find functions f such that $f(f(x))=xf(x)+1$, let $f:R\longrightarrow R$, and $f$ is continous,and such that $f(f(x))=xf(x)+1$, 
find all this $f$?
follow is my some idea:(but I don't have solution)
We have $f(f(0)) = 1$, so there is your $c = f(0)$, such that $f(c) = 1$. Assume there exists $v$ such that $f(v) = 0$. Then $f(0) = f(f(v)) = vf(v) + 1 = 1$, meaning $c=1$. Now, $f(0) = f(1) = 1$, so $1 = f(f(0)) = f(f(1)) = f(1)+1 = 2$, absurd.
So $f(x) \neq 0$ for all $x$, thus $f$ takes constant sign, being continuous. Assume now $f(x) = f(y) = t \neq 0$, so  $xt+1 = f(f(x)) = f(f(y))= yt + 1$, whence $(x-y)t=0$, thus $x=y$. This means $f$ is injective, therefore monotonous, being continuous. Moreover, assume $x=f(x)$, so $x = f(x) = f(f(x)) = xf(x) + 1 = x^2+1$, thus $x^2-x+1 = 0$, but this has no real roots, so $f(x) \neq x$ for all $x$. Then either $f(x) > x$ for all $x$, or $f(x) < x$ for all $x$, since $f(x)-x$ is continuous
mark: this problem is my found,come from this probelm,if when $f:N\longrightarrow N$,and add $f(1)=1$then this problem is equivalent follow problem
$$a_{n+1}=na_{n}+1,a_{1}=1$$
we can find $a_{n}=[e(n-1)!]$,
Thank you everyone can help
 A: From the OP and Zach L's comment to the OP, we can continuously extend $f$ to a function on the extended real numbers by setting $f(-\infty) = 0$ and $f(+\infty) = +\infty$.
Define a sequence $a_n$ of extended real numbers for all natural numbers by $a_0 = -\infty$, and $a_{n+1} = f(a_n)$. Observe that $f$ is a bijective function $[a_n, a_{n+1}] \to [a_{n+1}, a_{n+2}]$.
Let $g$ be the inverse of $f$ (with domain the non-negative extended real numbers)
The sequence $a_n$ is monotonic and increasing, and therefore has a limit $L$ in the extended real numbers. This satisfies
$$\begin{align} L &= \lim_{n \to +\infty} a_n 
\\&= \lim_{n \to +\infty} a_{n+1}
\\&= \lim_{n \to +\infty} f(a_n)
\\&= f(\lim_{n \to +\infty} a_n)
\\&= f(L)
 \end{align} $$
The OP has already shown that $f$ has no finite fixed points, so therefore $L = +\infty$.
This means the intervals $[a_n, a_{n+1}]$ cover the entire range $[-\infty, +\infty)$. 
The fact that $f(f(x)) = x f(x) + 1$ means that the value of $f$ on $[a_{n+1}, a_{n+2}]$ is determined by its values on $[a_n, a_{n+1}]$ (by considering $x \in [a_n, a_{n+1}]$).
Therefore, $f$ is completely determined by its values on $[a_0, a_1] = [-\infty, 0]$.
Conversely, I assert that if you choose any continuous, monotonically increasing function $f_0$ on $[-\infty, 0]$ such that $f_0(-\infty) = 0$ and $0 < f_0(0) < 1$, then 
we have a n increasing sequence (converging to $+\infty$) recursively defined by


*

*$a_0 = -\infty$

*$a_1 = 0$

*$a_2 = f_0(0)$

*$a_{n+2} = a_{n+1} a_n + 1$


and a sequence of invertible functions $f_n : [a_n, a_{n+1}] \to [a_{n+1}, a_{n+2}]$ recursively defined by


*

*$f_{n+1}(x) = f_n^{-1}(x) x + 1 $


and then the function
$$ f(x) = \begin{cases} f_n(x) & x \in [a_n, a_{n+1}]
\\ +\infty & x = +\infty \end{cases} $$
is (the continuous extension to the extended real numbers of) a solution to the problem.
