The Erdös-Strauss conjecture

Introduction. The greedy-algorithm result tells one that

  • a/b = 1/n[1]+a[1]/b[1] where a[1]/b[1] is a fraction whose numerator a[1] is less than a .
  • a[1]/b[1] = 1/n[2]+a[2]/b[2] where a[2]/b[2] is a fraction whose numerator a[2] is less than a[1]
  • a[2]/b[2] = 1/n[3]+a[3]/b[3] where a[3]/b[3] is a fraction whose numerator a[3] is less than a[2]

    etc

The sequence a[1], a[2], a[3], `...` is strictly monotonic decreasing, which eventually terminates when one has a[r] = 1 , for some r . For a given a , there will then be at most a terms in the sequence of unit fractions: 1/n[1], 1/n[2], 1/n[3], `...` , and clearly the maximum can happen in the event that the sequence a[1], a[2], a[3], `...` happens to be the natural numbers less than a : a-1, a-2, a-3, `...`, 2, 1 .

Thus, using the greedy algorithm :

  • 2/b will require at most 2 unit fractions
  • 3/b will require at most 3 unit fractions
  • 4/b will require at most 4 unit fractions
  • 5/b will require at most 5 unit fractions

    etc

Conjecture (Erdös-Strauss). For every natural number n , greater than 3, there is an Egyptian fraction representation that is better than the one ensured by the greedy algorithm representation; in other words: for every integer n (greater than 3) there are natural numbers x , y and z such that

4/n = 1/x+1/y+1/z

Interested readers may consult Chapter 30 of Mordell's Diophantine Equations .

Some elementary ideas (which could be at school level). Using

4/(3*x+2) = 1/(x+1)+1/(3*x+2)+1/((x+1)*(3*x+2))


Thus, setting
n = 3 x + 2 ( x = 0, 1, 2, 3, 4, 5, ... ), we have that the Erdös-Strauss conjecture is (trivially) true for all such n .

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This page was last updated 18 February 2005 15:07:06 -0000