take $x=y=0$ gives $f(0)=0$ take $y=0$ gives $f(f(x))=f(x)^2$ take $f(x)=y^2$ gives $0=f(x)^2-2f(x)^2+f(f(y))=-f(x)^2+f(y)^2$ thus $f(x)=f(y)$ or $f(x)=-f(y)$ then $f(y)=y^2$ or $f(y)=-y^2$ But $f(y)=-y^2$ is not indeed solution then $f(y)=y^2$

I have the same ! but not on the selection

After seeing the solution, it's seems easy but it's not

Can you say $f(x)=y^2$ without worrying about the sign? At this point, I think that all you can say is: If there is a value of $x$ with $f(x)=t>0$, then $f(t^{1/2})=t$ and $f(-t^{1/2})=t$ Shouldn't you prove explicitly why these values exist, and why there are no values of $x$ with $f(x)<0$.

I don't remember why but I think I could say that... I will look why

First of all welcome to the forum $LordWings$ Yes I agree with you and such $x$ exists Take $x=0$ in the original equation then $f(-y^2)=f(f(y))=f(f(-y))=f(y)^2=f(-y)^2 \geq 0$ so for all $x$ then $f(x)\geq 0$

Thank you, AYMANE! OK, now we know $f(x) \geq 0$. That answers the second part of my question. But, is that enough? Your solution assumes that for every $t=y^2>0$ there's a value of $x$ with $f(x)=t$.

Thanks AYMANE ^^ lordWings wrote: Your solution assumes that for every $t=y^2>0$ there's a value of $x$ with $f(x)=t$. I don't think that assume that... It only assume that there is value such that... I don't think we need to prove that the function is surjectiv on $\mathbb{R^+}$ but we can certainly do it !

That's it, you should prove that the function is surjective on $\mathbb{R^+}$. You have only proved: If there exists a value of $x$ with $f(x)=y^2$, then $f(y)=y^2$. You still should prove that it exists. The conclusion can't prove the premise.

No you are wrong and BogG is rihgt

Sorry , I'm usually wrong about these things. I just feel uncomfortable with the "take $f(x)=y^2$", wich is the key to your reasoning. If you can take any value of $y$, you're excluding there all the functions that aren't surjective on $\mathbb{R^+}$ (for instance, the bounded ones). And when you say that, the only things you know about $f(x)$ is that $f(0)=0$ and $f(f(x))=f(x)^2$. Maybe with $f(f(x))=f(x)^2$, we can prove that the function is surjective on $\mathbb{R^+}$ (or continuous and not bounded), but until we do, I think that the reasoning is not complete. Am I too complicated?

$f(x)=y^2$ is a just especial case like $x=y=0$ or $x=y$

What you cannot do with using $f(x)=y^2$ is to "change" the $y$ because you had "fixed" it. I cannot explain more or better

hi, here is my solution : lets assume that : $f(x) = u$ $y=0 \Rightarrow f(u) = u^2 + f(f(0)) (1)$ $x=y=0 \Rightarrow f(0) = 0 (2)$ $(1),(2) \Rightarrow f(x) = x^2$ .

You clearly have $f(u) = u^2$ with $u=f(x)$. To say $f(x)=x^2$ you must show that you can obtain all the $x$ with $f(u)$ -> you must prove that $f(u)$ is surjectiv . P.S Are you the same toetoe how is on the marocan forum "amateur de math" ?

AYMANE wrote: $f(x)=y^2$ is a just especial case like $x=y=0$ or $x=y$ $f(x)=y^2$ is not just a special case like $x=y=0$ or $x=y$. AYMANE has proved that $f(x) \geq 0$ for every $x$. Then, there will always will be a value of $y$ with $f(x)=y^2$ for every $x$. In this case you can use $f(x)=y^2$ as a special case to prove things about $x$ and they will be true for every $x$. The inverse is not true. We still don't know anything about the range of the function $f$, so there may be some values of $y$ that can't be reached with $f(x)=y^2$ (simply, such $x$ may not exist; for instance, $f(x)$ could be bounded or discontinuous). You can't use $f(x)=y^2$ as a special case to prove things about $y$, thinking that they will be true for every $y$. They will only be true if you prove, beforehand, that $y^2$ belongs to the range of the function $f$ (if you prove it later, it won't mean nothing, it just won't disprove your assumption). BogG noticed it looking at toetoe's solution. Exactly the same thing happens with toetoe's special case $u=f(x)$: it can only be used to prove things about $u$ if you prove, beforehand, that $u$ belongs to the range of the function $f$. So, in general, you can use the special case $s=h(t)$: - to prove things about $s$, only if you prove beforehand that $s$ belongs to the domain of the function $h$. - to prove things about $t$, only if you prove beforehand that $t$ belongs to the range of the function $h$ (this is more difficult, since we usually know the domain and codomain, but not the range). You can use special cases like $x=0$ or $x=k$ or $x=y$ to prove things about $x$ or $y$ without worries, because the domain and range of the functions $h(y)=0$, $h(y)=k$ and $h(y)=y$ are both $\mathbb{R}$. But if you use special cases like $x=f(y)$ or $x=-f(f(y))$ where $f$ is the function that you are trying to find out (remember, we're dealing with an equation), you should be more careful with its domain and range.

Maybe... I'm not yet really sure... but I have another solution ! $x=y=0$ $\rightarrow$ $f(0)=0$ $y=0$ $\rightarrow$ $f(f(x))=f(x)^2$ $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=f(f(x)-f(y)^2)^2$ $f(z)=z^2$ for all $z \in S = \{f(u)-f(v)^2 | u,v \in \mathbb{R}\}$. Choose $a$ such that $f(a) \not= 0$ and for exemple with $f(f(x))=f(x)^2$ easy show that $f(a) >0$ Then put $x=a$ $\rightarrow f(f(a)-y^2) - f(y)^2 = f(a)(f(a) - 2y^2)$. On the right we can have all the value $<0$ then on the left too $\rightarrow$ $\mathbb{R^-} \subset \mathbf{R}$ and then $f(z)= z^2$ for all $z\le 0$ $\rightarrow$ all the value can be reached in $\mathbb{R^+}$ by the non-negativ numbers $\rightarrow$ surjectivity of $f$ in $\mathbb{R^+}$ And now I think it's more clear

BogG wrote: $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=f(f(x)-f(y)^2)^2$ What happened here? It seems a typo and BogG wants to say $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$ but I think that's not right. I get $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2+(f(y)^2-y^4)$. And then the conclusion wouldn't hold. I hope I'm wrong, or we're back agin.

BogG wrote: P.S Are you the same toetoe how is on the marocan forum "amateur de math" ? yes,i am .

toetoe wrote: BogG wrote: P.S Are you the same toetoe how is on the marocan forum "amateur de math" ? yes,i am . Can you give me the link to this forum

Please, discuss this kind of things via PM!

yes,I agree with you mathmanman.

We're not advacing a lot here. Let's get back to the topic. Quote: Find all the functions $f : \mathbb{R} \to \mathbb{R}$ satisfying for all $x,y \in \mathbb{R}$ $f(f(x)-y^2) = f(x)^2 - 2f(x)y^2 + f(f(y))$. AYMANE has proved so far: 1) If we take $x=y=0$, then $f(0)=0$. 2) If we take $y=0$, then vor every $x$ we have $f(f(x))=f(x)^2$. 3) If we take $x=0$, then for every negative value $t=-y^2$ we have $f(t)\geq 0$. 4) If we take $y^2=f(x)$ ($y^2\in range(f)$), then $f(y)=y^2$. 5) $f(x)=x^2$ is one possible solution of the equation. More straightforwardly, toetoe proved that: 6) If we take $u=f(x)$ ($u\in range(f)$), then $f(u)=u^2$. We still don't know anything about the range of $f$ in general. BogG and I have been trying to prove that $range(f)=\mathbb{R^+}$, and that would make $f(x)=x^2$ the only solution. In fact, nobody noticed the next fact (compatible with facts 1 to 6): 7) $f(x)=0$ is one possible solution of the equation. To prove a point, in this case, $range(f)=\{0\}$, so we can't assume $range(f)=\mathbb{R^+}$ in general. Are there any other solutions? Perhaps we can begin with: Let's suppose there's a $x$ with $f(x)=a\neq 0$...

lordWings wrote: What happened here? It seems a typo and BogG wants to say $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$ but I think that's not right. I get $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ $\rightarrow$ $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2+(f(y)^2-y^4)$. And then the conclusion wouldn't hold. I hope I'm wrong, or we're back agin. Sorry I forgot to say something. Make this change $y$ become $f(y)$ then you have $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ who become $f(f(x) - f(y)^2)=f(x)^2 - 2f(x)f(y)^2+f(f(y))^2$ and with using $f(f(y))=f(y)^2$ you find $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$ Yes and of cours $f(x) = 0$ is also a solution

I think I finally understand BogG's reasoning, but it's really convoluted! (I'll change the name of some variables to avoid confusions.) With the last post, he's proved (taking $x=u$, $y=f(v)$ inside the equation, and using fact 2) that: 8) If there are $u,v\in\mathbb{R}$ with $s=f(u)-f(v)^2$, then $f(s)=s^2$. Then we go back to his previous post: We know from fact 7 that $f(x)=0$ is one possible solution. Any other solution will have at least one $x_0$ with $f(x_0)=z\neq 0$. If we take $x=z$ inside the equation (along with facts 2 and 6), we get $f(z^2-y^2)=z^4-2z^2y^2+f(y)^2$. This can be rewritten in this way: $f(z^2-y^2)-f(y)^2=z^2(z^2-2y^2)$. We must remember that $z$ is fixed, but $y$ can be any real number. So, there are $u=z^2-y^2$ and $v=y$ with $s=f(u)-f(v)^2=(u+v^2)(u-v^2)=u^2-v^4\le z^4$. Using fact 8 with this last result, we get: 9) If there is one $x_0$ with $f(x_0)=z\neq 0$, then for every $s\le z^4$, $f(s)=s^2$. This fact includes that for every $s\in\mathbb{R^-}$, $f(s)=s^2$. In this case, $f$ would be surjective on $\mathbb{R^+}$, because for every $k\in\mathbb{R^+}$, there is $s=-k^{1/2}\in\mathbb{R^-}$ with $f(s)=k$, and then, with case 4, $f(y)=y^2$ for every $y$. That's it: BogG has proved that, if the solution is not the trivial $f(y)=0$, then it must be $f(y)=y^2$ for every $y$.

There's one question left. At the Swiss IMO where this problem was proposed, were they supposed to solve it like BogG? Is it the most simple way to solve it? I think I would have never solved it that way in a limited time. Step 8 seems difficult and pointless until it's been used (deus ex machina) at step 9. Kudos to BogG. Of course, without these steps, we have already proved that $f(y)=0$ and $f(y)=y^2$ are solutions (maybe worthy of some points), but we haven't proved that these are the only solutions.

The last solution is not mine.. I only have copy it from the Swiss solution...

BogG wrote: Find all the functions $f : \mathbb{R} \to \mathbb{R}$ satisfying for all $x,y \in \mathbb{R}$ $f(f(x)-y^2) = f(x)^2 - 2f(x)y^2 + f(f(y))$. It's easy to show that, $f(f(x)) = f(x)^2$ Let, $S=\{x\ |\ f(x)=x^2\ \}$, then, $f(x)\in S$ for all $x$. Now, since $f(f(f(y)))=f(f(y))^2=f(y)^4$, $y\longrightarrow f(y)$ in given equation implies, $f(f(x)-f(y)^2)=(f(x)-f(y)^2)^2$ So, $f(x)-f(y)^2 \in S$ for all $x,\ y$. Now, from the given equation and $f(f(y))=f(y)^2$, we get, $f(f(x)-y^2)-f(y)^2=f(x)^2-2y^2f(x)$ So, $\{f(x)^2-2y^2f(x)\ |\ x,\ y\ \in\ \mathbb{R}\ \} = (-\infty, f(x)^2] \subseteq S$ for all $x$ So, $f(u)=u^2$ for all $u\leq f(x)^2$. Or, $\lim_{u\longrightarrow -\infty} f(u) = \infty$ Hence, $\lim_{u\longrightarrow -\infty} (-\infty, f(u)^2] = \mathbb{R} \subseteq S$ So, $S=\mathbb{R}$, Done!

Let $P(x,y)$ denote the given assertion. Since $f\equiv 0$ works, assume $f(u)\neq 0$ for some $u.$ Clearly $f(0)=0$ and $f(f(x))=f(x)^2.$ Note that $f(u)^4-2f(u)^2x^2$ can be any $z \leq 0$, so $P(f(u),x)$ implies $z=f(v)-f(x)^2$, for $v=f(u)^2-x^2.$ Then $P(v,f(x))$ implies $f(z)=z^2$ for all $z\leq 0.$ Then $P(z,y)$ shows $f(x)\equiv x^2;$ which works.