Problem 3. (Option 1)

Find the sum of the squares of the roots of the equation $\left(x^{2}+4 x\right)^{2}-2016\left(x^{2}+4 x\right)+2017=0$.
Solution: Let's make the substitution: $x^{2}+4 x+4=t$, then $x^{2}+4 x=t-4$ and the equation will take the form:

$(t-4)^{2}-2016(t-4)+2017=0$

$t^{2}-2024 t+10097=0$

The discriminant of the equation is greater than zero, therefore, the equation has two roots. By Vieta's theorem: $t_{1}+t_{2}=2024, t_{1} \cdot t_{2}=10097$, which means both roots of the quadratic equation are positive. Let's make the reverse substitution:

$$
\begin{array}{cc}
(x+2)^{2}=t_{1} & (x+2)^{2}=t_{2} \\
x+2= \pm \sqrt{t_{1}} & x+2= \pm \sqrt{t_{2}} \\
x_{1,2}=-2 \pm \sqrt{t_{1}} & x_{3,4}=-2 \pm \sqrt{t_{2}} \\
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=\left(-2+\sqrt{t_{1}}\right)^{2}+\left(-2-\sqrt{t_{1}}\right)^{2}+\left(-2+\sqrt{t_{2}}\right)^{2}+\left(-2+\sqrt{t_{2}}\right)^{2}= \\
=2\left(4+t_{1}\right)+2\left(4+t_{2}\right)=16+2\left(t_{1}+t_{2}\right)=16+2 \cdot 2024=4064
\end{array}
$$

Answer: 4064.