$4 \cdot 27$ Given that $a$ is a natural number, a quadratic trinomial with integer coefficients and $a$ as the leading coefficient has two distinct positive roots less than 1. Find the minimum value of $a$.


[Solution] Let $f(x)=a x^{2}+b x+c$
$$
=a\left(x-x_{1}\right)\left(x-x_{2}\right),
$$

where $0<x_{1}<x_{2}<1$.
From the given information, we know that $f(0)$ and $f(1)$ are both positive integers, so
$$
f(0) \cdot f(1) \geqslant 1 \text {, }
$$

which means $a^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geqslant 1$.
However, $\quad x_{1}\left(1-x_{1}\right) \leqslant \frac{1}{4}$,
$$
x_{2}\left(1-x_{2}\right) \leqslant \frac{1}{4}
$$

and since $x_{1} \neq x_{2}$, the equality in the above two inequalities cannot hold simultaneously. Therefore, we have
$$
x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right)a^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geqslant 1$,
thus $a^{2}>16$,
$$
a>4 \text {, }
$$

so $a \geqslant 5$.
On the other hand, when $a=5$, the equation
$$
5 x^{2}-5 x+1=0
$$

has two distinct roots in the interval $(0,1)$.
Therefore, the minimum value of $a$ is 5.