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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.

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    $\begingroup$ Can you prove it for the sum of two distinct prime reciprocals? $\endgroup$ – GEdgar Apr 8 '16 at 14:42
  • $\begingroup$ @GEdgar, not immediately. But you are right, I should've done this case before asking the general question $\endgroup$ – Yuriy S Apr 8 '16 at 14:44
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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.

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  • $\begingroup$ Thank you, this is completely clear to me $\endgroup$ – Yuriy S Apr 8 '16 at 14:50
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Apr 8 '16 at 14:51
  • $\begingroup$ @wythagoras: Thank you for noticing! Fixed. $\endgroup$ – André Nicolas Apr 9 '16 at 5:55
  • $\begingroup$ @AndréNicolas I also fixed it in the other places. $\endgroup$ – wythagoras Apr 9 '16 at 6:03
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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 $i$.

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