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 , with the numerators omitted for compactness:

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:

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