How to deal with binomial expansion within floor function as in $\lfloor{(a+\sqrt{b})^n\rfloor}$? In questions involving floor functions containing binomial coefficients, like example 368 in the posted image, where it asks

for $n$, a nonnegative integer, show that the integers $\lfloor{(1+\sqrt{2})^n\rfloor}$  are alternatively even and odd.  

The solution starts with "By the binomial theorem, $(1+\sqrt{2})^n + (1-\sqrt{2})^n$..."   
I would like some clarifications about the beginning of the solution steps. 
My questions are as follows: 
1) How does adding the term $(1-\sqrt{2})^n$ follow from the binomial theorem?   
2) Is it because the coefficients of the binomial expansions are in the form of $a+\sqrt{b}$ 
3)How did the author know to use $(1-\sqrt{2})^n$ to get the fractional part of $(1+\sqrt{2})^n$?   
4) If I change the terms to something that is different from $(a+\sqrt{b})$, to $(m + n)$, where m and n have some other kind of values, like transcendental functions evaluated at particular values, fractions of different values, n-th root of different values, etc, I don't think I could easily say that $(m-n)^n$ is the fractional part of $(m+n)^n$.  Basically, would the same technique work for them all.    Thank you in advance.


The point is that if you expand $(1+\sqrt 2)^n+(1-\sqrt 2)^n$ by the binomial theorem, the terms with $\sqrt 2$ raised to an odd power cancel while the ones with $\sqrt 2$ raised to an even power are equal in the two terms.  The $k$ in the summation is half the power of $\sqrt 2$ in the terms we are considering.  The leading factor of $2$ comes from the fact that the terms match.  The author uses $1-\sqrt 2$ because it is the conjugate of $1+\sqrt 2$ and makes the cancellation work.  To make this work with $(m+k)^n$ (please do not reuse $n$ in the expression when they are not the same) you need $(m+k)^n+(m-k)^n$ to be an integer and $|m-k| \lt 1$.  To get the sum to be an integer you want $m$ an integer and $k$ a square root so the cancellation gets rid of the square roots.  Then if $m$ is the integer one one side or the other of $\sqrt k$ the magic works.
