Let's determine the pairs of positive integers $\left(a_{1} ; a_{2}\right)$ for which the sequence defined by the recursion $a_{n+2}=\frac{a_{n}+a_{n+1}}{\left(a_{n}, a_{n+1}\right)}(n \geq 1)$ is periodic.
Solution. The terms of the sequence are clearly positive integers due to the formation rule. Let $(for all n)$ $d_{n}=(a_{n}, a_{n+1})$, then $d_{n+1}$ divides $a_{n+1}$ and $a_{n+2}$; therefore, it also divides $d_{n} a_{n+2}-a_{n+1}=a_{n}$. Thus, since $d_{n+1} \mid a_{n}$ and $d_{n+1} \mid a_{n+1}$, $d_{n+1}$ is a divisor of the greatest common divisor of $a_{n}$ and $a_{n+1}$, which is $d_{n}$. Specifically, we get that

$$
d_{n+1} \leq d_{n}
$$

for all $n$.

Suppose the sequence $\left(a_{k}\right)$ is periodic; then the sequence $\left(d_{k}\right)$ of the greatest common divisors of consecutive terms is also periodic. We have seen that this sequence is monotonically decreasing, so it must be constant:

$$
d_{1}=d_{2}=\ldots:=d
$$

If $d=1$, then $a_{n+2}=a_{n+1}+a_{n}>a_{n+1}$ for all $n$, so the sequence (from the second term onwards) is strictly monotonically increasing, and thus cannot be periodic. If $d \geq 3$, then

$$
\begin{aligned}
a_{n+3}+a_{n+2} & =\frac{a_{n+2}+a_{n+1}}{d}+\frac{a_{n+1}+a_{n}}{d}=\frac{\frac{a_{n+1}+a_{n}}{d}+a_{n+1}}{d}+\frac{a_{n+1}+a_{n}}{d}= \\
& =\left(\frac{1}{d^{2}}+\frac{2}{d}\right) a_{n+1}+\left(\frac{1}{d^{2}}+\frac{1}{d}\right) a_{n}<a_{n+1}+a_{n}
\end{aligned}
$$

According to this, the sequence $\left(a_{k+1}+a_{k}\right)$ has a subsequence that is strictly monotonically decreasing, so the sequence itself cannot be periodic. Therefore, the sequence $\left(a_{k}\right)$ cannot be periodic, contradicting our assumption. This proves that the value of $d$ can only be 2. The formation rule of the sequence is then as follows:

$$
a_{n+2}=\frac{a_{n+1}+a_{n}}{2}
$$

every term of the sequence is the arithmetic mean of the two preceding terms. From this, it is intuitively clear (or can be derived from the above formula with a simple rearrangement) that the difference between consecutive terms of the sequence is halved at each step:

$$
\left|a_{n+2}-a_{n+1}\right|=\left|\frac{a_{n+1}-a_{n}}{2}\right|
$$

If $a_{1} \neq a_{2}$, then $\left|a_{n+2}-a_{n+1}\right|=\frac{\left|a_{2}-a_{1}\right|}{2^{n}}$ shows that for sufficiently large values of $n$, $\left|a_{n+2}-a_{n+1}\right|$ cannot be an integer, contradicting the fact that the elements of the sequence $\left(a_{k}\right)$ are (positive) integers. Therefore, $a_{1}=a_{2}$, and since $\left(a_{1}, a_{2}\right)=2$, it follows that $a_{1}=a_{2}=2$. This implies that every element of the sequence is 2; in this one possible case, the (constant 2) sequence is indeed periodic.