Problem source

Let $a$ and $b$ be positive integers such that $b<2a-1$. For any given positive integer $n$, the integers $N$ and $M$ are defined by 
	\[
	[a+\sqrt{a^{2}-b}]^{n}=N-r,
	\]
	\[
	[a-\sqrt{a^{2}-b}]^{n}=M+s,
	\]
	where $0\leqslant r<1$ and $0\leqslant s<1$. Prove that 
\begin{questionparts}
	\item $M=0$, 
	\item $r=s$, 
	\item $r^{2}-Nr+b^{n}=0.$
\end{questionpart}
	Show that for large $n$, $\left(8+3\sqrt{7}\right)^{n}$ differs from an integer by about $2^{-4n}$.

Solution source

\begin{questionparts}
\item If we can show that $0 < a - \sqrt{a^2-b} < 1$ then we will be done, since raising a number in $[0,1)$ to a positive integer power will always remain in the same interval. 

Clearly $\sqrt{a^2-b} < \sqrt{a^2} = a$ so we have $a-\sqrt{a^2-b} > 0$

We also have that $\sqrt{a^2-b} > \sqrt{a^2-(2a-1)} = (a-1)$. Therefore $a - \sqrt{a^2-b} < a - (a-1) = 1$ as required.

\item If we can show that $\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n = N -r + s$ is an integer we will be done, since the only integer value $-r+s$ can be is $0$.

This is easy to see, since 
\begin{align*}
\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n &= \sum_{k=0}^n \binom{n}{k}a^{n-k}(\sqrt{a^2-b})^k +\sum_{k=0}^n \binom{n}{k}a^{n-k}(-\sqrt{a^2-b})^k \\
&=  \sum_{k=0}^n \binom{n}{k}a^{n-k}\l (\sqrt{a^2-b})^k +(-\sqrt{a^2-b})^k \r \\
\end{align*}

But every term where $k$ is odd in this sum is $0$ (since they cancel) and ever term where $k$ is even in this sum is an integer. Therefore the sum is an integer and we're done.

\item \begin{align*}
-r^2+rN &= -r(r-N) \\
&= s(r-N) \\
&=- \l a - \sqrt{a^2-b} \r^n \l a + \sqrt{a^2-b} \r^n \\
&= -\l \l a - \sqrt{a^2-b} \r \l a + \sqrt{a^2-b} \r\r^n \\
&= - \l a^2 - a^2+b\r^n \\
&= b^n
\end{align*}

Therefore $r^2-rN + b^n = 0$
\end{questionparts}

Looking at $\left(8+3\sqrt{7}\right)^{n}$ we have $a = 8, b = 1$ (since $8^2 - 1 = 9 \cdot 7$. So we can apply the result of the previous question to see that:

$\left(8+3\sqrt{7}\right)^{n}$ differs from an integer by $\left(8-3\sqrt{7}\right)^{n}$. 

\begin{align*}
8-3\sqrt{7} &= \frac{1}{8+3\sqrt{7}} \\
&\approx \frac{1}{8 + 8} \\
&\approx 2^{-4}
\end{align*}

Therefore it differs by approximation $(2^{-4})^n = 2^{-4n}$