16. The function $f(x)=a x^{2}+b x+c, a, b, c \in \mathbf{Z}$, and $f(x)$ has two distinct roots in $(0,1)$. Find the smallest positive integer $a$ that satisfies the above conditions.
16. Let the roots of the equation $f(x)=0$ be $r$ and $s(0<s<1<r)$. Then, $f(x)=a(x-r)(x-s)=a[x^2-(r+s)x+rs]$. Since $0<s<1<r$, we have $f(0)=c=ars>0, f(1)=a+b+c=a(1-r)(1-s)>$ 0. Given $0<s<1<r, c \in \mathbf{Z}$, then $c \geqslant 1$, i.e., $f(0) \geqslant 1$. Also, $a+b+c \geqslant 1$, i.e., $f(1) \geqslant 1$, so $1 \leqslant f(0) f(1)=a^{2} r(1-r) \cdot s(1-s) \leqslant \frac{a^{2}}{16}$. Since $r \neq s$, $r$ and $s$ cannot both be $\frac{1}{2}$, so the equality in the above inequality does not hold. Therefore, $a^{2}>16$, and the smallest positive integer $a=5$.