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# 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:

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

## Mathematics Subject Classification

11A67*no label found*

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