Find the 100th digit of $( 1 + \sqrt 2 )^{3000}$

Hints: $(1+\sqrt{2})^{3000}+(1-\sqrt{2})^{3000}$ is an integer and $(1-\sqrt{2})^{3000}$ is a small positive number.

Can you prove these two facts and use them to get the answer?

The way I understand it now is that since
$ (1+\sqrt{2})^{2} + (1 - \sqrt{2})^{2} = 6$
$\therefore$ $ (1+\sqrt{2})^{3000} + (1 - \sqrt{2})^{3000} = N$ (N is an integer)
since $(1 - \sqrt{2})^{3000}$ would be a very small number so $ (1+\sqrt{2})^{3000} = N - 0.0000...001$ so the 100th digit to the right of the decimal point should be 9