integral part of surd 
$(\sqrt{a}+b)^n=N+f$ where $f \in (0,1)$
$(\sqrt{a}+b)^{n+2} =M+g$ where $g \in (0,1)$
Given that $0<\sqrt{a}-b<1$ and $(a,b)$
   belongs to integers, then
  
  
*
  
*If $n$ is odd, $f>g$
  
*If $n$ is even,  $f<g$

How to prove/disprove it?
 A: Since $a$ and $b$ are integers (and $a > 0$ is not a square, otherwise $0 < \sqrt{a}-b < 1$ would not hold), we have
$$\begin{align}
(b+\sqrt{a})^n + (b-\sqrt{a})^n &= \sum_{k=0}^n \binom{n}{k}b^{n-k}a^{k/2} + \sum_{k=0}^n (-1)^k\binom{n}{k}b^{n-k}a^{k/2}\\
&= \sum_{k=0}^n \bigl(1 + (-1)^k\bigr)\binom{n}{k}b^{n-k}a^{k/2}\\
&= 2\sum_{m=0}^{\lfloor n/2\rfloor} \binom{n}{2m}b^{n-2m}a^m,
\end{align}$$
so $(b+\sqrt{a})^n + (b-\sqrt{a})^n$ is an integer. From $0 < \sqrt{a}-b < 1$ it follows that $0 < \lvert b-\sqrt{a}\rvert^n < 1$ and $(b-\sqrt{a})^n$ is positive when $n$ is even, and negative when $n$ is odd.
Hence for odd $n$, we have
$$(\sqrt{a}+b)^n = \underbrace{(b+\sqrt{a})^n + (b-\sqrt{a})^n}_{N(n)} + \underbrace{(\sqrt{a}-b)^n}_{f(n)}$$
and with the notations of the problem, $g = f(n+2) = (\sqrt{a}-b)^2\cdot f(n) = (\sqrt{a}-b)^2\cdot f < f$.
For even $n$ we have
$$(\sqrt{a}+b)^n = \underbrace{(b+\sqrt{a})^n + (b-\sqrt{a})^n - 1}_{N(n)} + \underbrace{1-(\sqrt{a}-b)^n}_{f(n)},$$
and in the notation of the problem $g = 1-(\sqrt{a}-b)^{n+2} > 1 - (\sqrt{a}-b)^n = f$.
