
4. Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer.


Solution: Let $x, y, z$ be three distinct positive integers satisfying the given conditions. We may assume that $xa=1$, we must have $b=2$. This gives

$$
\frac{1}{c+1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}
$$

or $c=5$. Thus $x: y: z=a+1: b+1: c+1=2: 3: 6$. Thus the required numbers with the least sum are $2,3,6$.
