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 $\frac{4}{8}$ , but for larger denominators it might not always be obvious that, for example, $\frac{2}{1729}=\frac{4}{3458}$ . If one is unsure, one can always enter the fraction $\frac{4}{n}$ 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

\frac{4}{2}=1+\frac{1}{2}+\frac{1}{2}.

Similarly, for $n=3$ we have

\frac{4}{3}=\frac{1}{2}+\frac{1}{2}+\frac{1}{3}.

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

\frac{4}{4}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6},

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 commutative, it does not matter in what order we list the unit fractions that add up to our desired $\frac{4}{n}$ . 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:

\frac{4}{42}=\frac{1}{12}+\frac{1}{126}+\frac{1}{252}.

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