Cute Diophantine equation (simplify the expression) Find the largest integer $n$ less than $1000$ of the form 
$n=(x+\sqrt{x^2-1})^{\frac{4}{3}}+(x+\sqrt{x^2-1})^{\frac{-4}{3}}$
for some positive integer $x$.
 A: Write $$u:=x+\sqrt{x^2-1}=e^{3t}\quad(t\geq0)\ .$$ 
Then
$$x={1\over2}\left(u+{1\over u}\right)=\cosh(3t)$$
and
$${n\over2}={1\over2}(u^{4/3}+u^{-4/3})=\cosh(4t)\ .$$
It follows that 
$$8x^4-8x^2+1=\cosh(12t)=4\left({n\over2}\right)^3-3\left({n\over2}\right)\ $$
(see the comment by san below), so that $x$ and $n$ are related by
$$4(2x^2-1)^2=n^3-3n+2=(n-1)^2(n+2)\ .\tag{1}$$
This implies that $n+2$ has to be a square: $n+2=m^2\geq4$. Introducing this into $(1)$ and taking the square root we obtain
$$2(2x^2-1)=(m^2-3)m\ ,$$
which then leads to
$$x^2={1\over4}(m^2-3m+2)={1\over4}(m-1)^2(m+2)\ .$$
This shows that $m+2$ has to be a square as well: $m+2=p^2\geq4$. This leads to
$$x={p(p^2-3)\over2},\qquad n=m^2-2=p^4-4p^2+2\ .$$
In this way we obtain for $p\geq2$ the pairs
$$(x,n)=\quad(1,2),\quad (9,47),\quad(26,194),\quad(55,527),\quad(99, 1154),\quad\ldots\ .$$
The answer to the original question therefore is $527$.
A: Consider the : 
$$a+\frac{1}{a} = n$$
Let $a=\frac{p}{q}$, then $$\frac{p^{2}+q^{2}}{pq}=n$$
$$\frac{(p-q)^{2}}{pq}+2 = n $$
Let $p=q+l$, so:
1) $l = 0$ then $p=q=a=1$, and you could finish it easy.
2) $l$ > $0$ then 
$$ \frac{l^{2}}{q(l+q)} = n - 2$$, but 
First summand not a integer number. So that's false.
