Problem source

\begin{questionparts}
\item Show that 
$\displaystyle \big( 5 + \sqrt {24}\;\big)^4 
+ \frac{1 }{\big(5 + \sqrt {24}\;\big)^4} \ $ is an integer.
Show also 
that 
\[\displaystyle 0.1 < \frac{1}{  5 + \sqrt {24}} <\frac 2 {19}< 0.11\,.\] 
Hence determine, with clear reasoning, 
the value of $\l 5 + \sqrt {24}\r^4$ correct to four decimal places.
\item If $N$ is an integer greater than 1, 
show that  $( N + \sqrt {N^2 - 1} \,) ^k$, where $k$ is a positive
integer,  differs from 
the integer nearest to it by less than $\big( 2N - \frac12 \big)^{-k}$.
\end{questionparts}

Solution source

\begin{questionparts}
\item First notice that $\frac{1}{5+\sqrt{24}} = \frac{5-\sqrt{24}}{25-24} = 5 - \sqrt{24}$, hence
\begin{align*}
&&  ( 5 + \sqrt {24})^4 
+ \frac{1 }{(5 + \sqrt {24})^4} &=  ( 5 + \sqrt {24})^4 + ( 5 - \sqrt {24})^4 \\
\end{align*}
where clearly all terms including $\sqrt{24}$ will cancel out, therefore it is an integer. 

\begin{align*}
&& 5 + \sqrt{24} &< 5 + 5 = 10 \\
\Rightarrow && \frac{1}{5+\sqrt{24}}& > \frac{1}{10} = 0.1 \\
&& 2(5 + \sqrt{24}) &=10 + \sqrt{96}  > 19 \\
\Rightarrow && \frac{1}{5+\sqrt{24}} & < \frac{2}{19} < \frac{2}{18} = \frac19 = 0.11111\ldots < 0.11
\end{align*}

Therefore, $10^{-4} < (5+\sqrt{24})^4 < 0.11^{-4} = 0.00014641$

\begin{align*}
&& (5+\sqrt{24})^4 +  (5-\sqrt{24})^4 &= 2(5^4+6\cdot5^2\cdot24+24^2) \\
&&&= 2\cdot (625 + 3600+576) \\
&&&= 9602 \\
\Rightarrow && (5+\sqrt{24})^4  &= 9602 - \epsilon, \epsilon \in (0.0001, 0.00014641) \\
\Rightarrow &&  (5+\sqrt{24})^4 &\in (9601.999854, ,9601.9999) \\
\Rightarrow &&  (5+\sqrt{24})^4 &= 9601.9998 \, (4 \text{ d.p.})
\end{align*}

\item Notice that $(N+\sqrt{N^2-1})^{k}+(N-\sqrt{N^2-1})^{k}$ is an integer for the same reason as before (sum of conjugates). Notice also that $\frac{1}{N+\sqrt{N^2-1}} = N - \sqrt{N^2-1}$ and that so it sufficies to show that 
\begin{align*}
&& N + \sqrt{N^2-1} &> 2N-\tfrac12 \\
\Leftrightarrow && \sqrt{N^2-1} &> N - \tfrac12 \\
\Leftrightarrow && N^2-1 &> N^2-N+1\\
\Leftrightarrow && N &> \tfrac32\\
\end{align*}
Which is true since $N > 1$ and $N$ is an integer.
\end{questionparts}