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? 
 A: Erdős and Graham mention in their monograph Old and New Problems and Results in Combinatiorial Number Theory (see here: http://www.math.ucsd.edu/~ronspubs/80_11_number_theory.pdf, bottom of page 38) that Barbeau notes that this is unknown, in the following paper:
Barbeau, E.J. Computer challenge corner: Problem 477: A brute force program. J. Rec. Math. 9 (1976/1977), p. 30.
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:
Barbeau, E.J. Expressing one as a sum of distinct reciprocals: comments and a bibliography. Eureka (Ottowa) 3 (1977), 178-181.
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: $S = \{6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 77, 82, 85, 86, 87, 91, 93, 95, 115, 119, 123, 133, 155, 187, 203, 209, 215, 221, 247, 265, 287, 299, 319, 323, 391, 689, 731, 901 \}$
