9.6. What is the smallest natural number $a$ for which there exist integers $b$ and $c$ such that the quadratic trinomial $a x^{2}+b x+c$ has two distinct positive roots, each not exceeding $\frac{1}{1000} ?$

(A. Khryabrov)
Answer. $a=1001000$.

First solution. We will prove that $a \geqslant 1001000$. Notice that if $y$ is a root of the quadratic polynomial $a x^{2}+b x+c$, then $1 / y$ is a root of the quadratic polynomial $c x^{2}+b x+a$. Therefore, in the problem, we need to find the smallest natural $a$ for which the roots $x_{1}$ and $x_{2}$ of some quadratic polynomial $c x^{2}+b x+a$ (with integer $b$ and $c$) are greater than 1000. Since $x_{1}$ and $x_{2}$ are positive and $x_{1} x_{2}=a / c$ (by Vieta's theorem), we have $c>0$.

If $c=1$, then $\left|x_{1}-x_{2}\right|=\sqrt{b^{2}-4 a} \geqslant 1$. Since the smaller root is at least 1000, the larger root is at least 1001, and thus $a=x_{1} x_{2} \geqslant 1001 \cdot 1000$. If $c \geqslant 2$, then $a=c x_{1} x_{2} \geqslant 2 x_{1} x_{2}>2000000$. In both cases, the required estimate is proven.

It remains to note that the quadratic polynomial $x^{2}-(1000+1001) x+1001 \cdot 1000$ has roots 1000 and 1001, so $a=1001000$ works.

Second solution. Let for brevity $n=1000$. Let $x_{1}$ and $x_{2}$ be two distinct roots of the quadratic polynomial $f(x)=a x^{2}+b x+c$, where $0-\frac{n b}{2}$.

Since the roots are distinct, the discriminant $D=b^{2}-4 a c$ is positive. Therefore, $b^{2}>4 a c>-2 n b c$ and, thus, $-b>2 n c$. Therefore, $a>(-b) \cdot \frac{n}{2}>2 n c \cdot \frac{n}{2}=n^{2} c$. Let $a=n^{2} c+d$, where $d$ is a natural number.

Suppose that $a<n^{2}+n$. Then $c=1$ and $d<n$. Therefore, $0 \leqslant f\left(\frac{1}{n}\right)=\frac{a}{n^{2}}+\frac{b}{n}+c=\frac{d}{n^{2}}+\frac{b}{n}+2<\frac{1}{n}+\frac{b}{n}+2$ and, thus, $-b<2 n+1$. Consequently, $-b \leqslant 2 n$ and $D=b^{2}-4 a c \leqslant 4 n^{2}-4\left(n^{2}+d\right)=-4 d<0$. This contradiction shows that $d \geqslant n$.

If $a=n^{2}+n$, then for $b=-2 n-1$ and $c=1$, the quadratic polynomial has roots $x_{1}=\frac{1}{n+1}$ and $x_{2}=\frac{1}{n}$.