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Solution for November 2007

The Problem:

Prove that for every n, the integers 1, 2, 3, ... , 2n – 1, 2n can be partitioned into pairs

{a1, b1}, {a2, b2}, ... , {an, bn}

so that ai + bi is prime for every i.

Correct responses:

Correct solutions were submitted by

Bojan Basic (Serbia)

Wolfgang Kais (Germany)

Gérard Billion (France)

Jeremy Lam (UK)

Lou Cairoli (USA)

Jacques Mertzeisen (France)

John Campbell (Alberta)

Viktor Pačnik  (Slovenia)

Adrien Delcour (Belgium)

Mark Pilloff (USA)

Philippe Fondanaiche (France)

Ananda Raidu (India)

Alenka Gabrovec (Slovenia)

John T. Robinson (USA)

Xavier Hecquet (France)

Heri Setiyawan (Indonesia)

Pierre Bornsztein (France) Luis Fabricio Tano (internet)
Jean-Luc Luyet (Switzerland)  


The solution:

Preliminary remarks.  The solution depends on a result that is known variously by the names Bertrand's Postulate (after Joseph Bertrand (1822-1900) who conjectured it in 1845) and Chebyshev's Theorem (after Pafnuty Chebyshev (1821-1894) who proved the conjecture in 1850):

For every integer n > 1, there is at least one prime number
n and 2n.

Or, in the words of mathematician N.J. Fine,

Chebyshev said it, but I'll say it again;
There's always a prime between n and 2n,

(which rhymes in American English, but probably sounds a bit strange to our British colleagues).  Details can be found on the web
or in standard number theory texts.
            All correspondents submitted essentially the same proof.  First, we look at a simple example to motivate our argument:  Let's partition the numbers from 1 to 8 into pairs, each of whose sum is prime.  We note that there are two primes between 8 and 15 that we can aim for, namely 11 and 13.  Let's choose 13; the corresponding pairs {5, 8}, {6, 7} sum to 13 and they use up all numbers in the subset from 5 to 8.  The task is now reduced to pairing the remaining numbers from 1 to 4.  This example should make it clear that an induction argument will be useful here.

Proof by induction.  To start the induction, the case n = 1 (with the given integers 1 and 2) is immediate — the numbers already form a pair whose sum is 3, a prime.  We next assume that the assertion is true for n running from 1 to k – 1; that is,

assume that for every n from 1 to k - 1, the integers 1, 2, 3, ..., 2n – 1, 2n can be partitioned into pairs {a1, b1}, {a2, b2}, ..., {an, bn} so that ai + bi is prime for every i.

We use the assumption for the case n = k; that is, we must pair the numbers from 1 to 2k.  By Bertrand's Postulate we are guaranteed a prime p between 2k and 4k.  Since
2k > p – 2k ≥ 1, we can pair
         2k with p – 2k,
         2k – 1 with p – 2k + 1,
          with .
Note that members of each pair sum to p, and all numbers from p – 2k to 2k have been used.  Because p – 2k – 1 is an even number that is at most 2(k – 1), the induction hypothesis can now be applied to pair the numbers from 1 to p – 2k – 1, and the proof is complete.




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