SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $P(x)=(x-a_1)(x-a_2)\cdots (x-a_{100})$ with $a_1,a_2,\dots ,a_{100}$ are real roots of polynomial $P(x)$. So $$P\left(2x^2-4x\right)=0\iff 2x^2-4x-a_i=0\text{ for }1\le i\le 100.$$ This equation cannot have $1$ root since $\Delta =16+8a_i\ne 0$. So each equation can have $0$ or $2$ roots. Note that $P\left(2x^2-4x\right)$ has $130$ roots so there are $\frac{130}{2}=65$ equations have $2$ roots, which mean there are $65$ numbers $a_i>-2$ and $35$ numbers $a_i<-2$.

By the same way, we have $$P\left(4x-2x^2\right)=0\iff 2x^2-4x+a_i=0$$ and $\Delta =16-8a_i$ for any $1\le i\le 100$. And there are $65$ numbers $a_i<2$ and $35$ numbers $a_i>2$.

By applying the principle of inclusion and exclusion, there are $$65+65-100=30\text{ numbers }{a}_i\in (-2;2).$$

Suppose that $a_1<a_2<\dots <a_{35}<-2<a_{36}<\dots <a_{64}<2<a_{65}<\dots <a_{100}$; and denote $M,N,K$ as the subsets of these numbers with the indices $1\to 35,36\to 64,65\to 100$.

Take $A_1(x)={\prod }_{m\in M}(x-m)$, $B(x)={\prod }_{n\in N}(x-n)$, $A_2(x)={\prod }_{k\in K}(x-k)$ then we will prove that $$|A_1(x_0)\cdot A_2(x_0)|>|B(x_0)|$$ for any number $x_0\in (-1;1)$.

Note that $\forall x_0\in (-1;1)$ then $|x_0-m|>1,\forall m<-2$ and $|x_0-k|>1,\forall k>2$. We can suppose that $x_0>0$ and for any $n\in N$, we have two cases:

1. If $n>0$ then $n\in (0;2)$ and $|x_0-n|<|x_0-2|<|x_0-k|$ with any $k>2$. 2. If $n<0$ then $n\in (-2;0)$ and $|x_0-n|<|x_0-(-2)|<|x_0-m|$ with any $m<-2$.

So in all cases of $n$ respect to the factor $x_0-n$ in $B(x_0)$, we can choose factor from $A_1(x_0)$ or $A_2(x_0)$ with absolute value greater than it, and since $|M|,|K|>|N|$, we always can do that. Hence $|A_1(x_0)\cdot A_2(x_0)|>|B(x_0)|,\forall x_0\in (-1;1)$ which mean this equation has no solution.

Therefore, we can choose $A(x)=A_1(x)A_2(x)$ and $B(x)$ to satisfy the given condition. $\square$