I was trying to solve the problem below but was completely unable to. My question is that which general category do questions like these belong to?. Once, I know that I would go learn the specific math concept and attempt at solving the question.
For each positive integer $n$, define $a_n$ and $b_n$ to be the positive integers such that
$$\left(\sqrt 3 + \sqrt 2\right)^{2n} = a_n + b_n\sqrt 6$$
and
$$\left(\sqrt 3 - \sqrt 2\right)^{2n} = a_n - b_n\sqrt 6$$
(a) Determine the values of $a_2$ and $b_2$.
(b) Prove that $2a_n - 1 < \left(\sqrt 3 + \sqrt 2\right)^{2n}<2a_n$ for all positive integers $n$.
(c) Let $d_n$ be the ones (units) digit of the number $\left(\sqrt 3 + \sqrt 2\right)^{2n}$ when it is written in decimal form. Determine, with justification, the value of $d_1+d_2+d_3+\cdots+d_{1865}+d_{1866}+d_{1867}$ (the given sum has 1867 terms.)
I would call all of these algebra, but this is definitely competition math algebra. These are solvable by expanding through binomial expansion (and I believe this is the intended solution).
If you need more hints, ask below.