8. Find the smallest positive integer $k$ such that the equation $2 x y-3 x-5 y=k$ has an odd number of positive integral solutions.
(1 mark)
求最小的正整數 $k$, 使方程 $2 x y-3 x-5 y=k$ 的正整數解數目為奇數。
8. 5
8. We have
$$
\begin{aligned}
2 x y-3 x-5 y & =k \\
x y-\frac{3}{2} x-\frac{5}{2} y & =\frac{k}{2} \\
\left(x-\frac{5}{2}\right)\left(y-\frac{3}{2}\right) & =\frac{k}{2}+\frac{15}{4} \\
(2 x-5)(2 y-3) & =2 k+15
\end{aligned}
$$

When $k$ is a positive integer, the number of positive integral solutions to the above equation is precisely the number of positive divisors of $2 k+15$. For this number to be odd, $2 k+15$ has to be a perfect square. The smallest such $k$ is 5 . Indeed, we can check that when $k=5$, the original equation has 3 solutions, namely, $(3,14) ;(5,4)$ and $(15,2)$.