SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Consider the equation $f(x)=1\iff x^4+a{x}^3+b{x}^2+cx=1$, since it has four roots $x_1,x_2,x_3,x_4$ then we can write it as $$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=0.$$ Note that $x_1+x_2=x_3+x_4=-\frac{a}{2}$ then $$\left(x^2-(x_1+x_2)x+x_1{x}_2\right)\left(x^2-(x_3+x_4)x+x_3{x}_4\right)=0$$ or $$\left(x^2+\frac{ax}{2}+x_1{x}_2\right)\left(x^2+\frac{ax}{2}+x_3{x}_4\right)=0.$$ Continue with the equation $f(x)=2\iff \left(x^2+\frac{ax}{2}+x_1{x}_2\right)\left(x^2+\frac{ax}{2}+x_3{x}_4\right)=1$. By putting $X=x^2+\frac{ax}{2}$, we get $$(X+x_1{x}_2)(X+x_3{x}_4)=1$$ Since $f(x)=2$ has four real roots then this equation has two roots like $X=\alpha ,X=\beta$. This implies that $$\{\begin{array}{l}{x}^2+\frac{ax}{2}=\alpha \\ x^2+\frac{ax}{2}=\beta \end{array}$$ Then four roots $x_1^{\prime },x_2^{\prime },x_3^{\prime },x_4^{\prime }$ of the $f(x)=2$ can be divided into two pairs that have the sum equal to $-\frac{a}{2}$ which means $x_1^{\prime }+x_2^{\prime }=x_3^{\prime }+x_4^{\prime }$. This finishes the proof.