Does there exists a function $f:\Bbb R\to \Bbb R$ such that $f(f(x))=x^2-2$  I don't know where to start?
50981
Find all $f(x)$ satisfying $f(f(x)) = x^2 - 2$.   Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further.   Possibly Helpful Links: Information on similar problems can be found  and .  Source:  It is about to be closed for containing too many problems in one question. I'm posting each problem separately.
100826
Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.  My try:  If $ x=y=0$ then $f(0)=0$ and  if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then  $f(x)=f\left( \frac{x+1}{2}  \right)+f\left( \frac{x-1}{2}  \right)$.  But how to find all functions?
25649
Is there a function $f\colon \mathbb{R} \to\mathbb{R} $ such that $ f(f(x)) = -x$ ?
24377
find  all functions $f:\Bbb R\to \Bbb R$ and such for any $x,y\in\Bbb R$,and such $$f(x+f(y)+yf(x))=y+f(x)+xf(y)$$     I have proven that $$f(f(x))=x$$   proof:Let $y=0$,we have $$f(x+f(0))=f(x)+xf(0)$$ let $x=0,y\to x$,we have $$f(f(x)+xf(0))=x+f(0)$$ so we have $$f(f(x+f(0))=x+f(0)$$ then $$f(f(x))=x$$  and $f(x)=x$ is one solution but I can't argue this to be the only solution. It doesn't seem easy.Thank for you help!
20382
This is a problem I was casually discussing with friends:     Given any function $g(x): \mathbb{R} \rightarrow \mathbb{R}$, find a   function $f(x): \mathbb{R} \rightarrow \mathbb{R}$ such that $f(f(x)) = g(x)$.   Is it possible to find a solution for $f(x)$ or prove that there isn't a solution for $f(x)$ such that $f(f(x)) = g(x)$ for any given $g(x)$?   For example, if $g(x) = x + 2$, then $f(x)$ could easily be $f(x) = x + 1$. But if $g(x) = x^2 - 1$, then it's not clear to see what $f$ can satisfy such solution.   What approach might be promising? My first gut reaction would be some analytical method but can't think of a way still...Discussions are welcomed!
48713
Let $f:\Bbb{R}\to\Bbb{R}$ be a function given by  $$f(x)=\begin{cases} \exp\left(\frac{1}{x^2-1}\right) &amp; \text{if }\vert x\vert\lt 1\\  0 &amp; \text{if }\vert x\vert\geqslant 1  \end{cases}$$  I would like to prove that $f\in C^\infty$, that is, $f\in C^k$ for all $k\in \mathbb{N}$. I think that it can be done by induction on $k$. If $\vert x\vert\gt1$, the problem is trivial. On other points, the base case is the simplest and the only that I'm be able to do. Can someone help me?  Thanks.
11705
I am looking for an example of a function $f:\mathbb R^2\to\mathbb R$ such that there is a point $x\in\mathbb R^2$ with the following properties:  1) All partial derivatives of second order exist in a neighborhood of $x$.  2) At least one of those partial derivatives is not continuous in $x$.  3) The Hessian matrix of $f$ in $x$ is symmetric.  I think it should be possible to find such a function but I wasn't very successful in finding one. If we drop the first property it is easy, with it however, I didn't find any example yet. Any help appreciated.
6913
Problem: A function $f: \mathbb R \to \mathbb R$ satisfies equation $f(x+y)=f(x)f(y)$ for all $x,y\in \mathbb R$ and $f(x)\neq 0$ for all $x \in \mathbb R$. If $f'(0)=2$, find $f'(x)$ in terms of $f(x)$ Solution: $f(x+y)=f(x)f(y)$ Differentiating partially w.r.t $x$ we get $f'(x+y)=f'(x)f(y)$ At x=0 $f'(y)=f'(0)f(y)$ $f'(y)= 2f(y)$ $f'(x)= 2f(x)$  Can we solve  question like above ? I mean,$f'(x)$ represent ordinary diff. while I used it for partial diff.
25227
I've been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:     Find a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)) = -x$ for all $x \in \mathbb{R}$.   I am also wondering, can we find $f$ so that is continuous?  I was thinking of letting $f$ be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.  Source:
45702
Be $f:\mathbb R^ +\mapsto\mathbb R$ a function that satisfies the following conditions:  a)$ f(f(f(x)))+2x=f(3x)$ for every $x\gt 0$;  b)  $\lim_{x \to \infty} (f(x)-x)=0$.  This was proposed by Gabriel Dospinescu  at a romanian competitions about a decade ago. No idea how to approach it. Thank you!
97607
The problem states:     Let $\Bbb R$ denote the set of all real numbers. Find all functions $f : \Bbb R \rightarrow \Bbb R$ such that $$f(x^{2}+f(y))=y+(f(x))^{2} \space \space \space \forall x, y \in \Bbb R.$$   My progress: 1. If we substitute $x=y=0$ in the given equation, then we get $$f(f(0))=(f(0))^{2}.$$ 2. We then substitute $x=0$ in the given equation and find out that $\forall y \in \Bbb R$, $$f(f(y))=y+(f(0))^{2}.$$ 3. Now, we observe that, for all $x,y \in \Bbb R$, $$y+(f(x))^{2}=f(x^{2}+f(y))=f((-x)^{2}+f(y))=y+(f(-x))^{2}.$$ Hence $\forall x \in \Bbb R,$ $$f(-x)=f(x) \space \space or \space \space f(-x)=-f(x).$$ 4. But, if for some $x \in \Bbb R$, $f(-x)=f(x)$, then we would have, $$x+(f(y))^{2}=f(y^{2}+f(x))=f(y^{2}+f(-x))=-x+(f(y))^{2}$$ for any $y \in \Bbb R$ implying $x=0$. So, for any $x \neq 0$, $f(-x)=-f(x).$ 5. Now if there exists any $y \neq 0$ such that $f(y) \neq 0$ then, $$y+(f(0))^{2}=f(f(y))=f(-f(-y))=-f(f(-y))=y-(f(0))^{2}$$ which implies $f(0)=0$. So, if $f(0) \neq 0$, then $f(x)=0$ whenever $x \neq 0$. But then $f(0) \neq 0$ would imply $(f(0))^{2}=f(f(0))=0$ which cannot happen and hence $f(0)=0$. 6. So summing up all that we have got so far, we see that $f$ has the following properties:   (i)$f$ is an odd function.(ii)$f(x^{2})=(f(x))^{2}.$ (iii)$f(f(x))=x.$  This is where I am stuck. I have observed that using the above mentioned three properties, we can write the given equation as $$f(x^{2}+f(y))=f(x^{2})+f(f(y))$$ which is almost $f(a+b)=f(a)+f(b)$, but that is of no help. Any hints would be welcome.
61787
Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $f(f(f(x)))=x,\forall x \in \mathbb R$ , then is it true that $f(x)=x , \forall x \in \mathbb R$ ? I can see that $f$ is bijective , hence strictly monotone . But I cannot progress further . Please help. Thanks in advance
83131
Is the cardinality of  $$X = \{f: \Bbb R \to \Bbb R \;|\; f \text{ is differentiable everywhere}\}$$ the same as $\Bbb R$? How to prove it?
26822
Question     Let the function $f:\mathbb R\to\mathbb R$,and such   $$f(xf(y)+x)=xy+f(x)$$      Find all $f(x)$   Let $x=1,y=1$,then $$f(f(1)+1)=1+f(1)$$ let $f(1)=t$,then $$f(t+1)=1+t$$  So I guess $$f(x)=x$$ is such it, and I found $f(x)=-x$ is also such it?   Can you help?  I found this simaler problem：(BMO 1997, 2000) $$f(xf(x)+f(y))=y+(f(x))^2,f:R\to R$$ this answer $f(x)=x$ or $f(x)=-x$,  the full solution can see
111707
Suppose $f$ is a function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$, then we have to show that $\forall z\in \mathbb C f(z)+f(-z)=0 $.   Actually I have solved this problem and have no doubt about it but actually this question came in MADHAVA-2020. So I posted it on this site.   I am not answering the question and inviting all users to answer it.  It is a nice problem.   I will answer it within a week, but I am letting everyone try.
72804
I am wondering what is the class of functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that $f(f(x))=f(x)$?  I think it should be:   Constant Value functions the identity function absolute value function $|x|$   But I don't know if this is right or how to show it rigorously.  Any suggestions?
47973
Let , $f:\Bbb C \to \Bbb C$ be an entire function and $\displaystyle \int_{\Bbb R^2}|f(x+iy)|\,dx\,dy&lt;\infty$. Then prove that $f(z)=0$ for all $z\in \Bbb C$.   Put $x=r\cos \theta$ and $y=r\sin \theta$. Then , integration becomes $\displaystyle \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi}|f(re^{i\theta})|r\,d\theta\,dr&lt;\infty$. But from here I'm unable to conclude nothing.! How I proceed? Any hint please.