Proving that if $x+\frac{1}{x}$ is an integer, then $x^n+\frac{1}{x^n}$ is also an integer for all even $n$ using induction 
Prove that if $x+\frac{1}{x}$ is an integer, then  $x^n+\frac{1}{x^n}$
is also an integer.

I am aware this question has already been answered previously here and here, but I need help in proving it in an alternative way.
I am first trying to prove this for the case where $n$ is even.
Defining $$I_n:=x^n+\frac{1}{x^n},$$
I noticed that $$\\{\left(x+\frac{1}{x} \right)^n}=x^n+\frac{1}{x^n}+\sum_{k=1}^{n/2}{{n\choose k}{\left(x^{n-2k}+\frac{1}{x^{n-2k}}\right)}}$$
$$\Rightarrow I_n=(I_1)^n-\sum_{k=1}^{n/2}{{n\choose k}}\cdot I_{n-2k}$$
I initially planned on proving through induction, i.e., by proving that if $I_n$ is integer for some value of $n$, then $I_{n+2}$ must also be an integer, but for $I_{n+2}$ to be an integer, you also need $I_{n-2},I_{n-4}\ ,...I_{4},I_{2}$ to be integers. How do I mathematically prove that $I_{n}$ being an integer also implies all of these are also integers?
Moreover, would to be correct to instead assume that all $I_n, I_{n-2},I_{n-4}\ ,...I_{4},I_{2}$ are integers for some value of $n$ and then prove that $I_{n+2}$ is an integer? I ask this because I haven't seen induction proofs assuming the assumption to be true for more than 1 value of the variable.
 A: For $n$ even, observe: $I_n = I_{n-2}\cdot I_2 - I_{n-4}$, [$1$]. Thus you can prove the claim using induction on $n$ being even. Starting with $n = 2$, $I_2 = \left(x+\frac{1}{x}\right)^2 - 2$ is an integer since $x+\dfrac{1}{x}$ is by assumption. Assume the claim is true for even $n$ up to $n-2$. This means $I_{n-2}, I_{n-4}$ are integers.Then [$1$] says that $I_n$ is also an integer.  Thus by strong induction on even $n$, $I_n$ is an integer for all even $n$.
A: $$x^{n+1}+\frac1{x^{n+1}}=\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)-\left(x^{n-1}+\frac1{x^{n-1}}\right).$$
A: Generalizing,
let
$u_n
=a^n+b^n
$.
Then
$\begin{array}\\
u_1u_n
&=(a+b)(a^n+b^n)\\
&=a^{n+1}+ab^n+ba^n+b^{n+1}\\
&=a^{n+1}+b^{n+1}+ab(b^{n-1}+a^{n-1})\\
&=u_{n+1}+abu_{n-1}\\
\text{so}\\
u_{n+1}
&=u_1u_n-abu_{n-1}\\
\end{array}
$
Therefore
if $ab$, $u_1$ and $u_2$
are integers,
then
$u_n$ is an integer
for $n \ge 1$.
Since
$u_2
=a^2+b^2
=a^2+2ab+b^2-2ab
=(a+b)^2-2ab
=u_1^2-2ab
$,
if
$a+b$ and $ab$ are integers,
then
$a^n+b^n$ is an integer
for $n \ge 1$.
If
$a+\frac1{a}$ is an integer,
then
$b = \frac1{a}$
so
$ab=1$ is an integer,
so
$a^n+\frac1{a^n}$
is an integer for
$n \ge 1$.
