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.