### 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 cosine function is equal to 1 only when the angle is a multiple of 360 degrees.
So I want to find values of $k$ such that $(k^2 + 6^2)^\circ$ is a multiple of 360 degrees.
I can rewrite this as $k^2 + 36 = 360n$, where $n$ is a positive integer.
This is equivalent to $k^2 = 360n - 36$, or $k = \sqrt{360n - 36}$.
Now I need to find the smallest values of $n$ that make $k$ a positive integer.
I can try some values of $n$ starting from 1 and see if they work.
If $n = 1$, then $k = \sqrt{324} = 18$, which is a valid solution.
If $n = 2$, then $k = \sqrt{684}$, which is not an integer.
If $n = 3$, then $k = \sqrt{1044}$, which is not an integer either.
If $n = 4$, then $k = \sqrt{1404}$, which is still not an integer.
If $n = 5$, then $k = \sqrt{1764} = 42$, which is another valid solution.
I don't need to try any larger values of $n$, because they will only make $k$ larger.
So the two smallest solutions are $k = 18$ and $k = 42$.