4. 4.1. For the quadratic trinomial \( f(x) = a x^{2} + b x + c \), where \( a > 0 \), the condition \( |f(1)| = |f(2)| = |f(3)| = 1 \) is satisfied. What can the coefficients be?
Answer: $a=2, b=-8, c=7$ or $a=1, b=-3, c=1$ or $a=1, b=-5, c=5$.

## Solution.

Note that a quadratic trinomial can take the same value at no more than two different points. Therefore, the numbers $f(1), f(2), f(3)$ cannot all have the same sign. Thus, exactly two of these numbers must take the same value. If a quadratic trinomial $a x^{2}+b x+c$ takes the same value at different points, then these points are symmetric with respect to $x=\frac{-b}{2 a}$ - the axis of symmetry of the parabola. Therefore, there are 3 possible cases for the location of the axis of symmetry: $x=2, x=1.5, x=2.5$. Given that the branches of the parabola are directed upwards, only the following 3 cases are possible: 1) $f(1)=1, f(2)=-1, f(3)=1$, 2) $f(1)=-1, f(2)=-1, f(3)=1$, 3) $f(1)=1, f(2)=-1, f(3)=-1$. Considering each case reduces to solving a system of 3 equations.