\section*{Problem 5}

What is the smallest positive integer a such that we can find integers \(\mathrm{b}\) and \(\mathrm{c}\) so that \(\mathrm{ax}^{2}+\mathrm{bx}\) \(+\mathrm{c}\) has two distinct positive roots less than 1 ?


\section*{Solution}

\(4 \mathrm{x}^{2}-4 \mathrm{x}+1=(2 \mathrm{x}-1)^{2}\), which has the double root \(1 / 2\). So it remains to consider \(\mathrm{a}=1,2,3\).

-b/a is the sum of the roots, so \(\mathrm{b}\) is negative. c/a is the product of the roots, so \(\mathrm{c}\) is positive. If \(\mathrm{a}=1\), then the product of the roots is \(\mathrm{c}\), which is at least 1 , so both roots cannot lie strictly between 0 and 1 . If \(\mathrm{a}=2\), then the sum of the roots is less than 2 , so \(\mathrm{b}\) must be \(-1,-2\), or -3 . The roots are real so \(\mathrm{b}^{2}>4 \mathrm{ac}=8 \mathrm{c}\). Hence \(\mathrm{b}=-3\) and \(\mathrm{c}=1\). But \(2 \mathrm{x}^{2}-3 \mathrm{x}+1=(2 \mathrm{x}-1)(\mathrm{x}-1)\) and one root is not less than 1. If \(a=3\), then \(b\) must be \(-1,-2, \ldots\), or -5 . But \(b^{2}>4 a c=12 c\), so (b, c \()=(-4,1),(-5,1)\) or \((-5,2)\). In the first and last case, the equation has a root 1 . In the middle case it has a root \(5 / 6+\sqrt{ } 13 / 6=1.434>1\). Thus there are no solutions for \(\mathrm{a}=1,2,3\) and so the smallest value of a is 4 .

