Problem:
There are real numbers a,b,c, and d such that β20 is a root of x3+ax+b and β21 is a root of x3+cx2+d. These two polynomials share a complex root m+nββ i, where m and n are positive integers and i=β1β. Find m+n.
Solution:
Let the shared complex root be z=m+nββ i, with conjugate zΛ=mβnββ i. Since the coefficients of both polynomials are real, both z and zΛ must appear as roots.
The roots of x3+ax+b are β20, z, and zΛ, so by Vietaβs formula, the sum of the roots is β20+z+zΛ=0, which gives z+zΛ=20 and hence 2m=20, so m=10.
Now consider the polynomial x3+cx2+d, whose roots are β21, z, and zΛ. The sum of the products of the roots taken two at a time is equal to the coefficient of x1 (which is zero) by Vietaβs formula:
(β21)z+(β21)zΛ+zzΛ=0, or β21(z+zΛ)+zzΛ=0.
We already know z+zΛ=20, and zzΛ=(m+nβi)(mβnβi)=m2+n=100+n.
Substitute into the equation: β21(20)+100+n=0, which gives β420+100+n=0, so n=320.