2013 AMC 12B Problem 25

Problem 25 of 25HarderAlgebra Number Theory

Let

G

be the set of polynomials of the form

\begin{aligned} &P(z) = z^n + c_{n-1}z^{n-1} + \cdots \\ &\quad {}+ c_2 z^2 + c_1 z + 50, \end{aligned}

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?

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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.

For

d=1,2,5,10,25,50,

the numbers of conjugate pairs with

a^2+b^2=d

and

b\ne0

are

1,2,4,4,5,6,

respectively. Adding the two linear choices

z-d,z+d

gives

|B_1|=3,

|B_2|=4,

|B_5|=6,

|B_{10}|=6,

|B_{25}|=7,

and

|B_{50}|=8.

The factor-magnitude partitions of

50

using values greater than

1

are

50,25\cdot2,10\cdot5,

and

5\cdot5\cdot2.

Distinct roots require choosing two different

B_5

factors in the last case. Finally, account for the free presence of

z+1

and

z^2+1

(with

z-1

forced by the sign of the remaining product), gives

\begin{aligned} &2^2\left(8 + 7\cdot 4 + 6\cdot 6 + 4\binom{6}{2}\right) \\ &\quad = 4(8 + 28 + 36 + 60) = 528. \end{aligned}

Thus, the correct answer is B.