gobathegreat wrote: If $a$ is real number such that $x_1$ and $x_2$, $x_1\neq x_2$ , are real numbers and roots of equation $x_2-x+a=0$. Prove that $\mid {x_1}^2-{x_2}^2 \mid =1$ iff $\mid {x_1}^3-{x_2}^3 \mid =1$ Conditions $x_1\ne x_2$ real numbers solutions of $x^2-x+a$ mean $a<\frac 14$ Then $|x_1^2-x_2^2|=\sqrt{1-4a}$ and $|x_1^3-x_2^3|=(1-a)\sqrt{1-4a}$ It is immediate to get then $|x_1^2-x_2^2|=1$ $\iff$ $a=0$ And easy to get $|x_1^3-x_2^3|=1$ $\iff$ $a(4a^2-9a+6)=0$ $\iff$ $a=0$ Q.E.D.