## Problem 6.

It is known that the polynomial $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-\mathrm{x}+\mathrm{k}$ has three roots that are integers.

Determine the number k.
For $\mathrm{k}=0$ we have $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-\mathrm{x}=\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)$, which has roots $0, -1 \text{ and } 1$.

It is shown that this is the only value of $k$ for which $\mathrm{p}(\mathrm{x})$ has three integer roots. Indeed, if $a, b, c$ are integers, and $p(x)=(x-a)(x-b)(x-c)$, we get:

$$
\left.\begin{array}{c}
a+b+c=0 \\
ab+ac+bc=-1 \\
abc=-k
\end{array}\right\}
$$

Then,

$$
(a+b+c)^{2}=0=a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=a^{2}+b^{2}+c^{2}-2
$$

That is, $a^{2}+b^{2}+c^{2}=2$, where $a^{2}, b^{2}, c^{2}$ are non-negative integers. Necessarily, one of the values $a, b$ or $c$ must be zero, so $\mathrm{k}=-\mathrm{abc}=0$.

They can also represent $q(x)=y \quad$ and observe that $q(x)+k$ cannot have three integer roots, as there are no integers in $(-1,0)$ or in $(0,1)$.