could be more.

14. Dec 3, 2011

Dick

Treat y as a constant.

15. Dec 3, 2011

mtayab1994

I don't know that's what i'm trying to reach but i also got this one:

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

16. Dec 3, 2011

micromass

Yes, but you got to watch out. f is not necessarily invertible. So the function $f^{-1}$ might not exist!!

17. Dec 3, 2011

mtayab1994

Yes but is it correct?

18. Dec 3, 2011

mtayab1994

That's the question, its the second one if you understand some french micromass.

19. Dec 3, 2011

micromass

Not until you show f to be invertible. What happens if x=0 ?

20. Dec 3, 2011

mtayab1994

for x=0 you get fof(y)=y and f(y)=f^-1(y)