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.