Let $\mathcal{F}$ be the set of all Fibonacci numbers (defined by $ F_n=F_{n-1}+F_{n-2}$ with $F_1=F_2=1$), and put $D:=\mathcal{F}-\mathcal{F}$.
(a) Is it true that if $B$ is a finite subsetset of integers with $D\cap (B-B)=\{ 0\}$, then $D+B\neq \mathbb{Z}$?
(b) If the answer is positive, then is it also true that $D+B\neq \mathbb{Z}$, for every finite subsetset $B$ of integers?
Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$, and see A question about Fibonacci numbers
Here's a proof of (b):
It suffices to show that for any $\epsilon >0$ we can find an $n$ such that $D$ only hits $\epsilon n$ congruence classes when reduced modulo $n$. Since if you take $\epsilon < 1/|B|$ then $D+B$ misses a congruence class mod $n$ so $D+B \ne \mathbb{Z}$.
Now to show this, by the Chinese remainder theorem it suffices to construct relatively prime integers $m, m'$ such that $\mathcal{F}$ only hits $\delta m$ residue classes mod $m$ and $\delta'm'$ with $\delta \delta' < \epsilon$.
I claim I can take $m, m'$ of the form: $m = 11\cdot 31 \cdot 61 \cdot ... \cdot p_i$ and $m' = 19\cdot 29 \cdot 59 \cdot ... \cdot p_j$. That is, $m$ is the product of the first $i$ primes congruent to $1$ mod $5$ and $m'$ is the product of the first $j$ primes congruent to $-1$ mod $5$.
Clearly these are relatively prime, so it's enough to show that we can choose $i,j$ large enough so that the Fibonacci numbers have arbitrarily small density modulo $m$ and $m'$.
If $p$ is a prime congruent to $\pm 1$ mod $5$ I claim the Fibonacci numbers miss at least one congruence class mod $p$. In fact, the Fibonacci numbers will be periodic mod $p$ with period dividing $p-1$. Why? Quadratic reciprocity ensures there is a square root of $5$ mod $p$ and we can write $F_k \equiv ab^k+cd^k \mod p$ for some integers $a,b,c,d$ (i.e. we have a mod $p$ Binet formula), and then Fermat's little theorem gives the periodicity.
Applying the Chinese remainder theorem again we see that the Fibbonacci numbers hit at most $(11-1)(31-1)...(p_i - 1)$ congruence classes mod $m$ (and similar for m'). We just need to check that we can choose $i$ large enough so that this product is less than $\delta m$. This is equivalent to the statement:
$$\prod_{p\equiv 1\mod5} \frac{p-1}{p} =0$$
Which is equivalent to the statement that the sum of the reciprocals of primes congruent to $1$ mod $5$ diverges, and a well known theorem of Dirichlet says that the sum of the reciprocals of all primes congruent to $a$ mod $b$ (with a,b relatively prime) always diverges.
