Suppose that $b$ and $c$ are real numbers such that the equation

$$
x^{2}+b x+c=0
$$

has two different solutions $x_{1}, x_{2}$. Suppose that

(a) The (positive) difference between $x_{1}$ and $x_{2}$ is 1 ;

(b) The (positive) difference between $b$ and $c$ is also 1.

Find all possible values of $b$ and $c$.
Given the two solutions $x_{1}$ and $x_{2}$, we know that the equation factors:

$$
x^{2}+b x+c=\left(x-x_{1}\right)\left(x-x_{2}\right)=0 \text {. }
$$

From this we derive that

$$
b=-x_{1}-x_{2} \text { and } c=x_{1} x_{2} .
$$

In condition (a), we switch the labels on $x_{1}$ and $x_{2}$, if necessary, to assume that $x_{2}=x_{1}+1$. However, for condition (b) we still have the two cases $c-b=1, c-b=-1$.

We compute

$$
\begin{aligned}
c-b & =x_{1} x_{2}+x_{1}+x_{2} \\
& =x_{1}\left(x_{1}+1\right)+x_{1}+\left(x_{1}+1\right) \\
& =x_{1}^{2}+3 x_{1}+1
\end{aligned}
$$

Setting $c-b=1$ gives $x_{1}^{2}+3 x_{1}=0$, which has two solutions:
(a) $x_{1}=0$, giving $x_{2}=1, b=-1, c=0$;

(b) $x_{1}=-3$, giving $x_{2}=-2, b=5, c=6$.

Likewise, setting $c-b=-1$ gives $x_{1}^{2}+3 x_{1}+2=0$, which also has two solutions:

(a) $x_{1}=-1$, giving $x_{2}=0, b=1, c=0$;

(b) $x_{1}=-2$, giving $x_{2}=-1, b=3, c=2$.