$31 \cdot 7$ The non-negative integer solutions of the equation $5 x+3 y=71$ are
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = \alpha _ { 1 } , } \\
{ y _ { 1 } = \beta _ { 1 } ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = \alpha _ { 2 } , } \\
{ y _ { 2 } = \beta _ { 2 } ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x _ { 3 } = \alpha _ { 3 } , } \\
{ y _ { 3 } = \beta _ { 3 } ; }
\end{array} \ldots \left\{\begin{array}{l}
x_{k}=\alpha_{k}, \\
y_{k}=\beta_{k} .
\end{array}\right.\right.\right.\right.
$$

Then
(A) $\alpha_{1}, \alpha_{2}, \alpha_{3}, \cdots, \alpha_{k}$ are the samples of non-negative integer solutions of this equation.
(B) $\alpha_{1}, \alpha_{2}, \alpha_{3}, \cdots, \alpha_{k}$ are all the positive integer solutions of this equation.
(C) Triangles can be formed with $\alpha_{i}, \beta_{i}, \alpha_{i}+\beta_{i}(i=1,2, \cdots, k)$ as the three sides.
(D) When $k$ is a non-negative integer not greater than 4, $\left\{\begin{array}{l}x=13-3 k \\ y=5 k+2,\end{array}\right.$ are the non-negative integer solutions of the equation.
(Shanxi Province, China Junior High School Mathematics Competition, 1990)
[Solution] Substituting $\left\{\begin{array}{l}x=13-3 k, \\ y=5 x+2,\end{array}\right.$ into $5 x+3 y=71$ gives
$$
65-15 k+15 k+6=71 \text {, }
$$

Therefore, when $k \leqslant 4$, and $k$ is a non-negative integer, $x=13-3 k, y=5 k+2$ are non-negative integer solutions to the equation. Hence, the answer is $(D)$.