Problem:
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 $\left\lvert z \right\rvert = 20$ or $\left\lvert z \right\rvert = 13$.
Solution:
In the first case, the polynomial has $3$ real roots. These roots could be any from the options $13, -13, 20, -20$. There are $\binom{4}{3}+4+(4)(3) = 20$ polynomials in this case.
In the second case, there is one real root and two complex roots.
There are $4$ options for the one real root, namely $13, -13, 20, -20$. WLOG, let this root be $20$. The two complex roots are conjugates, so let them be $x+yi$ and $x-yi$. Expand the polynomial $$P(z) = (P-20)(P-(x+yi))(P-(x-yi))$$ to get $$z^3 + (-2x-20)z^2 + (z)(x^2+y^2+40x)+(-20x^2-20y^2).$$ Note that in this expanded version of $P(z)$, the coefficient $c$ is already an integer. For the coefficient $a$ to be an integer, we need $2x$ to be an integer. Note that this automatically makes $b=x^2+y^2+40x$ an integer. So, let $x = \frac{m}{2}$ for an integer $m$. Now, we have
$$\sqrt{x^2+y^2}=20, 13$$
$$\implies x^2+y^2 = 400, 169$$
$$\implies m^2 + 4y^2 = 1600, 676$$
In the first case, where it's equal to $1600$, the set of possible values for $m$ are $-39, \dots, 0, \dots, 39$, and these are $79$ values. Similarly, when it's equal to $676$, the set of possible values for $m$ are $-25, \dots, 0, \dots, 25$, and these are $51$ values. In total, we have $130$ solutions. Multiplying by $4$ for the options for the real root, we have $130 \cdot 4 = 520$ polynomials in this case.
So there are $\boxed{540}$ total polynomials.
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