Problem:
Let S be the set of all polynomials of the form z3+az2+bz+c, where a,b, and c are integers. Find the number of polynomials in S such that each of its roots z satisfies either ∣z∣=20 or ∣z∣=13.
Solution:
Let f∈S only have roots with modulus (absolute value) 20 or 13. If f has all real roots, then they must come from the set {−20,−13,13,20}. There are 20 possible ways to choose three elements from this set, with replacement, where order is not important.
Suppose f has a nonreal root z0=r+si, where r and s are real. Then r−si is another root, and there are four possibilities for the third (integer) root, k. Because (z−(r+si))(z−(r−si))(z−k)= z3−(2r+k)z2+(2kr+r2+s2)z−(r2+s2)k, it follows that 2r must be an integer. If ∣z0∣=20, then there are 79 possible values for r:0,±21,±22,…,±239. If ∣z0∣=13, then there are 51 possible values for r:0,±21,±22,…,±225. Therefore there are 79⋅4+51⋅4+20=540 polynomials in S that only have roots with modulus 20 or 13.
The problems on this page are the property of the MAA's American Mathematics Competitions