5 The number of positive integers $k$ that make the equation $a^{2}+b^{2}=k \cdot a b$ have positive integer solutions $(a, b)$ is
A. 0
B. 1
C. More than 1, but only a finite number
D. Infinitely many
5 B. Let the greatest common divisor of $a$ and $b$ be $d, a=d \cdot a_{1}, b=d \cdot b_{1}\left(a_{1}\right.$ and $b_{1}$ are coprime), substituting into $a^{2}+b^{2}=k \cdot a b$ yields $a_{1}^{2}+b_{1}^{2}=k \cdot a_{1} b_{1}$. Therefore, we get $a_{1} \mid b_{1}$ and $b_{1} \mid a_{1}$, hence $a_{1}=b_{1}$, at this time $k=2$. When $k=2$, we can take $a_{1}=b_{1}$, which satisfies the requirement.