Problem:
There are nonzero integers a,b,r, and s such that the complex number r+si is a zero of the polynomial P(x)=x3−ax2+bx−65. For each possible combination of a and b, let pa,b be the sum of the zeros of P(x). Find the sum of the pa,b 's for all possible combinations of a and b.
Solution:
Because P(x) has real coefficients, the number r−si must also be a zero of P(x), and the third zero must be a real number q. The sum of the zeros is a=q+2r, so q is an integer. The product of the zeros is 65=q(r2+s2), so r2+s2 is a factor of 65, and therefore must be 1,5,13, or 65. Note that 5=12+22,13=22+32, and 65=12+82=42+72. Therefore {∣r∣,∣s∣} is one of the sets {1,2},{2,3},{1,8},{4,7}, and each set corresponds to 4 distinct polynomials P(x). For the set {1,2},q=565=13, for {2,3},q=1365=5, and for the last two sets q=6565=1. The requested sum is then
a,b∑pa,b=4⋅13+4⋅5+8⋅1=80
The problems on this page are the property of the MAA's American Mathematics Competitions