, 2020

Solution:

By the condition of the problem statement, we have
$$a_{n}^{2} - a_{n-1} a_{n+1} = 1 = a_{n-1}^{2} - a_{n-2} a_{n}$$
This is equivalent to
$$\frac{a_{n-2} + a_{n}}{a_{n-1}} = \frac{a_{n-1} + a_{n+1}}{a_{n}}$$
Let $k = \frac{a_{0} + a_{2}}{a_{1}}$. Then we have
$$\frac{a_{n-1} + a_{n+1}}{a_{n}} = \frac{a_{n-2} + a_{n}}{a_{n-1}} = \frac{a_{n-3} + a_{n-1}}{a_{n-2}} = \cdots = \frac{a_{0} + a_{2}}{a_{1}} = k.$$
Therefore we have $a_{n+1} = k a_{n} - a_{n-1}$ for all $n \geq 1$. We know that $k$ is a positive rational number because $a_{0}, a_{1}$, and $a_{2}$ are all positive integers. We claim that $k$ must be an integer. Suppose that $k = \frac{p}{q}$ with $\operatorname{gcd}(p, q) = 1$. Since $k a_{n} = a_{n-1} + a_{n+1}$ is always an integer for $n \geq 1$, we must have $q \mid a_{n}$ for all $n \geq 1$. This contradicts $a_{2}^{2} - a_{1} a_{3} = 1$. Conversely, if $k$ is an integer, inductively all $a_{i}$ are integers.

Now we compute $a_{2} = \frac{t^{2} - 1}{15}$, so $k = \frac{t^{2} + 224}{15 t}$ is an integer. Therefore $15 k - t = \frac{224}{t}$ is an integer. Combining with the condition that $a_{2}$ is an integer limits the possible values of $t$ to $1, 4, 14, 16, 56, 224$. The values $t < 15$ all lead to $a_{n} = 0$ for some $n$ whereas $t > 15$ leads to a good sequence. The sum of the solutions is
$$16 + 56 + 224 = 296$$