Problem 7. For what least value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to $-1540?$
Answer: -19.

Solution. Applying the polynomial formula, we get

$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_{2}} \cdot a^{n_{3}} \cdot x^{n_{2}+2 n_{3}}
$$

To determine which terms in the sum contain $x^{4}$, we need to solve the system of equations in non-negative integers:

$$
\left\{\begin{array}{l}
n_{1}+n_{2}+n_{3}=8 \\
n_{2}+2 n_{3}=4
\end{array}\right.
$$

From the second equation, it follows that $n_{2}$ is even. Due to the non-negativity of the variables, $n_{2}$ can take the values 0, 2, and 4. Solving the system for each of these $n_{2}$, we will have three cases: