Find the smallest positive integer n such that we can find a polynomial nx^{2} + ax + b with integer coefficients and two distinct roots in the interval (0, 1).

**Solution**

Answer: n = 5 with equation 5x^{2} - 5x + 1 with roots (1 ± √(1/5) )/2.

The product of the roots lies in the interval (0, 1), so b must be 1, 2, 3, ... or n-1 (1). The larger root is (-a + √(a^{2} - 4bn) )/(2n). This must be less than 1, so -a < b + n (2). The roots are real and distinct, so a^{2} > 4bn (3).

Putting (2) and (3) together we get: (n + b - 1)^{2} ≥ a^{2} ≥ 4bn + 1. So if b = 1, then n ≥ 5. If b = 2, then n ≥ 6. If b = 3, then n ≥ 8. If b = 4, then n ≥ 10. Thus there are no solutions for n < 5 (which requires b < 4 by (1) ). If n = 5, then the only possible solution is 5x^{2} - 5x + 1, which is easily verified to be a solution.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002