11.5. For a certain quadratic trinomial, the following information is known: its leading coefficient is one, it has integer roots, and its graph (parabola) intersects the line $y=2017$ at two points with integer coordinates. Can the ordinate of the vertex of the parabola be uniquely determined from this information?
Answer: It is possible. Solution. Let $n_{1}, n_{2}\left(n_{1}<n_{2}\right)$ be the roots of the given quadratic trinomial $f(x)$. Then $f(x)=\left(x-n_{1}\right)\left(x-n_{2}\right)$. From the condition about the integer roots of the equation $\left(x-n_{1}\right)\left(x-n_{2}\right)=2017$, we obtain (due to the simplicity of the number 2017) that for the root $x_{0}$ of this equation, we will have either $x_{0}-n_{1}=2017, x_{0}-n_{2}=1$, or $x_{0}-n_{1}=-1, x_{0}-n_{2}=-2017$. In both cases, $n_{2}-n_{1}=2016$. Knowing the distance between the roots, we can uniquely determine the ordinate of the vertex (see problem 9.2). It is equal to $-\left(\frac{2016}{2}\right)^{2}=-1008^{2}$.