2. Find all quadratic trinomials $a x^{2}+b x+c$ such that if any of the coefficients $a, b, c$ is increased by 1, the resulting new quadratic trinomial will have a double root.
2. For the coefficient $a$, it must hold that $a \neq 0$ and $a \neq -1$ so that all considered trinomials are indeed quadratic trinomials. We know that a quadratic trinomial has a double root if and only if its discriminant is zero. Therefore, let's construct the discriminants of all three trinomials with increased coefficients:

$$
\begin{array}{lll}
(a+1) x^{2}+b c+c & \text { has discriminant } & D_{1}=b^{2}-4(a+1) c, \\
a x^{2}+(b+1) x+c & \text { has discriminant } & D_{2}=(b+1)^{2}-4 a c, \\
a x^{2}+b x+(c+1) & \text { has discriminant } & D_{3}=b^{2}-4 a(c+1) .
\end{array}
$$

We are looking for real numbers $a, b, c$ such that $a \neq 0, a \neq -1$ and $D_{1}=D_{2}=D_{3}=0$.
From the equality $D_{1}=D_{3}$, it follows that $c=a$, so $D_{2}=(b+1)^{2}-4 a^{2}=(b+1-2 a)(b+1+2 a)$; the equality $D_{2}=0$ then means that $b= \pm 2 a-1$, and thus $D_{1}=( \pm 2 a-1)^{2}-4(a+1) a=$ $=4 a^{2} \mp 4 a+1-4 a^{2}-4 a=1 \mp 4 a-4 a$, so $D_{1}=-8 a+1$ or $D_{1}=1$. Therefore, from the equality $D_{1}=0$ it follows that $a=1 / 8, b=2 a-1=-3 / 4$ and $c=a=1 / 8$ (the verification is easy, but not necessary, as our method guarantees the equalities $D_{1}=D_{3}, D_{2}=0$ and $D_{1}=0$).

Answer: The problem is satisfied by the unique trinomial $\frac{1}{8} x^{2}-\frac{3}{4} x+\frac{1}{8}$.

For a complete solution, award 6 points, of which 2 points for correctly setting up the discriminants $D_{1}, D_{2}, D_{3}$. For omitting the conditions $a \neq 0, a \neq -1$ in an otherwise complete solution, deduct 1 point.