Taking these type of problems to a new level

From the given condition, we see that if $x=\alpha$ satisfies, then $x=\dfrac{1}{\alpha}$ also satisfies.

Think of a quadratic equation $x^2-3x+1=0$ .

See that $\alpha$ and $\dfrac{1}{\alpha}$ are roots of this equation.

From this quadratic equation, we can form a recurrence relation $b_n-3b_{n-1}+b_{n-2}=0$ . See that if you design it's initial terms as $b_0=2$ and $b_1=3$ , then we get each term of this recurrence as $b_n=\alpha^n+\dfrac{1}{\alpha^n}$ .

Now, as we have $b_n=3b_{n-1}-a_{n-2}$ and $b_0$ and $b_1$ were defined integers, so we have each term of this sequence to be an integer.

But see that originally , we were given for $x^2+\dfrac{1}{x^2}=3$ , so we conclude that $b_n$ are the terms with even power of the roots, i.e. $b_i$ are ranging in $a_{2n}$

This means that only terms in the sequence $\{a_i\}$ which are integers, are for $i=2k$ for $k\in \mathbb{N}$ .

Thus we have $\boxed{500}$ integers in the sequence.

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