ALG-B3-M02-P024 Small roots of an integer quadratic

Hint 1. Pass to reciprocal roots.

Hint 2. If the reciprocal roots are at least \(50\), then for integer coefficients their difference is controlled by the discriminant.

Let \(00\). If \(c=1\), then \(R_2-R_1=\sqrt{b^2-4a}\), the positive square root of an integer; for distinct roots, \(R_2-R_1\ge1\). Hence \(R_1\ge50\), \(R_2\ge51\), and \(a=R_1R_2\ge50\cdot51=2550\). If \(c\ge2\), then \(a=cR_1R_2\ge2\cdot50^2>2550\). Thus \(a\ge2550\). This bound is attained because \(2550x^2-101x+1\) has roots \(\frac1{50}\) and \(\frac1{51}\).

A. Source analysis. Main objects: an integer quadratic and two small positive roots.

B. Insufficient first move. Working directly with small roots gives inconvenient fractions.

C. Hidden observation. Reciprocal roots become large, and integer coefficients create a discrete gap.

D. Required move. Replace roots by reciprocals and split into c=1 and c>=2.

E. Number of ideas. Three ideas: reciprocal substitution, discriminant gap, and construction using neighboring integers.

F. Difficulty justification. Final level 8: the solution requires a non-obvious reciprocal move and a careful lower bound.

G. Why it is not one-step. It is not one-step: after substitution one must use integrality and then construct an exact example.