Team selection tests for IMO 2018

We will prove that $x_0\in \left(F_{3029},F_{3030}\right)$ by showing that for all $x^{\ast }\in \left(F_1,F_{3029}\right]$, there is some $x^{\ast \ast }\in \left(F_{3029},F_{3030}\right)$ for which $\left|f\left(x^{\ast \ast }\right)\right|>\left|f\left(x^{\ast }\right)\right|$.

Indeed, if $$x^{\ast }\in \left\{F_1,F_2,\dots ,F_{3029}\right\}$$ then $\left|f\left(x^{\ast }\right)\right|=0$ which is obvious.

Suppose that $x^{\ast }\in \left(F_k,F_{k+1}\right)$ for $1\le k\le 3028$ then put $m=x^{\ast }-F_k$, we choose $x^{\ast \ast }=F_{3030}-m$. We need $$\left|f\left(x^{\ast \ast }\right)\right|>\left|f\left(x^{\ast }\right)\right|\iff \left|\prod _{i=1}^{3030}\left(x^{\ast \ast }-F_i\right)\right|>\left|\prod _{i=1}^{3030}\left(x^{\ast }-F_i\right)\right|.$$ Each side has 3030 positive factors then we will make pair one of the left and one of the right such that the value of the right is bigger (except the case $i=k$ then $\left|x^{\ast }-F_k\right|=m=\left|x^{\ast \ast }-F_{3030}\right|$ ). For details:

1. If $i=1,2,\dots ,k$, it is clearly that $x^{\ast \ast }-F_i>x^{\ast }-F_i>0$.

2. If $1\le i\le 3030-k$, we have $\left|x^{\ast }-F_{k+i}\right|<\left|x^{\ast \ast }-F_{3030-i}\right|$. Note that we just consider the separated ranges, i.e. $k+i\le 3030-i$. Otherwise, some ranges are overlap then we remove that part, then $$\begin{array}{rl}& \\ & F_{k+i}-x^{\ast }=\left(F_{k+1}-x^{\ast }\right)+\left(F_{k+2}-F_{k+1}\right)+\cdots +\left(F_{k+i}-F_{k+i-1}\right)\\ & <\left(x^{\ast \ast }-F_{3029}\right)+\left(F_{3029}-F_{3028}\right)+\cdots +\left(F_{3030-(i-1)}-F_{3030-i}\right)\\ & =x^{\ast \ast }-F_{3030-i}\end{array}$$ Hence, the statement is proved. From here, we have $$\begin{array}{rl}{x}_0& >F_{3029}=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{3030}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{3030}\right]\\ & >\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^{3029}>{\left(\frac{1+\sqrt{5}}{2}\right)}^{3027}.\end{array}$$ It is easy to check that $\frac{1+\sqrt{5}}{2}>2^{\frac{2}{3}}$ then substitute into the above inequality, we get $x_0>2^{2018}$.