How should one approach the floor function of a binomial expansion mod an integer? For instance, what is $$\left\lfloor\left(\sqrt{2}+\sqrt{3}\right)^{2020}\right\rfloor$$ mod $101$?
I tried to expand using the Binomial Theorem, and then taking the each term of the resulting expansion mod 101, but stumbled upon cases like $\binom{2020}{101k}$. I don't really know how to go from there. Any help would be appreciated!
Let $x_n=(\sqrt{2}+\sqrt{3})^{2n}, y_n=(\sqrt{3}-\sqrt{2})^{2n}$.
We have $x_n+y_n=2((\sqrt{3})^{2n}+\tbinom{2n}{2}(\sqrt{3})^{2n-2}(\sqrt{2})^2+...+\tbinom{2n}{2n-2}(\sqrt{3})^2(\sqrt{2})^{2n-2}+(\sqrt{2})^{2n})\in \mathbb N$.
Also $y_n\in (0,1)$, so $\lfloor x_n\rfloor = 2((\sqrt{3})^{2n}+\tbinom{2n}{2}(\sqrt{3})^{2n-2}(\sqrt{2})^2+...+\tbinom{2n}{2n-2}(\sqrt{3})^2(\sqrt{2})^{2n-2}+(\sqrt{2})^{2n})-1$.
Let $n=1010$, then $\lfloor x_{1010}\rfloor = 2((\sqrt{3})^{2020}+\tbinom{2020}{2}(\sqrt{3})^{2018}(\sqrt{2})^2+...+\tbinom{2020}{2018}(\sqrt{3})^2(\sqrt{2})^{2018}+(\sqrt{2})^{2020})-1\equiv 2(3^{1010}+2^{1010})-1\pmod {101}$
Then you can finish it by yourself.