4. Given that for a positive integer $a$, there exists an integer-coefficient quadratic trinomial with $a$ as the coefficient of the quadratic term, which has two distinct positive roots less than 1, then the minimum value of $a$ is $\qquad$
4. Solution: Let the quadratic trinomial be $f(x)=a\left(x-x_{1}\right)(x-$ $\left.x_{2}\right), a \in \mathbf{N}^{+}$. According to the problem, we have $00, f(1)=a\left(1-x_{1}\right)\left(1-x_{2}\right)>$ 0

Also, since $f(x)=a x^{2}-a\left(x_{1}+x_{2}\right) x+a x_{1} x_{2}$ is a quadratic trinomial with integer coefficients,
$\therefore f(0), f(1)$ are both integers, thus $f(0) \geqslant 1, f(1)$ $\geqslant 1$, hence $f(0) f(1) \geqslant 1$.
$$
\begin{array}{l} 
\text { Also, } \because x_{1}\left(1-x_{1}\right) \leqslant\left[\frac{x_{1}+\left(1-x_{1}\right)}{2}\right]^{2}=\frac{1}{4}, x_{2}(1 \\
\left.-x_{2}\right) \leqslant\left[\frac{x_{2}+\left(1-x_{2}\right)}{2}\right]^{2}=\frac{1}{4},
\end{array}
$$

From $x_{1} \neq x_{2}$, we know that the equality in the two inequalities cannot hold simultaneously,
$$
\begin{array}{l}
\therefore x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right)1, \therefore a>4 . \\
\quad \text { Also, } a \in \mathbf{N}_{+}, \therefore a \geqslant 5 .
\end{array}
$$