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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.)

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  • $\begingroup$ For a) take n=2 and calculate. You get two equations from it. Solve this system of $a_2$ and $b_2$ $\endgroup$ – Cornman Nov 19 '17 at 1:46
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    $\begingroup$ (a) is arithmetic. (b) might be induction. (c) could be Number Theory. $\endgroup$ – Gerry Myerson Nov 19 '17 at 1:47
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    $\begingroup$ It is important that $\sqrt 3 - \sqrt 2 \lt 1$ so when you raise it to a power it gets small. Then note that $(\sqrt 3 + \sqrt 2)^{2n}+(\sqrt 3 - \sqrt 2)^{2n}$ is an integer because the terms with square roots in them cancel. This is a recurring theme in olympiad problems. $\endgroup$ – Ross Millikan Nov 19 '17 at 3:55
  • $\begingroup$ Look at the units digits of both $a_n$ and $b_n$ for small $n$ (say, through $n=5$) and see if you notice anything useful. $\endgroup$ – Alexander Burstein Nov 19 '17 at 4:07
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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.

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