2023 AIME II Problem 13

With $s = \sec A$ and $t = \tan A,$ the condition says $st = 2,$ and always $s^2 - t^2 = 1,$ so $s^2 + t^2 = \sqrt{17}$

$s^n + t^n$ is an integer only when $4$ divides $n,$ and $x_j = s^{4j} + t^{4j}$ satisfies $x_{j+1} = 9x_j - 16x_{j-1};$ track the units digits

Let $s = \sec A$ and $t = \tan A.$ The hypothesis $\tan A = 2\cos A$ says $\tan A \sec A = 2,$ i.e. $st = 2,$ and always $s^2 - t^2 = 1.$ Then $(s^2 + t^2)^2$ $= (s^2 - t^2)^2 + 4s^2t^2$ $= 17,$ so $u = s^2$ and $v = t^2$ satisfy $u + v = \sqrt{17},$ $uv = 4.$ The sums $w_m = u^m + v^m$ obey $w_{m+1} = \sqrt{17}\,w_m - 4w_{m-1}$ with $w_0 = 2,$ $w_1 = \sqrt{17};$ by induction, when $m$ is even, $w_m$ is a positive integer, and when $m$ is odd, it is an integer times $\sqrt{17}.$

For even $n = 2m,$ $s^n + t^n = w_m,$ an integer exactly when $m$ is even, i.e. $4$ divides $n.$ For odd $n,$ $(s^n + t^n)^2$ $= w_n + 2 (st)^n$ $= w_n + 2^{n+1}$ is irrational, so $s^n + t^n$ is not an integer. Thus write $n = 4j$ and $x_j = w_{2j}.$ Since $u^2 + v^2 = 9$ and $u^2 v^2 = 16,$ the integers $x_j$ satisfy $$\begin{gathered} x_{j+1} = 9x_j - 16x_{j-1}, \\ x_0 = 2, \\ x_1 = 9, \end{gathered}$$ giving $9, 49, 297, 1889, \ldots$ whose units digits repeat with period three: $9, 9, 7.$ The units digit is $7$ when $3$ divides $j$ and $9$ otherwise.

The valid $n \le 1000$ are $n = 4j$ with $1 \le j \le 250$ and $j$ not divisible by $3:$ there are $250 - 83 = 167$ of them.