Question

Let $G$ be the set of polynomials of the form

$P(z) = z^n + c_{n-1}z^{n-1} + \cdots + c_2 z^2 + c_1 z + 50,$

where $c_1, c_2, \ldots, c_{n-1}$ are integers and $P(z)$ has $n$ distinct roots of the form $a + ib$ with $a$ and $b$ integers. How many polynomials are in $G?$

$288$

$528$

$576$

$992$

$1056$

Solution:

Since the coefficients are real, nonreal roots occur in conjugate pairs, so $P(z)$ factors into distinct linear factors $(z - c)$ with $c \in \mathbb{Z}$ and quadratics $(z - (a+ib))(z - (a-ib)) = z^2 - 2az + (a^2 + b^2).$ Each factor's constant term divides $50.$ Counting basic factors of magnitude $d$ (the solutions of $a^2 + b^2 = d,$ plus the two linear $z \pm d$ ) gives $|B_1| = 3,$ $|B_2| = 4,$ $|B_5| = 6,$ $|B_{10}| = 6,$ $|B_{25}| = 7,$ $|B_{50}| = 8.$ Building the constant term $50$ as a single factor or a product over complementary divisors, and accounting for the free presence of $z + 1$ and $z^2 + 1$ (with $z - 1$ forced by the sign of the remaining product), gives $2^2\left(8 + 7\cdot 4 + 6\cdot 6 + 4\binom{6}{2}\right) = 4(8 + 28 + 36 + 60) = 528.$ Thus, the correct answer is B.

2013 AMC 12B Problem 25

Solution:

Problem 25 in Other Years