Saudi Arabia Mathematical Competitions

If $f=(X-x_1)(X-x_2),x_1,x_2\in \mathbb{Z}$, then take $u_n=n+x_1$ and get $f\left(u_n\right)=n\left(n+x_1-x_2\right)$, hence $n\mid f\left(u_n\right),n\ge 1$.

Conversely, assume that $f\left(u_n\right)=k_n\cdot n$, for some integer $k_n$, $n=1,2,\dots$ Then, the quadratic equation $$u^2+a\nu +b-k_n\cdot n=0$$ has integer zeros, hence its discriminant ${\Delta }_n$ is a perfect square, that is ${\Delta }_n=t_n^2,n=1,2,\dots$. This is equivalent to $$a^2-4\left(b-k_n\cdot n\right)=t_n^2$$ hence we have $$\Delta +4k_n\cdot n=t_n^2,n=1,2,\dots \text{\hspace{1em}(1)}$$ where $\Delta =a^2-4b$ is the discriminant of equation $f(n)=0$. In (1) we take $n={\Delta }^2$ and obtain $$\Delta \cdot \left(1+4k_{{\Delta }^2}\cdot \Delta \right)=t_{{\Delta }_2}^2.\text{\hspace{1em}(2)}$$ Since $\Delta$ and $1+4k_{{\Delta }^2}\cdot \Delta$ are relatively prime, from (2) it follows that $\Delta$ is a perfect square, hence the zeros of $f$, $x_1,x_2$ are rational. We have $\Delta =a^2-4b=c^2,c\in \mathbb{Z}$, and $$x_{1,2}=\frac{1}{2}(-a\pm c).\text{\hspace{1em}(3)}$$ It is clear that $a$ and $c$ are of the same parity, and from (3) we get that $x_1,x_2\in \mathbb{Z}$.