Let $a x^{2}+b x+c$ be an integer-coefficient polynomial with two distinct roots that are greater than 0 and less than 1. How small can $|a|$ be?
Since the polynomial has exactly two roots, the coefficient of the quadratic term is not zero. If $a>0$.

The polynomial has roots, and these roots are distinct, so its discriminant is positive:

$$
b^{2}>4 a c
$$

We express the condition on the magnitude of the roots, $x_{1}$ and $x_{2}$, using the relationships between the roots and coefficients. Both roots are positive if and only if their sum and product are positive. Both roots are less than 1 if and only if $\left(1-x_{1}\right)$ and $\left(1-x_{2}\right)$ are positive numbers, i.e., their sum and product are positive.

Together with the condition $b^{2}>4 a c$, both roots will be greater than 0 and less than 1 if and only if the following quantities are all positive:

$$
\begin{gathered}
x_{1}+x_{2}=-\frac{b}{a} \\
x_{1} x_{2}=\frac{c}{a} \\
\left(1-x_{1}\right)+\left(1-x_{2}\right)=2-\left(x_{1}+x_{2}\right)=2+\frac{b}{a}, \text { and } \\
\left(1-x_{1}\right)\left(1-x_{2}\right)=1-\left(x_{1}+x_{2}\right)+x_{1} x_{2}=1+\frac{b}{a}+\frac{c}{a}
\end{gathered}
$$

If $a>0$, then from the first two conditions,

$$
b<0 \quad \text{and} \quad c>0
$$

and from the other two conditions,

$$
\begin{gathered}
2 a+b>0, \text { and } \\
a+b+c>0
\end{gathered}
$$

We seek the smallest positive integer $a$ for which there exist integers $b$ and $c$ such that (1), (2), (3), and (4) are satisfied.

From (4), we get $a+c>-b=|b|$. Since both sides of this inequality are integers, we also have $a+c-1 \geqq|b|$. Comparing this with $|b|>2 \sqrt{a c}$ from (1),

$$
a+c-1>2 \sqrt{a c}
$$

we obtain, after rearranging,

$$
(\sqrt{a}-\sqrt{c})^{2}>1
$$

Since the product of the roots, $x_{1} x_{2}=\frac{c}{a}$, is less than 1, we have $a>c$ (which can also be obtained from (1) and (3)). From (5),

$$
\begin{gathered}
\sqrt{a}-\sqrt{c}>1, \text { hence } \\
\sqrt{a}>1+\sqrt{c} \geqq 2
\end{gathered}
$$

Thus, $a>4$.

If $a=5$, then $0-b>\sqrt{20}$, so $b=-5$.

The polynomial $5 x^{2}-5 x+1$ satisfies all conditions (1) through (4) for its coefficients, and indeed it has two distinct roots between 0 and 1.

Remarks. 1. If $a>4$, then the roots of the equation $a x^{2}-a x+1=0$ are distinct and lie between 0 and 1.

2. If we do not require the coefficients of the polynomial to be integers, then the absolute value of the coefficient of the quadratic term can be arbitrarily small.
3. Conditions (2) through (4) can also be obtained as follows: If $a>0$, then the roots of the polynomial $f(x)=a x^{2}+b x+c$ lie in the interval $(0 ; 1)$ if and only if $f(0)=c>0, f(1)=a+b+c>0, 0<-\frac{b}{2 a}<1$, and $f\left(-\frac{b}{2 a}\right)<0$.