6.15 Let $k$ be a positive integer, and the quadratic equation
$$
(k-1) x^{2}-p x+k=0
$$

has two positive integer roots. Find the value of $k^{k p}\left(p^{p}+k^{k}\right)$.
(China Beijing Junior High School Grade 2 Mathematics Competition, 1984)
［Solution］Since $k$ is a positive integer, and the equation
$$
(k-1) x^{2}-p x+k=0
$$

is a quadratic equation, then $k \geqslant 2$.
Let the two positive integer roots of equation (1) be $x_{1}$ and $x_{2}$. By Vieta's formulas, we have
$$
x_{1} x_{2}=\frac{k}{k-1} \text {. }
$$

If $k-1 \neq 1$, then $(k-1, k)=1$, in this case $\frac{k}{k-1}$ cannot be an integer.
Therefore, it must be that $k-1=1, k=2$.
At this point,
$$
x_{1} x_{2}=2=1 \cdot 2 .
$$

Since $x_{1}$ and $x_{2}$ are positive integers, $x_{1}$ and $x_{2}$ can only be one 1 and one 2.
By Vieta's formulas again, $\quad x_{1}+x_{2}=3=\frac{p}{k-1}=p$,

Thus,
$$
k^{k p}\left(p^{p}+k^{k}\right)=2^{2 \cdot 3}\left(3^{3}+2^{2}\right)=1984 .
$$