Does there exist an infinite sequence of differentiable functions $f_0,f_1,f_2,\ldots:[0,1]\to\Bbb R$, not all the zero function, such that, for all $i$,
- $f_{i+1}'=f_i$, and
- $f_i(0)$ and $f_i(1)$ are both integers?
I conjecture the answer is no, though I'm unsure of how to prove it. I can answer some variations of the question, though.
It is natural to ask what happens if the $1$ in the above question is replaced with other numbers. If $1$ in the above question is replaced with $\ln2$ (meaning we want $f_i(0)$ and $f_i(\ln2)$ to be integers), then there is such a sequence, namely $(e^x,e^x,e^x,\ldots)$. Similarly, if $1$ is replaced with $\pi$, then $(\cos x,\sin x,-\cos x,-\sin x,\cos x,\dotsb)$ is such a sequence.
A little thought will reveal a generalization. Consider again the above question with $1$ replaced with some other number $\alpha$. If $e^\alpha\in\Bbb Q$ then there is such a sequence of functions, and if both $\sin\alpha,\cos\alpha\in\Bbb Q$ then there is such a sequence of functions (in each case take some constant multiple of the sequences mentioned above). It is interesting to note that the sets $\{x:\sin x,\cos x\in\Bbb Q\}$ and $\{x:e^x\in\Bbb Q\}$ are both closed under addition and subtraction.
I conjecture that this gives us all solutions to these variations. That is, I conjecture that if there exists an infinite sequence of differentiable functions $f_0,f_1,f_2,\ldots:[0,\alpha]\to\Bbb R$, not all the zero function, such that, for all $i$, $f_{i+1}'=f_i$, and in addition $f_i(0)$ and $f_i(\alpha)$ are both integers for all $i$, then in fact $\alpha\in\{x:\sin x,\cos x\in\Bbb Q\}\cup\{x:e^x\in\Bbb Q\}$.
The answer of the original question is indeed no.
Suppose we have an infinite sequence of differentiable functions satisfying the conditions. Let $a_i = f_i(0)\in\mathbb{Z}$ then $f_{i+1}(x)=\int_0^x f_i(t)dt+a_{i+1}$.
Thus, we get $$f_{k}(1) = \frac{1}{(k-1)!}\int_0^1(1-s)^{k-1}f_0(s)ds +\frac{1}{(k-1)!}a_1+\dots+a_{k}$$
Multiply both sides by $(k-1)!$ , we have that $\int_0^1(1-s)^{k-1}f_0(s)ds\in\mathbb{Z}$. Since $f_0$ is differentiable (hence continuous), there exists $M\in\mathbb{R}$ such that $\vert f_0(x)\vert \leq M$ for $x\in[0,1]$. Therefore $\int_0^1(1-s)^{k-1}f_0(s)ds\leq M\int_0^1(1-s)^{k-1}ds = \frac{M}{k}$, so the integral is zero after finite terms.
Let $N$ be an integer such that $\int_0^1(1-s)^{k-1}f_0(s)ds = 0$ if $k\geq N$. Then $(1-x)^{N-1}f_0(x)$ is the zero function using the Weierstrass approximation theorem, therefore $f_0(x)$ is the zero function. The argument above applying on subsequences$\{f_i,f_{i+1},\dots\}$ shows $f_i$ are also zero functions, so there is a contradiction.