Answer: 5 Let $x$ and $y$ be the roots. Then:
$$
\begin{array}{c}
\frac{b}{a}=x+y1 \\
b^{2}>4 a c>4 c^{2} \Rightarrow b>2 c
\end{array}
$$

Evaluated at 1 , the polynomial must be greater than 0 , so $a+c>b$. Then:
$$
\begin{array}{c}
2 c<b<a+c \\
2 c+1 \leq b \leq a+c-1 \\
a \geq c+2 \geq 3
\end{array}
$$

If $a=3$, then $c=1$ and $b=3$, by the above bounds, but this polynomial has complex roots. Similarly, if $a=4$, then $c=1$ and $b$ is forced to be either 3 or 4 , again giving either 0 or 1 distinct real roots. So $a \geq 5$. But the polynomial $5 x^{2}-5 x+1$ satisfies the condition, so 5 is the answer.
33. [15] The polynomial $a x^{2}-b x+c$ has two distinct roots $p$ and $q$, with $a, b$, and $c$ positive integers and with $0<p, q<1$. Find the minimum possible value of $a$.