Solution:
First, note that for $k \geq 1$,
$$a_{k+1}^2=\frac{k(k+1) a_k^2}{k(k+1)+2 a_k^2} \Longleftrightarrow \frac{1}{a_{k+1}^2}-\frac{1}{a_k^2}=\frac{2}{k(k+1)}=\frac{2}{k}-\frac{2}{k+1}$$
and summing the second equation from $k=1$ to $k=n-1$ with $n \geq 2$, we get
$$\frac{1}{a_n^2}-\frac{1}{a_1^2}=\sum_{k=1}^{n-1}\left(\frac{1}{a_{k+1}^2}-\frac{1}{a_k^2}\right)=\sum_{k=1}^{n-1}\left(\frac{2}{k}-\frac{2}{k+1}\right)=2-\frac{2}{n}$$
Since $a_1=1$, we see that
$$\frac{1}{a_n^2}=3-\frac{2}{n} \Longleftrightarrow a_n=\sqrt{\frac{n}{3 n-2}}$$
for all integers $n \geq 1$ and we wish to find the sum of all positive integers $n<1000$ such that $\frac{n}{3 n-2}$ is a square of some rational number. To help us look for such integers $n$, we use the following lemma that provides integer solutions to the generalized Pell equation.

Lemma. ${ }^{1}$ Let $d$ be a squarefree positive integer, and let $a$ and $b$ be positive integers such that $a^2-d b^2=1$. Set $u=a+b \sqrt{d}$. Then for each nonzero integer $n$, every solution of $x^2-d y^2=n$ is a power of $u$ times $x+y \sqrt{d}$ where $(x, y)$ is an integer solution of $x^2-d y^2=n$ with $|x| \leq \sqrt{|n|}(\sqrt{u}+1) / 2$ and $|y| \leq \sqrt{|n|}(\sqrt{u}+1) /(2 \sqrt{d})$.

We now let $g=\operatorname{gcd}(n, 3 n-2)$. Then $g \in\{1,2\}$ since $g \mid 3 n-(3 n-2)=2$. We now consider the following cases:

- Suppose $g=1$. Then $n=y^2$ and $3 n-2=x^2$ for some relatively prime positive integers $x$ and $y$. This leads us to the generalized Pell equation $x^2-3 y^2=-2$. Set $u=2+\sqrt{3}$, with $(2,1)$ being a solution of $x^2-3 y^2=1$ in positive integers. We now look for the positive integer solutions $(x, y)$ of $x^2-3 y^2=-2$ with $x \leq \sqrt{2}(\sqrt{u}+1) / 2 \approx 2.07$ and $y \leq \sqrt{2}(\sqrt{u}+1) /(2 \sqrt{3}) \approx 1.2$. We obtain $(x, y)=(1,1)$ as the only such integer solution, so by the above lemma, we see that all positive integer solutions $\left(x_k, y_k\right)$ of $x^2-3 y^2=-2$ are given by $x_k+y_k \sqrt{3}=(1+\sqrt{3})(2+\sqrt{3})^k$ for all integers $k \geq 0$. We now compute this product for small values of $k$ :

As $n<1000$, we require that $y_k \leq 31$, so the positive integer solutions $\left(x_k, y_k\right)$ of $x^2-3 y^2=$ -2 with $y_k \leq 31$ are $\left(x_k, y_k\right)=(1,1),(5,3),(19,11)$ (which indeed have relatively prime coordinates). These correspond to the values of $n: n=1,9,121$.

- Suppose $g=2$. Then $n=2 y^2$ and $3 n-2=2 x^2$ for some relatively prime positive integers $x$ and $y$. This leads us to the generalized Pell equation $x^2-3 y^2=-1$. Again, we set $u=2+\sqrt{3}$. We now look for the positive integer solutions $(x, y)$ of $x^2-3 y^2=-1$ with $x \leq(\sqrt{u}+1) / 2 \approx 1.47$ and $y \leq(\sqrt{u}+1) /(2 \sqrt{3}) \approx 0.85$. It turns out that there are no such solutions on these bounds, so by the above lemma, we conclude that $x^2-3 y^2=-1$ has no solutions in positive integers.

Hence, the only positive integers $n<1000$ for which $a_n$ is a rational number are $n=1,9,121$ and the sum is $1+9+121=131$.

[^0]: ${ }^{1}$ For proof, see Theorem 3.3 from https://kconrad.math.uconn.edu/blurbs/ugradnumthy/pelleqn2.pdf.