Show that $(1-\beta)(p+\beta)=1$ 
If $$(7+4\sqrt{3})^n = p+\beta,$$ where $n$ and $p$ are positive integers and $\beta$ is a proper fraction, then show that $$(1-\beta)(p+\beta)=1.$$

I cant even understand how to express the term in a positive number and a proper fraction. I would appreciate any hint.

Fun question! The key realization is that
$$
(7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n \in \mathbb{Z}
$$
(do you see why?) and moreover, that
$$
0 < (7 - 4\sqrt{3}) < 1,
$$
so that
$$
0 < (7 - 4\sqrt{3})^n < 1,
$$
for all natural numbers $n$.
It follows from here that
\begin{align*}
p &= (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n - 1 \\
\beta &= 1 - (7 - 4\sqrt{3})^n.
\end{align*}
Now to finish, we see directly from the above that
\begin{align*}
p + \beta &= (7 + 4\sqrt{3})^n \\
1 - \beta &= (7 - 4\sqrt{3})^n.
\end{align*}
Multiply them together and see what you get.
