Sum of two functions taking integer values infinitely often

Question

Let $x \in Z^+$ and $n \in Z^+$. Let $f(x)$ be a function that takes integer values infinitely often. Then it follows that $f(x+n)$ also takes integer values infinitely often. Let $f(x)$ be continuous and differentiable. I think that if $p(x)=f(x)+f(x+n)$ takes integer values more than once than it takes integer values infinitely often.

Is this a known property? Can anybody give a counter example to this?

Edit: $f(x)$ is not a piece-wise function. I am extremely sorry that I didn't mention these things before. I have been thinking about this all day and I didn't realize that I had assumed these.

Answer 1

Try $f(n)=\log_2(n)$, and various values of $c$.

Answer 2

Suppose $f(x) = 0$ at infinitely many integer values $x = n_1, n_2, \ldots$, and $f(x) = \pi$ otherwise. Then $f(x) + f(x + 1)$ is an integer iff $x$ and $x + 1$ are both in the set $\lbrace n_1, n_2, \ldots\rbrace$. By choosing the $n_i$ suitably, you can make this happen for as many $x$ as you like.

Answer 3

OK, I think the idea can be modified. Let $f(x) = 0$ at certain integer values $n_1, n_2, \ldots$ as before. Let $f(x) = 1/3$ away from these values, but make $f(x)$ continuous by having narrow spikes going down to the points where $f(x) = 0$. Then the same argument will still work.

Answer 4

I found an example such that what I considered was wrong. Infinite product formula like this one $f(x) = (x-1) \prod_{p \in P} (1 - \frac{e^x}{e^p})^2$ is equal to 0 when $x$ is prime or $x=1$.