Does a finite sum of distinct prime reciprocals always give an irreducible fraction?

Question

If we add any finite number of any distinct prime reciprocals, will the result always be an irreducible fraction?

If not, is there any bound on the value of a greatest common divisor for the numerator and demoninator of such fraction?

This boils down to the question about the greatest common divisor for:

$$Q=p_1p_2 \cdots p_n~~~~\text{and}~~~~P=\sum_{k=1}^{n} \frac{Q}{p_k}$$

Here $\{p_k\}$ is any finite subset of primes.

If this question is trivial, I apologise in advance. I checked this quickly for small primes and got only irreducible fractions. Elementary number theory is not my strong point.

Answer 1

For distinct primes $p_1,\dots,p_n$ the fraction $S=\frac{1}{p_1}+\cdots+\frac{1}{p_n}$ is always irreducible. For if it were reducible, some $p_i$ would "cancel", without loss of generality $p_1$. So we could express $S$ as $\frac{A}{p_2p_3\cdots p_n}$.

Multiply both sides of the equation $$\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_n}=\frac{A}{p_2p_3\cdots p_n}$$ by $p_2p_3\cdots p_n$. On the right-hand side we obtain an integer. On the left-hand side we do not.

Answer 2

As you've shown, this comes down to the question of whether $$P=\sum_{k=1}^n p_1p_2\ldots\hat p_k\ldots p_n$$ and $p_1\ldots p_n$ are coprime, where $\hat a$ means that the product excludes $a$.

The only candidates for common prime factors are the $p_i$. But

$$S\equiv p_1p_2\ldots\hat p_i\ldots p_n\pmod {p_i}$$which is nonzero as the primes are distinct and not equal to $p_i$. Hence $p_i\nmid S$ for each $i$.