Two properties of the set of Bernoulli numbers Let $\mathcal{B}$ be the set of all Bernoulli numbers. We are looking for an answer for
one of the following (or both) questions.
(a) Is there an infinite subset $S$ of rationals  such that
$(\mathcal{B}-\mathcal{B})\cap (S-S)=\{ 0\}$?
(b)  Is it true that $(\mathcal{B}-\mathcal{B})+F\neq \mathbb{Q}$, for every finite subset  $F$ of rationals?
Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$ and  $A+B=\{a+b: a\in A , b\in B\}$.
Also, $(\mathcal{B}-\mathcal{B})\cap (S-S)=\{ 0\}$ is equivalent to the property that every element $x$ of  $\mathcal{B}+S$ has a unique
representation $x=b+s$ with $b\in \mathcal{B}$ and $s\in S$.
Thanks in advance
 A: The answer to both questions is yes.

The von Staudt-Clausen theorem implies that the denominators of Bernoulli numbers $B$ are square-free. Notice that Rational numbers with square-free denominators make a subgroup of $\mathbb Q$, so a group generated by $B$ would be a subgroup of that subgroup, which means denominators of nonzero elements of $(B- B)$ are square-free.
Then for example, $S=\{1/p^2:p\text{ is odd prime}\}$ answers (a) because denominators of nonzero elements of $(S-S)$ are odd squares (of semiprimes).

For (b) assume the opposite, that there exists finite $F$ such that $(B-B)+F=\mathbb Q$.
Since $F$ is finite there exits a $k$ such that for all primes $p$, prime power $p^k$ does not divide any denominators in $F$. Recall that $(B-B)$ is square-free (denominators are not divisible by $p^n,n\gt 1$). It follows that denominators of elements in $(B-B)+F$ are divisible by at most $p^{k-1}\cdot p^1=p^k$, i.e. are not divisible by $p^{k+1}$.
That is, $1/p^{k+1}\not\in((B-B)+F)\subset\mathbb Q$, which is a contradiction.
