Test: 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 $a$.

---

The above text translated into English, preserving the original text's line breaks and format.
Answer: $a=5$.
Solution: Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), 00$.

Since $f(0)$ and $f(1)$ are positive integers, $f(0) f(1) \geqslant 1$, which means $a^{2} x_{1}(1-$ $\left.x_{i}\right) x_{2}\left(1-x_{2}\right) \geqslant 1$.

It is particularly noted that $x(1-x) \leqslant \frac{1}{4}$ always holds. The equality holds only when $x=\frac{1}{2}$. Since the numbers $x_{1}$ and $x_{2}$ are not equal, and $x_{1}\left(1-x_{1}\right)$ and $x_{2}\left(1-x_{2}\right)$ are both positive, we have $x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right)16$, which means $a>4$.
When $a=5$, we get the quadratic equation $5 x^{2}-5 x+1=0$, which has two distinct roots in the interval $[0,1]$.