mathschallenge.net

Problem

Given that P = {p₁, p₂, ... , p_k} is a set of distinct, not necessarily consecutive primes, prove that 1/p₁ + 1/p₂ + ... 1/p_k is never integer.

Solution

S	=	1/p1 + 1/p2 + ... 1/pk
	=	(p2p3...pk)/(p1p2...pk) + (p1p3...pk)/(p1p2...pk) + ...
	=	(p2p3...pk + p1p3...pk + ... + p1p2...pk1)/(p1p2...pk)

Therefore, p₂p₃...p_k + p₁p₃...p_k + ... + p₁p₂...p_k1 = S p₁p₂...p_k.

As all but the last term on the left hand side contains the factor, p_k, we can write, p_kQ + p₁p₂...p_k1 = S p₁p₂...p_k.

Dividing both sides by p_k: Q + p₁p₂...p_k1/p_k = S p₁p₂...p_k1.

But as none of p₁, p₂, ..., p_k1 divide by p_k, the RHS, and more specifically, S, cannot be integer. Q.E.D.

What can you determine about the possible values of the fraction, S = a/b?