A1. Arbitrarily color the edges of the complete graph $K_{v}$ with two colors (red or blue). What is the minimum number of monochromatic triangles (triangles with all three sides the same color) in the graph?
The solution to $\mathrm{A1}$ adopts the method of calculating the number of bichromatic angles discussed in problem B. First, let $v$ be even $(v=2 n)$. For a vertex $x$ that connects to $r$ red edges and $b$ blue edges, the number of bichromatic angles with $x$ as the vertex is
$$
b r \leqslant n(n-1) \text {. }
$$

Thus, the total number of bichromatic angles is less than or equal to $2 n \times n(n-1)$. The number of triangles that are not monochromatic is less than or equal to
$$
2 n \times n(n-1) \div 2=n^{2}(n-1) \text {. }
$$

The number of monochromatic triangles is greater than or equal to
$$
\begin{array}{l}
\mathrm{C}_{2 n}^{3}-n^{2}(n-1) \\
=\frac{1}{3} n(n-1)(n-2)=2 \mathrm{C}_{n}^{3} .
\end{array}
$$

On the other hand, consider the sets
$$
\begin{array}{l}
X=\{1,2, \cdots, n\}, \\
Y=\{n+1, n+2, \cdots, 2 n\} .
\end{array}
$$
Connect every pair of points between $X$ and $Y$ with blue lines. Connect every pair of points within $X$ and within $Y$ with red lines. This results in a bichromatic complete graph $K_{2 n}$ with exactly $2 \mathrm{C}_{n}^{3}$ monochromatic (all red) triangles. Therefore, the minimum number of monochromatic triangles is $2 \mathrm{C}_{n}^{3}$.

Now, let $v$ be odd $(v=2 n+1)$. In this case,
$$
b r \leqslant n \times n.
$$
Thus, the total number of bichromatic angles is less than or equal to $(2 n+1) n^{2}$. The number of monochromatic triangles is greater than or equal to
$$
\begin{array}{c}
C_{2 n+1}^{3}-\frac{1}{2}(2 n+1) n^{2} \\
=\frac{(2 n+1) n(n-2)}{6} \\
=C_{n}^{3}+C_{n+1}^{3}-\frac{n}{2} .
\end{array}
$$

Therefore, the number of monochromatic triangles is greater than or equal to
$$
\mathrm{C}_{n}^{3}+\mathrm{C}_{n+1}^{3}-\left[\frac{n}{2}\right],
$$

where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
Consider the sets
$$
\begin{array}{l}
X=\{1,2, \cdots, n\}, \\
Y=\{n+1, n+2, \cdots, 2 n+1\} .
\end{array}
$$

When $n$ is odd, all edges within $X$ are colored red, and within $Y$,
$$
(n+1, n+2),(n+3, n+4), \cdots,(2 n-2,2 n-1)
$$

are colored blue, and the rest are colored red; between $X$ and $Y$,
$$
(1, n+1),(2, n+2), \cdots,(n-1,2 n-1)
$$

are colored red, and the rest are colored blue.
In this case, there are $\mathrm{C}_{n}^{3}$ red triangles within $X$, $\mathrm{C}_{n+1}^{3}-\frac{n-1}{2}(n-1)$ red triangles within $Y$, and $\frac{n-1}{2}(n-2)$ blue triangles between $X$ and $Y$.
Thus, there are a total of $\mathrm{C}_{n}^{3}+\mathrm{C}_{n+1}^{3}-\frac{n-1}{2}$ monochromatic triangles.
When $n$ is even, the internal edges of $X$ are colored as above, and within $Y$,
$$
(n+1, n+2),(n+3, n+4), \cdots,(2 n-1,2 n)
$$

are colored blue, and the rest are colored red; between $X$ and $Y$,
$$
(1, n+1),(2, n+2), \cdots,(n, 2 n)
$$

are colored red, and the rest are colored blue.
In this case, there are $\mathrm{C}_{n}^{3}$ red triangles within $X$, $\mathrm{C}_{n+1}^{3}-\frac{n}{2}(n-1)$ red triangles within $Y$, and $\frac{n}{2}(n-2)$ blue triangles between $X$ and $Y$.
Thus, there are a total of $\mathrm{C}_{n}^{3}+\mathrm{C}_{n+1}^{3}-\frac{n}{2}$ monochromatic triangles.
Therefore, the minimum number of monochromatic triangles is
$$
\left\{\begin{array}{ll}
2 \mathrm{C}_{n}^{3}, & v=2 n ; \\
\mathrm{C}_{n}^{3}+\mathrm{C}_{n+1}^{3}-\left[\frac{n}{2}\right], & v=2 n+1 .
\end{array}\right.
$$