Problem 1813

To determine whether there exists a finite set $A$ of positive integers of at least two elements and an infinite set $B$ of positive integers such that any two distinct elements in $A+B$ are coprime, and for any coprime positive integers $m, n$, there exists an element $x$ in $A+B$ satisfying $x \equiv n \pmod{m}$, we proceed as follows:

### Proof by Contradiction

Assume, for the sake of contradiction, that such sets $A$ and $B$ exist, with $A = \{a_1, a_2, \ldots, a_k\}$.

#### Lemma 1
If $(A, B)$ satisfy the conditions of the problem with $|A| = k$, then there exist infinitely many $b \in B$ such that the number $a_i + b$ has at least $k$ distinct prime divisors.

Proof of Lemma 1:
Let $m_1, m_2, \ldots, m_k$ be pairwise coprime numbers with at least $k$ prime divisors each. Let $m = m_1 m_2 \ldots m_k$, and let $n$ be chosen such that $n \equiv a_i - a_{i+1} \pmod{m_i}$ for all $i$ (indices are modulo $k$). Since $(A, B)$ satisfy the conditions of the problem, there exists $a_i, b$ such that
$$a_i + b \equiv n \pmod{m}.$$
Therefore,
$$a_i + b \equiv a_i - a_{i+1} \pmod{m_i}.$$
Now,
$$a_{i+1} + b \equiv 0 \pmod{m_i},$$
which proves the statement.

#### Main Proof
Let $b$ be an arbitrarily large number satisfying the condition of Lemma 1. Without loss of generality (after suitable relabeling), let the respective $a_i$ be $a_1$.

Let $p_1, p_2, \ldots, p_k$ be distinct primes dividing $b + a_1$. Choose $m = p_1 p_2 \ldots p_k$, and $n$ such that
$$n \equiv a_i - a_{i+1} \pmod{p_i}$$
for all $1 \le i \le k$, where indices are considered modulo $k$. By the Chinese Remainder Theorem, $n$ corresponds to some residue modulo $m$. By assuming that $p_i$ are large enough (that is, larger than $|a_{i+1} - a_i|$ for all $i$), we can guarantee $(m, n) = 1$. To get $p_i$ large, just increase $b$ if necessary: as the elements of $A+B$ are coprime, only finitely many $b' \in B$ may be such that $b' + a_i$ has a "small" prime divisor dividing $|a_{i+1} - a_i|$ for some $i$.

Now, assume there exists some $c \in B$ such that $x = a_i + c$ ($1 \le i \le n$) satisfies $x \equiv n \pmod{m}$. This, in particular, implies
$$a_i + c \equiv a_i - a_{i+1} \pmod{p_i}.$$
This implies
$$a_{i+1} + c \equiv 0 \pmod{p_i},$$
with $a_{i+1} + c \in A + B$. Since the elements of $A+B$ are pairwise coprime, this implies $c = b$, with the $b$ described as above. Additionally,
$$a_{i+1} + b \equiv 0 \pmod{p_1},$$
and
$$a_1 + b \equiv 0 \pmod{p_1}$$
by definition, so $a_{i+1} = a_1$, and therefore $i \equiv 0 \pmod{k}$. That is,
$$a_1 + b \equiv n \pmod{m}.$$
Now,
$$a_1 + b \equiv n \equiv a_1 - a_2 \pmod{p_1},$$
and also
$$a_1 + b \equiv 0 \pmod{p_1}.$$
This means that $a_1 \equiv a_2 \pmod{p_1}$, which cannot happen for big enough $p_1$. As explained before, we can force the $p_i$ to be as large as we want. This, finally, is a contradiction.

Therefore, the answer is: \boxed{\text{No}}.