Computing $x^{2017}+\frac1{x^{2017}}$ given the value of $x+\frac1x$. If $x+\dfrac{1}{x}=2017$, then the value of $x^{2017}+\dfrac{1}{x^{2017}}$ is,
A) $2017^{2016(\sqrt{5}-\sqrt{3})}$
B) $2017^{2016(\sqrt{3}-\sqrt{2})}$
C) $2017^{2\sqrt{3}}$
D)  $2017^{3\sqrt{2}}$
E) None of the above
Progress: We could make use of the identity $$x^{m+n}+\dfrac{1}{x^{m+n}}=\left( x^m+\dfrac{1}{x^{m}}\right)\left(x^{n}+\dfrac{1}{x^{n}}\right)-\left(x^{m-n}+\dfrac{1}{x^{m-n}}\right), $$
however this is quite tedious since partitioning $2017$ and going on further takes a lot of time. Any solutions or hints are welcome. 
 A: Note that the equation $x+1/x=2017$ reduces to a quadratic equation, and clearly the two roots must be $x_1=\alpha$ and $x_2=1/\alpha$, since if $\alpha$ is a root, so is $1/\alpha$. Now, the coefficients of this quadratic equation are integers, which means that all symmetric functions in the two roots $x_1,x_2$ with rational coefficients are rational numbers, because all such functions are expressible as rational functions of the elementary symmetric functions $x_1+x_2$ and $x_1x_2$. 
Thus, $x_1^{2017}+x_2^{2017}$ (the expression you're looking at) is also a rational number, and in particular it can't be any of the numerical answers given.
A: A special case of the formula above has $n = 1$, thus $$x^{m+1} + x^{-(m+1)} = (x+x^{-1})(x^m + x^{-m}) - (x^{m-1} + x^{-(m-1)}).$$  Define $a_m = x^m + x^{-m}$, so that we now have the linear recurrence $$a_{m+1} = 2017 a_m - a_{m-1},$$ and $a_0 = 2$, $a_1 = 2017$.  
This immediately tells us that $a_{2017}$ is an integer, being a term in a linear recurrence that begins with integers.  None of the given answer choices are integers.
A: Let $s_m=x^m+\frac{1}{x^m}$
Verify the identity, $$s_{m+1}=s_1 s_m-s_{m-1}$$
Since you already know $s_1$ it is only a matter of solving a second order recurrence relation which can be done using generating functions.
