I can't believe this is real

It is given that:

$\begin{aligned} x & = \frac 1{\frac 1{\frac 1{\frac 1{\frac 1{cdots} - 1} - 1} - 1} - 1} - 1 \\ \implies x & = \frac 1x - 1 \\ x - \frac 1x & = -1 \\ \left(x - \frac 1x\right)^2 & = (-1)^2 = 1 \\ x^2 - 2 + \frac 1{x^2} & = 1 \\ x^2 + \frac 1{x^2} & = 3 \\ \left(x+\frac 1x\right)^2 & = x^2 + 2 + \frac 1{x^2} = 5 \end{aligned}$

Let $P_n = x^n + \dfrac 1{x^n}$ ; then $P_2 = 3$ and $P_1^2 = 5$ .

Let us prove by induction the claim that $P_n = x^n + \dfrac 1{x^n} = \begin{cases} F_n \left(x + \dfrac 1x \right) & \text{when }n \text{ is odd} \\ F_{n+1} + F_{n-1} & \text{when }n \text{ is even} \end{cases}$ , where $F_n$ is the $n$ th Fibonacci number .

When $n=1$ , then $P_1 = x + \dfrac 1x = F_1 \left(x + \dfrac 1x\right)$ and $P_2 = 3 = F_3 + F_1$ , therefore the claim is true for $n=1$ and $n=2$ .

Assuming the claim is true for $n$ and $n+1$ , where $n$ is odd. By Newton's sum , we have:

$\begin{aligned} P_{n+2} & = P_1P_{n+1} - P_n \\ & = (F_{n+2}+F_n)P_1 - F_nP_1 = F_{n+2}\left(x + \dfrac 1x\right) \\ P_{n+3} & = P_1P_{n+2} - P_{n+1} \\ & = F_{n+2}P_1^2 - (F_{n+2}+F_n) = 5F_{n+2} - F_{n+2} - F_n \\ & = F_{n+4} - F_{n+3} + F_{n+3} - F_{n+1} + F_{n+2} + F_{n+1} + F_n - F_n \\ & = F_{n+4} + F_{n+2} \end{aligned}$

Therefore the claim is also true for $n+2$ and $n+3$ , hence true for all $n \ge 1$ . And $P_{10} = F_{11} + F_9 = 89+34 = \boxed{123}$ .

Since this is a recurrent relation we can rewrite it:

$x = \frac { 1 } { x } - 1 \\ x - \frac { 1 } { x } = - 1$

Let's look at a series of even degree binoms:

${ \left( x-\frac { 1 }{ x } \right) }^{ 2 }={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } -2={ \left( -1 \right) }^{ 2 }=1\\ { \left( x-\frac { 1 }{ x } \right) }^{ 4 }={ x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } -4{ x }^{ 2 }-\frac { 4 }{ { x }^{ 2 } } +6={ \left( -1 \right) }^{ 4 }=1\\ { \left( x-\frac { 1 }{ x } \right) }^{ 6 }={ x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } -6{ x }^{ 4 }-\frac { 6 }{ { x }^{ 4 } } +15{ x }^{ 2 }+\frac { 15 }{ { x }^{ 2 } } -20={ \left( -1 \right) }^{ 6 }=1\\ { \left( x-\frac { 1 }{ x } \right) }^{ 8 }={ x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } -8{ x }^{ 6 }-\frac { 8 }{ { x }^{ 6 } } +28{ x }^{ 4 }+\frac { 28 }{ { x }^{ 4 } } -56{ x }^{ 2 }-\frac { 56 }{ { x }^{ 2 } } +70={ \left( -1 \right) }^{ 8 }=1\\ { \left( x-\frac { 1 }{ x } \right) }^{ 10 }={ x }^{ 10 }+\frac { 1 }{ { x }^{ 10 } } -10{ x }^{ 8 }-\frac { 10 }{ { x }^{ 8 } } +45{ x }^{ 6\ }+\frac { 45 }{ { x }^{ 6 } } -120{ x }^{ 4 }-\frac { 120 }{ { x }^{ 4 } } +210{ x }^{ 2 }+\frac { 210 }{ { x }^{ 2 } } -252={ \left( -1 \right) }^{ 10 }=1$

Now we can derive the sums:

${ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =1+2\\ { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } =1-6+4\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \\ { x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } =1+20-15\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) +6\left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) \\ { x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } =1-70+56\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -28\left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) +8\left( { x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } \right) \\ { x }^{ 10 }+\frac { 1 }{ { x }^{ 10 } } =1+252-210\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) +120\left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) -45\left( { x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } \right) +10\left( { x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } \right) \\$

${ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =3\\ { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } =-5+4\cdot 3=7\\ { x }^{ 6 }+\frac { 1 }{ { x }^{ 6 } } =21-15\cdot 3+6\cdot 7=18\\ { x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } =-69+56\cdot 3-28\cdot 7+8\cdot 18=47\\ { x }^{ 10 }+\frac { 1 }{ { x }^{ 10 } } =253-210\cdot 3+120\cdot 7-45\cdot 18+10\cdot 47=\boxed { 123 } \\$

im not sure if this is the right way to solve it buut i just wrote 1/x-1=x as x²+x-1=0 and solved it, both roots give the same answer

From the definition of $x$ , we can see that $x-\frac{1}{x}=-1$ , or alternatively $x^2+x-1=0$ . We could just solve and calculate here, but this would get messy.

Let $a_n=x^n+\left( -\frac{1}{x} \right)^n$ . Note that this has the form of a solution to a recurrence relation - and this is what we'll use.

The quadratic equation for $x$ translates to the following recurrence for $a_n$ :

$a_{n+2}+a_{n+1}-a_n=0$

We want to find $a_{10}$ . We started with $a_1=-1$ . Trivially, $a_0=1+1=2$ . Together with the recurrence, these two values are enough to calculate the remaining $a_n$ values we need; $a_2=a_0-a_1=3$ , $a_3=a_1-a_0=-4$ , etc., up to $a_{10}=\boxed{123}$ .