2005 AIME I Problem 6

Question

Let

P

be the product of the nonreal roots of

x^4 - 4x^3 + 6x^2 - 4x = 2005.

Find

\lfloor P\rfloor.

(The notation

\lfloor P\rfloor

denotes the greatest integer that is less than or equal to

P.

)

Answer

The left side is one term short of the binomial expansion of

(x - 1)^4;

add

1

to both sides

From

(x-1)^4 = 2006,

the nonreal roots are

1 \pm i\sqrt[4]{2006};

multiply this conjugate pair

Adding

1

to both sides turns the left side into a perfect fourth power:

\begin{aligned} &(x - 1)^4 = x^4 - 4x^3 + 6x^2 \\ &\quad {}- 4x + 1 = 2006. \end{aligned}

So

x - 1

is a fourth root of

2006:

the four roots are

x = 1 \pm \sqrt[4]{2006}

(real) and

x = 1 \pm i\sqrt[4]{2006}

(nonreal).

The product of the conjugate pair of nonreal roots is

\begin{aligned} &\left(1 + i\sqrt[4]{2006}\right) \\ &\quad {}\cdot \left(1 - i\sqrt[4]{2006}\right) \\ &= 1 + \sqrt{2006}. \end{aligned}

Since

44^2 = 1936

\lt 2006

\lt 2025 = 45^2,

we have

45 \lt P \lt 46,

so

\lfloor P\rfloor = 45.

Problem 6 in other years