Prove $ \frac{1}{x^{2n}}+x^{2n}$ is an integer for all $n$. Question:
Suppose $ \frac{1}{x^2}+x^2$ is an integer. Prove that $ \frac{1}{x^{2n}}+x^{2n}$ is an integer for all natural $n$.
Hint: Use Strong Induction 
My attempt:
Base Case is trivial.
I.H: Assume the result is true for $n = 1,2, ...., k.$
Consider $n = k+1$.
$ \frac{1}{x^{2\left(k+1\right)}}+x^{2\left(k+1\right)}\ =\ \frac{1}{x^{\left(2k+2\right)}}+x^{2k+2}$.
I am not sure what to do from here and how to use the induction hypothesis. 

Observe that
$(\dfrac{1}{x^2} + x^2)(\dfrac{1}{x^{2k}} + x^{2k}) = \dfrac{1}{x^{2(k + 1)}} + x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}} + x^{2(k + 1)}$
$= (\dfrac{1}{x^{2(k + 1)}} +  x^{2(k + 1)}) + (x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}}), \tag{1}$
whence
$\dfrac{1}{x^{2(k + 1)}} +  x^{2(k + 1)} = (\dfrac{1}{x^2} + x^2)(\dfrac{1}{x^{2k}} + x^{2k}) - (x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}}) \tag{2}$
is an integer if
$(\dfrac{1}{x^2} + x^2)(\dfrac{1}{x^{2k}} + x^{2k})  \tag{3}$
and
$(x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}}) \tag{4}$
are integers.
From here, induction will take off and fly on its own.
