2013 AIME II 第 12 题

Nonreal roots come in conjugate pairs, so either all three roots are among $\pm 20$ and $\pm 13,$ or one real root is joined by a conjugate pair on one of the two circles

For roots $r \pm s\mathrm{i}$ and real $k,$ expanding shows integer coefficients hold exactly when $2r$ is an integer; count the half-integers $r$ with $|r|$ less than the modulus

A cubic with real coefficients has either three real roots or one real root and a conjugate pair. The only real numbers with modulus $20$ or $13$ are $\pm 20$ and $\pm 13,$ so in the all-real case the roots form a multiset of size $3$ from those $4$ values: $\binom{6}{3} = 20$ polynomials.

Otherwise the roots are $k \in \{\pm 20, \pm 13\}$ and a conjugate pair $r \pm s\mathrm{i}$ with $s \ne 0$ and $r^2 + s^2 = 400$ or $169.$ Expanding $$(z - k)\bigl(z^2 - 2rz + (r^2 + s^2)\bigr)$$ shows the coefficients are $-(2r + k),$ $r^2 + s^2 + 2rk,$ and $-(r^2 + s^2)k,$ which are all integers exactly when $2r$ is an integer. On the circle of radius $20$ we need $|r| \lt 20,$ allowing $2r \in \{-39, \ldots, 39\}:$ $79$ choices; on the circle of radius $13,$ $2r \in \{-25, \ldots, 25\}:$ $51$ choices. With $4$ choices of $k,$ that gives $4(79 + 51) = 520$ polynomials, each distinct since the roots determine the polynomial.

In total there are $20 + 520 = 540$ such polynomials.