All-Soviet-Union Mathematical Olympiad

Solution:
$4x^2 - 4x + 1 = (2x - 1)^2$, which has the double root $1/2$. So it remains to consider $a = 1,2,3$.

- $b/a$ is the sum of the roots, so $b$ is negative. $c/a$ is the product of the roots, so $c$ is positive. If $a = 1$, then the product of the roots is $c$, which is at least $1$, so both roots cannot lie strictly between $0$ and $1$.

If $a = 2$, then the sum of the roots is less than $2$, so $b$ must be $-1$, $-2$, or $-3$. The roots are real so $b^2 > 4ac = 8c$. Hence $b = -3$ and $c = 1$. But $2x^2 - 3x + 1 = (2x - 1)(x - 1)$ and one root is not less than $1$.

If $a = 3$, then $b$ must be $-1$, $-2$, ..., or $-5$. But $b^2 > 4ac = 12c$, 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 $a = 1,2,3$ and so the smallest value of $a$ is $4$.