examples supporting the Erdős-Straus conjecture


As with any conjecture, a million examples are not enough to prove the Erdős-Straus conjecture, but a single counterexample is enough to disprove. But if these examples at least provide a slightly better understanding of the problem at hand, the effort is not entirely wasted.

Most users are well aware that a computer algebra system (and even fraction-capable scientific calculators) will automatically express a fraction by the lowest common denominator. This presents no problem for smaller instances, such as 48, but for larger denominators it might not always be obvious that, for example, 21729=43458. If one is unsure, one can always enter the fraction 4n by itself and the CAS will dutifully respond with the LCD expression, which will hopefully match the sum of three unit fractions entered earlier.

I say we start with n=2 if for no other reason than to start at the beginning. The only possible solution is

42=1+12+12.

Similarly, for n=3 we have

43=12+12+13.

It is with n=4 that solutions with distinct denominators become available:

44=12+13+16,

easily suggested by the study of perfect numbers. Some may consider solutions with distinct denominators more elegant, some are just happy to find any solution for a given n.

Since addition is commutativePlanetmathPlanetmathPlanetmathPlanetmath, it does not matter in what order we list the unit fractions that add up to our desired 4n. However, by tradition, they are listed in descending order: the biggest fraction (the one with the smallest denominator) is listed first, the smallest last.

The following table lists some distinct denominator solutions for 4<n<21, with the numerators omitted for compactness:

5 2, 5, 10
6 2, 8, 24; 3, 4, 12
7 2, 21, 42; 3, 6, 14
8 3, 6, 42; 3, 8, 24; 4, 6, 12
9 3, 12, 36; 4, 6, 36
10 3, 20, 60; 5, 6, 30; 4, 10, 20
11 3, 44, 132; 4, 11, 44; 4, 12, 33
12 4, 18, 36; 4, 16, 48; 6, 8, 24
13 4, 26, 52
14 4, 42, 84; 5, 70, 140; 6, 14, 21; 7, 8, 56; 6, 12, 28
15 4, 90, 180; 5, 18, 90; 6, 15, 30; 7, 10, 42
16 5, 30, 60; 6, 12, 84; 8, 12, 24; 6, 20, 30
17 5, 30, 510; 6, 17, 102
18 6, 24, 72; 8, 12, 72
19 6, 38, 57
20 7, 20, 140

As the table shows, solutions for prime n are harder to come by than for composite n. When n=pq, with p prime and q any other integer, solutions for n can be simply derived from those for p by multiplying those denominators by q. For example, for n=42, we can take the solutions for n=6, multiply those denominators by 7 and voilà:

442=112+1126+1252.
Title examples supporting the Erdős-Straus conjecture
Canonical name ExamplesSupportingTheErdHosStrausConjecture
Date of creation 2013-03-22 16:54:10
Last modified on 2013-03-22 16:54:10
Owner PrimeFan (13766)
Last modified by PrimeFan (13766)
Numerical id 9
Author PrimeFan (13766)
Entry type Example
Classification msc 11A67