Two Subsets of Squares of Reciprocals of Primes with Equal Sums Let $$A=\{\frac{1}{2^2},\frac{1}{3^2},\frac{1}{5^2},...\}$$ be the set of squares of the reciprocals of prime numbers.
We have
$$\sum_{x\in A}x < \infty$$
Do there exist $B \subset A$, $C \subset A$, $B \cap C = \emptyset$, such that 
$$\sum_{x\in B}x = \sum_{x\in C}x  \ \ \ \ \ ?$$
It is important that we deal with primes and not with all natural numbers, otherwise we would have infinitely many solutions with both $B$ and $C$ finite, as described in Wikipedia, a type of solution which cannot solve this problem by W-t-P's answer below.
 A: The answer is negative. Let $C=\sum_{p}\frac{1}{p^2}$. This constant is approximately $\frac{1}{2}\log\frac{5}{2}$ due to Euler's product, leading to:
$$ \sum_{p}\frac{1}{p^2}\approx \sum_p\frac{1}{2}\log\left(\frac{1+\frac{1}{p^2}}{1-\frac{1}{p^2}}\right)=\frac{1}{2}\log\frac{\zeta(2)^2}{\zeta(4)}. $$
Assume that the set of prime numbers can be partitioned as $U\cup V$ with $U\cap V=\emptyset$ and $U,V\neq\emptyset$. 
We may assume without loss of generality that $2\in U$, hence
$$ \sum_{p\in U}\frac{1}{p^2}\geq \frac{1}{4} > \frac{1}{2}\sum_{p\in U\cup V}\frac{1}{p^2}=\frac{C}{2} $$
which contradicts
$$ \sum_{p\in U}\frac{1}{p^2}=\sum_{p\in V}\frac{1}{p^2}.$$
A: Assuming that $B$ and $C$ (are finite and) have the required property, denote by $P$ the product of all primes $p$ with $1/p^2\in B$, and by $Q$ the product of all primes $p$ with $1/p^2\in C$. Then the LHS of $\sum_{x\in B}x=\sum_{x\in C}x$ would be a rational number which, reduced to its lower terms, has its denominator equal to $\prod_{p\in P} p^2$, while the RHS has the denominator equal to $\prod_{p\in Q}p^2$. By the uniqueness of prime decomposition, we would then have $P=Q$, a contradiction.
