$31 \cdot 29$ makes the equation $a^{2}+b^{2}=k \cdot a b$ have positive integer solutions $(a, b)$ for the number of positive integers $k$ is
(A) 0.
(B) 1.
(C) More than 1, but only a finite number.
(D) Infinitely many.
(Anhui Province Wuhu City Mathematics Competition, 1993)
[Solution 1] Let the greatest common divisor of $a$ and $b$ be $d$, then $a=d \cdot a_{1}, b=d \cdot b_{1}$ ($a_{1}$ and $b_{1}$ are coprime), substituting into $a^{2}+b^{2}=k \cdot a b$ gives $a_{1}^{2}+b_{1}^{2}=k \cdot a_{1} b_{1}$,

thus we get $a_{1} \mid b_{1}$ and $b_{1} \mid a_{1}$, hence $a_{1}=b_{1}$, at this point $k=2$. When $k=2$, we can take $a=b$, which satisfies the condition.
Therefore, the answer is $(B)$.
[Solution 2] From $a^{2}+b^{2}=k a b$, we get $\left(\frac{a}{b}\right)^{2}-k \cdot \frac{a}{b}+1=0$, i.e., the equation $x^{2}-k x+1=0$ has a rational solution, so its discriminant $\Delta=k^{2}-4$ should be a perfect square.
Let $k^{2}-4=c^{2}$ (where $c$ is a non-negative integer), then
$$
(k-c)(k+c)=4=2 \cdot 2=1 \cdot 4 .
$$

Thus, $k-c=2, k+c=2$, or $k-c=1, k+c=4$. From this, it is easy to see that $k=2$.
Therefore, the answer is $(B)$.