Computing $x^{2017}+\frac1{x^{2017}}$ given the value of $x+\frac1x$.

Question

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.

Without obtaining a closed form expression, we know that the two roots of $x+\frac1x=2017$ will be "close" to $2017$ and $1/2017$. Therefore $x^{2017}+x^{-2017}$ will be "close" to $2017^{2017}$. None of the answers is "close" to this approximation. — Mark Viola, Apr 16 '17 at 06:36
Here is a weird solution: It is clear that $x^{2017} + x^{-2017}$ is an algebraic number. Now by the Gelfond-Schneider theorem, A)-D) are all transcendental. Therefore E) is the answer. — Sangchul Lee, Apr 16 '17 at 06:53
It is near to $2017^{2016*2(\sqrt5-\sqrt3)}$ in those choice. — Takahiro Waki, Apr 16 '17 at 16:04

score 2 · Accepted Answer · answered Apr 16 '17 at 06:30

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.

score 2 · Answer 2 · answered Apr 16 '17 at 06:38

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.

score 1 · Answer 3 · answered Apr 16 '17 at 06:31

1

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.

answered Apr 16 '17 at 06:31

Shaswata

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Computing $x^{2017}+\frac1{x^{2017}}$ given the value of $x+\frac1x$.

3 Answers3

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