Problem (Algebra):
Compute the number of ordered pairs of integers $(b, c)$, with $-20 \leq b \leq 20,-20 \leq c \leq 20$, such that the equations $x^{2}+b x+c=0$ and $x^{2}+c x+b=0$ share at least one root.

Solution:
Let $r$ be the common root. Then $r^{2}+b r+c=r^{2}+c r+b \Rightarrow b r-c r=b-c$. So either $b=c$ or $r=1$. In the latter case, $1+b+c=0$, so $c=-1-b$.

There are 41 ordered pairs where $b=c$. If $c=-1-b$ and $-20 \leq b \leq 20$, then $-21 \leq c \leq 19$. Therefore there are 40 ordered pairs $(b,-1-b)$ where both terms are in the required intervals. Thus there are $41+40=\mathbf{8 1}$ solutions.

Final answer: 81