### 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 know that the cosine function is 1 when the angle is a multiple of 360 degrees, or 0, 360, 720, and so on.
So I need to find the values of $k$ that make $(k^2 + 6^2)^\circ$ a multiple of 360.
I can rewrite this as $k^2 + 36 = 360n$ for some positive integer $n$.
Then I can solve for $k$ by subtracting 36 from both sides and taking the square root: $k = \sqrt{360n - 36}$.
Now I need to find the smallest values of $n$ that make $360n - 36$ a perfect square.
One way to do this is to try different values of $n$ and see if the square root is an integer.
I can start with $n = 1$, which gives $k = \sqrt{360 - 36} = \sqrt{324} = 18$.
That's a valid solution, so I'll write it down: 18.
I can continue with $n = 2$, which gives $k = \sqrt{720 - 36} = \sqrt{684}$.
That's not an integer, so I'll skip it.
I can try $n = 3$, which gives $k = \sqrt{1080 - 36} = \sqrt{1044}$.
That's also not an integer, so I'll skip it too.
I can try $n = 4$, which gives $k = \sqrt{1440 - 36} = \sqrt{1404}$.
That's also not an integer, so I'll skip it as well.
I can try $n = 5$, which gives $k = \sqrt{1800 - 36} = \sqrt{1764} = 42$.
That's another valid solution, so I'll write it down: 42.
I could keep going, but the problem only asked for the two smallest solutions, so I'm done.