2013 AIME II #12

Problem:

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

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 $(\genfrac{}{}{0ex}{}{4}{3})+4+(4)(3)=20$ 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 

$$\sqrt{x^2+y^2}=20,13$$

$$⟹x^2+y^2=400,169$$

$$⟹m^2+4y^2=1600,676$$

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 $\boxed{{\displaystyle 540}}$ total polynomials.