# Product of sum of reciprocals of prime numbers

For any positive integers $$k$$ and $$\ell$$, does the equation $$\left(\sum_{i=1}^k \frac{1}{p_i}\right) \left(\sum_{j=1}^\ell \frac{1}{q_j}\right) = 1$$ have solutions in distinct primes, that is, $$p_1, p_2, \dots, p_k, q_1, q_2, \dots, q_\ell$$ are distinct?

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– Seva
Jan 14, 2019 at 6:36
• math.stackexchange.com/questions/3064588/… Jan 14, 2019 at 14:32
• The requirement that $\{p_1, \ldots, p_k\}$ and $\{q_1, \ldots, q_l\}$ are disjoint is not needed. The $p_i$-adic order of $\sum_i 1/p_i$ is $-1$, so if $p_i$ is also one of the $q_j$, the $p_i$-adic order of the product of the two sums is $-2$, not $0$. Jan 14, 2019 at 18:22
• If the product of two positive numbers is $1$, the sum of those two numbers must be at least $2$. The sum of the reciprocals of the first $58$ primes is less than $2$, so if a solution exists, it must have $k+l\geq 59$. Jan 15, 2019 at 2:39

On the other hand, also mentioned by Erdős and Graham, Barbeau does exhibit a set $$\{x_1, .., x_{101}\}$$ such that $$1 = \displaystyle \sum_{i=1}^{101} \dfrac{1}{x_i}$$ where all $$x_i$$ are the product of two distinct primes, in the following paper:
Edit: A.W. Johnson also found a set $$S$$ of $$x_i$$ (with $$|S| = 48$$) such that the sum of the reciprocals of the $$x_i$$ equals $$1$$ and the $$x_i$$ are the product of two distinct primes. The set $$S$$ is as follows: $$S = \{6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 77, 82, 85, 86, 87, 91, 93, 95, 115, 119, 123, 133, 155, 187, 203, 209, 215, 221, 247, 265, 287, 299, 319, 323, 391, 689, 731, 901 \}$$