How to find a differentiable map $T : \mathbb{R} \rightarrow \mathbb{R}$ whose fixed points are exactly integers How to find a differentiable map $T : \mathbb{R} \rightarrow \mathbb{R}$ whose fixed points are exactly integers?
I have to
a)Find points $|T'(x)| > 1$ (if any)
and prove that
b)There must exist at least one $x$ such that $|T'(x)| > 1$.
I think the functions 
$$x - Tx = x(x^2 - 1)(x^2 - 4)(x^2 - 9)...$$
and
$$x - Tx = x(e^x - e^1)(e^{-x} - e^1)(e^x - e^2)(e^{-x} - e^2)(e^x - e^3)...$$ will satistfy the criteria.
Is there any closed form expressions for $T'(x)$?
How can I proceed further?
 A: You're on the right track but that product does not work. Find a function that has zeros exactly at the integers. Then add $x$ to it.
A: Since you are interested in infinite products, I recommend taking a look at the Wikipedia article for a little orientation on these.  I believe you will find it useful in finding the example you want.  
(b) is a consequence of the mean value theorem. Consider $T(\frac{1}{2})$, for example.  It is either less than or greater than $\frac{1}{2}$.  If less, then the average rate of change from $\frac{1}{2}$ to $1$ is greater than $1$.  If greater, then the average rate of change from $0$ to $\frac{1}{2}$ is greater than $1$.
The difficulty of (a) could depend on the example, and that must wait until you have an explicit example.  Of course by (b) you know that there will be such points.
A: $f(x)=\sin(2\pi x)+x$ satisfies $f(n)=n$ for all $n\in\mathbb{Z}$  I don't know what you want with the rest...
A: I think this problem is equivalent to the problem of finding a curvature on a plane which intersect with $y=x$ only at interger points .Of course ,this curvature must be the graph of a function.but it is not difficult to find.For example,you can piece some circle curves together.(I don't know how to post the picture.)
