Let S be the set of all polynomials of the form \(z^3+az^2+bz+c\), where a, b, and c are integers. Find the number of polynomials in S such that each of its roots z satisfies either |z|=13 or |z|=20.

Guest Nov 29, 2018

#1**+1 **

There are either 3 real roots to this polynomial or 1 real root and 2 complex conjugate roots.

The conjugates will have the same magnitude.

So we end up with factorizations that have the following form

\(p(z) = (z-a)(z-b)(z-c) \\ \text{where }a,b,c \text{ are chosen from the set }\{-20,-13,13,20\}\\ p(z) = (z-a)\left(z-b e^{i \theta}\right)\left(z-b e^{-i\theta}\right) \\ \text{where }a \text{ is chosen from the above set and }b \text{ is chosen from }\{13,20\}\)

\(\text{in the second case we are restricted to certain values of }\theta\\ \text{in order to ensure integer coefficients}\\ \text{I recommend that you work this out but in general what is required is }\\ \cos(\theta) \in \mathbb{Z}\\ \text{this restricts }\theta \text{ to the following set of values}\\ \theta \in \left \{0, \dfrac{\pi}{3}, \dfrac{\pi}{2}, \dfrac{2\pi}{3},\pi, \dfrac{4\pi}{3}, \dfrac{3\pi}{2}, \dfrac{5\pi}{3}\right\} \text{however the values of }0 \text{ and }\pi \\ \text{are already covered by the groups of purely real roots above}\\ \text{so the set of allowable angles is reduced to }\\ \theta \in \left \{\dfrac{\pi}{3}, \dfrac{\pi}{2}, \dfrac{2\pi}{3},\dfrac{4\pi}{3}, \dfrac{3\pi}{2}, \dfrac{5\pi}{3}\right\}\)

Now let's start counting.

To determine the number of unique polynomials generated by the first triplets of roots we note

That we can pick 4 triplets where all 3 roots are identical

4x3=12 triplets where there is 1 pair of roots and then an additional root

4x3x2/3! = 4 triples where all three roots are distinct

Again you should work out why this is so.

These cases sum to 20 distinct polynomials

Now for the cases with a pair of conjugate roots we note

We have 4 choices for the real root. Then we have 2 choices for the magnitude of the complex roots

and 6 choices for the angle of the complex roots. That gets us a total of 4x2x6 = 48 distinct polynomials

So in total we have 48 + 20 = 68 distinct polynomials with integer coefficients.

Someone verifying all this would be nice.

Rom
Nov 29, 2018