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:
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.
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.
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 \((2^{-4})^n = 2^{-4n}\)