Toe 1. The three sides of a triangle are three consecutive integers, and one angle is twice that of another. Find the lengths of the sides of this triangle.

Note: The original text seems to have some typographical errors or unconventional notations, such as "趾" (toe) and "笛" (flute), which do not make sense in the context. The translation assumes these are errors and attempts to convey the mathematical problem as clearly as possible.
Let $\triangle A B C$ have three sides $A C=n, A B=n-1, B C=n+1$, then we have $A>B>C$. Since one angle is twice another, we have $4 C\cos 45^{\circ}=\frac{\sqrt{2}}{2}$,

which means
$$
\begin{aligned}
\cos C & =\frac{n^{2}+(n+1)^{2}-(n-1)^{2}}{2 n(n+1)} \\
& =\frac{n+4}{2(n+1)}>\frac{\sqrt{2}}{2} .
\end{aligned}
$$

From $\frac{n+4}{2(n+1)}>\frac{\sqrt{2}}{2}$, squaring both sides and simplifying, we get
$$
n^{2}-4 n-142$. We can determine that $n$ can only take the values $3,4,5,6$.
We calculate $\cos A=\frac{n-4}{2(n-1)}$,
$$
\cos B=\frac{n^{2}+2}{2\left(n^{2}-1\right)} \text { (see table) }
$$

From the table below, we can see that only when $n=5$, $n-1=4$, $n+1=6$, $\cos A=\cos 2 C=\frac{1}{8}$, i.e., $A=2 C$ satisfies the problem's conditions. Therefore, the only solution for the triangle's sides is $4,5,6$. There is only one solution.
The table for verification is as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline n value & 3 & 4 & 5 & 6 \\
\hline $\cos \mathrm{A}$ & \begin{tabular}{l}
$-\frac{1}{4}$
\end{tabular} & 0 & \begin{tabular}{l}
$\frac{1}{8}$
\end{tabular} & \begin{tabular}{l}
$\frac{1}{5}$
\end{tabular} \\
\hline $\cos \mathrm{B}$ & \begin{tabular}{l}
$\frac{11}{16}$
\end{tabular} & \begin{tabular}{l}
$\frac{3}{5}$
\end{tabular} & \begin{tabular}{l}
$\frac{9}{16}$
\end{tabular} & \begin{tabular}{l}
$\frac{19}{35}$
\end{tabular} \\
\hline $\cos C$ & \begin{tabular}{l}
$\frac{7}{8}$
\end{tabular} & \begin{tabular}{l}
$\frac{4}{5}$
\end{tabular} & \begin{tabular}{l}
$\frac{3}{4}$
\end{tabular} & \begin{tabular}{l}
$\frac{5}{7}$
\end{tabular} \\
\hline $\cos 21$ & \begin{tabular}{l}
$-\frac{i}{128}$
\end{tabular} & $<0$ & $<0$ & $<0$ \\
\hline $\cos 2 \mathrm{C}$ & \begin{tabular}{l}
$\frac{17}{32}$
\end{tabular} & \begin{tabular}{l}
$\frac{7}{25}$
\end{tabular} & \begin{tabular}{l}
$\frac{1}{8}$
\end{tabular} & $\frac{1}{49}$ \\
\hline
\end{tabular}