### Instruction: There are infinitely many positive integers $k$ which satisfy the equation
\[\cos^2 (k^2 + 6^2)^\circ = 1.\]Enter the two smallest solutions, separated by commas.### Response: I notice that the equation involves a cosine squared, and I know that $\cos^2 x = 1$ only when $x$ is a multiple of $180^\circ$.
So I need to find values of $k$ such that $(k^2 + 6^2)^\circ$ is a multiple of $180^\circ$.
I can write this as $(k^2 + 36)^\circ = 180n^\circ$ for some integer $n$.
Dividing both sides by $180^\circ$, I get $(k^2 + 36) / 180 = n$.
Multiplying both sides by $180$, I get $k^2 + 36 = 180n$.
Rearranging, I get $k^2 = 180n - 36$.
Now I need to find the smallest values of $n$ that make $180n - 36$ a perfect square.
I can try some small values of $n$ and see if they work.
If $n = 1$, then $180n - 36 = 144 = 12^2$, so that's a possible value of $k^2$.
Taking the square root, I get $k = \pm 12$, but I only want positive values of $k$, so $k = 12$ is one solution.
If $n = 2$, then $180n - 36 = 324 = 18^2$, so that's another possible value of $k^2$.
Taking the square root, I get $k = \pm 18$, but again, I only want positive values of $k$, so $k = 18$ is another solution.
I don't need to try any larger values of $n$, because they would give larger values of $k$, and I only want the smallest solutions.
Therefore, the two smallest solutions are $k = 12$ and $k = 18$.