Tomi's Challenges - #3

We want to make this huge expression as helpful as possible, first of all. We'll do that by trying to get rid of the minus sign on the right side of the equation. So, multiplying by $-1$ , we get:

$-\frac{1}{\frac{1}{\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2+\frac{1}{x^{21}-2}}}}}}}-2}=x^{21}$ $\newline$ $\frac{1}{2-\frac{1}{\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2+\frac{1}{x^{21}-2}}}}}}}}=x^{21}$

Next, we'll factor $-1$ from the denominator of the bottommost fraction. We get:

$\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-x^{21}}}}}}}}}=x^{21}$

Let $t=x^{21}$ .

$\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-t}}}}}}}}=t$

This equation is nothing but one of the ways to find the recursive fraction $\frac{1}{2-\frac{1}{2-...}}$ . But, we can also find the value of this fraction using $t=\frac{1}{2-t}$ . So that means that for the value of t satisfying this equation, the equation we talked about before and this equation are equivalent. Solving this equation with respect to $t$ , we get $t=1$ , plugging back in we get $x^{21}=1$ , $x=\boxed{1}$ .