Four different positive integers whose reciprocals sum to 1

Question

The 6 solutions for n=4 are 2,3,7,42; 2,3,8,24; 2,3,9,18; 2,3,10,15; 2,4,5,20; 2,4,6,12.

How would one prove 42 is indeed the largest here? And/or this list is exhaustive?

It's a finite problem. At least one of the integers must be $3$ or less, so just break it into cases. A bit tedious, but not actually hard. — lulu, Oct 19 '23 at 16:00

score 2 · Accepted Answer · answered Oct 19 '23 at 16:20

The smallest integer must be $2$, since otherwise the highest the sum can be is $$\frac{1}{3} + \frac{1}{4} + \frac{1}{5}+\frac{1}{6} < 1.$$

By similar reasoning, the second smallest integer must be in $\{3,4,5\}$, since

$$\frac{1}{6} + \frac{1}{7} + \frac{1}{8} < \frac{1}{2}.$$

Proceeding in this way, the third smallest integer can be no higher than 11, since

$$\frac{1}{12}+ \frac{1}{13} < 1 - \frac{1}{2} - \frac{1}{3}.$$

As the fourth fraction is determined by the first three, this gives a total of 21 cases to check.

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