# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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)