Problem:
For how many real numbers a does the quadratic equation x2+ax+6a=0 have only integer roots for x?
Solution:
Suppose x2+ax+6a=0 has integer roots m and n, with m≤n. Since
x2+ax+6a=(x−m)(x−n)=x2−(m+n)x+mn,
we must have a=−(m+n) and 6a=mn. This implies that a must be an integer and that −6(m+n)=mn. This last equation is equivalent to mn+6m+6n+36=36, or
(m+6)(n+6)=36.
It is not hard to see that the only integer solutions with m≤n are the ten pairs (−42,−7),(−24,−8),(−18,−9),(−15,−10),(−12,−12),(−5,30),(−4,12),(−3,6), (−2,3),(0,0). The corresponding values of a=−(m+n) are 49,32,27,25,24,−25−8,−3,−1, and 0. Thus there are 10 values of a for which x2+ax+6a=0 has integer roots.
The problems on this page are the property of the MAA's American Mathematics Competitions