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Prime Reciprocals

Problem

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


Solution

S = 1/p1 + 1/p2 + ... 1/pk
  = (p2p3...pk)/(p1p2...pk) + (p1p3...pk)/(p1p2...pk) + ...
  = (p2p3...pk + p1p3...pk + ... + p1p2...pkminus1)/(p1p2...pk)

Therefore, p2p3...pk + p1p3...pk + ... + p1p2...pkminus1 = S p1p2...pk.

As all but the last term on the left hand side contains the factor, pk, we can write, pkQ + p1p2...pkminus1 = S p1p2...pk.

Dividing both sides by pk: Q + p1p2...pkminus1/pk = S p1p2...pkminus1.

But as none of p1, p2, ..., pkminus1 divide by pk, 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?

Problem ID: 238 (02 Aug 2005)     Difficulty: 4 Star

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