Problem
ALG-B1-M04-P017 Three Consecutive Constant Terms
#17
★★★★★ Level 5 of 5
Suppose both trinomials \(x^2+px+q\) and \(x^2+px+q+1\) have integer roots. Prove that \(x^2+px+q+2\) has no real roots.
Inspired by regional olympiad method · 2019 · Grade 11 · Problem 2
Compare the discriminants of the three equations.
The discriminants are \(D_1=p^2-4q\), \(D_2=p^2-4q-4\), \(D_3=p^2-4q-8\).
Since the first two trinomials have integer roots, \(D_1=m^2\), \(D_2=n^2\) for non-negative integers \(m,n\). Thus \(m^2-n^2=4\), so \((m-n)(m+n)=4\). The two factors have the same parity, so only \(m-n=m+n=2\) is possible, giving \(n=0\).
Then \(D_3=D_2-4=-4<0\). Therefore the third trinomial has no real roots.
Key: integer roots turn discriminants into squares.