Why is this the nearest integer? I was trying to prove that $(N+\sqrt{N^2-1})^k$, where k is a positive integer, differs from the integer nearest to it by less than $(2N-\frac{1}{2})^{-k}$. Note: N is an integer greater than 1. 
So, I tried to look for the answer of the question, which I have taken it from an exam paper. 
It said that:
We let T=$(N+\sqrt{N^2-1})^K +(N+\sqrt{N^2-1})^{-k}$= $(N+\sqrt{N^2-1})^k + (N-\sqrt{N^2-1})^k=2(N^k + kC2 N^{K-2}(N^2-1)+...$ which is clear it is an integer.  
We know that: 
$(N-\frac{1}{2})^2 =N^2-N+\frac{1}{4}=\frac{5}{4}-N$ This is <0 when N > 1, since N is an integer.  
So, 
$N-\frac{1}{2}<\sqrt{N^2-1}$.
$2N-\frac{1}{2} < N+\sqrt{N^2-1}$
$(2N-\frac{1}{2})^{-k}>(N+\sqrt{N^2-1}^{-k}$
Let $|T-(N+\sqrt{N^2-1})^{-k}$| we would be able to prove the question.
However, my question is that, we need to have the nearest integer, so, the integer that the answer used is
 $(N+\sqrt{N^2-1})^k + (N + \sqrt{N^2-1})^{-k}$.
Why is this the nearest integer? 
Thank you so much for your reply. 
 A: By induction on $k$, we have
$$ (N\pm \sqrt{N^2-1})^k=a\pm b\sqrt{N^2-1}$$
with $a,b\in\Bbb Z$.
And of course
$(N-\frac12)^2=N^2-N+\frac14<N^2-1 $ implies
$$ 2N-\frac12<N+\sqrt{N^2-1}<2N.$$
Now from 
$$ (N+\sqrt{N^2-1})^k(N-\sqrt{N^2-1})^k=((N+\sqrt{N^2-1})(N-\sqrt{N^2-1}))^k=1^k=1,$$
we conclude that $(N+\sqrt{N^2-1})^k=a+b\sqrt{N^2-1}$ differs from the integer $2a$ by
$$a-b\sqrt{N^2-1}=\frac1{(N+\sqrt{N^2-1})^k}<\frac1{(2N-\frac12)^k}.$$ 
A: Given that $\,N>1\,$ is an integer, 
define the real numbers 
$\, u := (N+\sqrt{N^2-1}),\,$
$\, v := (N-\sqrt{N^2-1}),\,$
and thus $\,u+v=2N, u\,v=1.\,$
Define the sequence
$\, a(n) := u^n+v^n.\,$
This sequence satisfies $\,a(n) = a(-n)\,$ and a
 linear recurrence
$\, a(n+1) = 2Na(n)-a(n-1) \,$
both for all integer $\,n\,$ with
$\,a(0) = 2\,$ and $\,a(1) =2N.\,$ Thus all
$\,a(n)\,$ are positive even integers.
Read the Wikipedia article Lucas sequence for many details 
about such sequences.
Now notice that $\,0<v<\frac12\,$
and so $\,0<v^n<\frac12\,$ for all positive integer $\,n.\,$ Thus
$\,a(n)\,$ is the closest integer to
$\,u^n\,$ and $\,u^n=a(n)-v^n
.\,$
