3A. 4B. Determine all integer values of \(a\) for which the quadratic trinomial \(x^{2} - 3ax + 2a^{2} + 1\) can be written in the form \((x + b)(x + c)\) where \(b\) and \(c\) are integers.
Solution. From the condition of the problem, we have $x^{2}-3 a x+2 a^{2}+1=(x+b)(x+c)$, from which it follows that $x^{2}-3 a x+2 a^{2}+1=x^{2}+(b+c) x+b c$, so

$$
\left\{\begin{array}{l}
-3 a=b+c \\
2 a^{2}+1=b c
\end{array}\right.
$$

From the first equation, we have $c=-3 a-b$, and if we substitute this into the second equation, we get

$$
2 a^{2}+1=-3 a b+b^{2}, \text { i.e. }(a+b)(2 a+b)=-1
$$

However, $a$ and $b$ are integers, so from the last equation, we get

$$
\left\{\begin{array} { l } 
{ a + b = 1 } \\
{ 2 a + 3 b = - 1 }
\end{array} \text { or } \left\{\begin{array}{l}
a+b=-1 \\
2 a+3 b=-1
\end{array}\right.\right.
$$

From the first system, we get $a=-2, b=3$ and thus $c=3$, and from the second system, we get $a=2, b=-3$ and thus $c=-3$. Finally, the solution to the problem is $a= \pm 2$.