, 2018

$n = 3$

It's clear that the only solution when $n = 1$ is $a_1 = 1$. Now we show that:

(1) There are only finitely many $(x, y) \in \mathbb{Q}_{>0}^2$ such that $x + y$ and $\frac{1}{x} + \frac{1}{y}$ are both integers.

Write $x = a/b$ and $y = c/d$ in the standard form. Then, $x + y \in \mathbb{Z}$ and $\frac{1}{x} + \frac{1}{y} \in \mathbb{Z}$ is equivalent to the following two conditions

(i) $bd \mid ad + bc$.

Notice that this leads to $d \mid ad + bc \Rightarrow d \mid bc \Rightarrow b \mid c$ since $(c, d) = 1$.

Symmetrically, we get $b \mid d$. This leads to $b = d$.

(ii) $ac \mid ad + bc$.

A same procedure as above give us $a = c$.

Hence, we must have $x = y$, so the problem becomes finding $x \in \mathbb{Q}_{>0}$ such that $2x \in \mathbb{Z}$ and $\frac{2}{x} \in \mathbb{Z}$. It's clear that the only solutions are $x = 1/2, 1, 2$.

(2) There are infinitely many $(x, y, z) \in \mathbb{Q}_{>0}^3$ such that $x + y + z$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ are all integers.

In particular, let us show that there are infinitely many such solutions satisfying $x + y + z = 1$. Note that in this case, we can rewrite
$$(x, y, z) = \left( \frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c} \right)$$
with $a, b, c \in \mathbb{Z}$. We want this to satisfy
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{a+b+c}{a} + \frac{a+b+c}{b} + \frac{a+b+c}{c} \in \mathbb{Z},$$
or equivalently,
$$\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} \in \mathbb{Z}.$$
We further fix $a = 1$, then we only need to show that there exists infinitely many integers $(b, c)$ such that
$$\frac{1}{b} + \frac{1}{c} + \frac{c}{b} + \frac{b}{c} = 3 \Leftrightarrow b^2 + c^2 - 3bc + b + c = 0. \quad (1)$$
To show that there are infinitely many solutions to Eq. (1), we use the Vieta jumping (a.k.a. root flipping): starting with $(b, c) = (2, 3)$. The following algorithm generates infinitely many solutions. Let $c \ge b$, and view Eq. (1) as a quadratic equation in $b$ of a fixed $c$ as
$$b^2 - (3c - 1) \times b + (c^2 + c) = 0. \quad (2)$$
Then there exists another root $b_0 \in \mathbb{Z}$ of Eq. (2) which satisfies $b+b_0 = 3c - 1$ and $b \times b_0 = c^2 + c$. Since $c \ge b$ by assumption,
$$b_0 = \frac{c^2 + c}{b} \ge \frac{c^2 + c}{c} > c.$$
Hence, from solution $(b, c)$, we obtain another solution $(c, b_0)$ with $b_0 > c$. We can then do this jump again, but this time treat $c$ as the variable in the quadratic equation. Continue this process, and we get infinitely many solutions.