2023 AIME II Problem 8

Question

Let

\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},

where

i = \sqrt{-1}.

Find the value of the product

\prod_{k=0}^{6} \left(\omega^{3k} + \omega^k + 1\right).

Answer

The

k = 0

factor equals

3;

the rest is

\prod P(z)

over the primitive seventh roots of unity, where

P(x) = x^3 + x + 1

\prod_{z^7 = 1} P(z) = \prod (1 - \beta^7)

over the roots

\beta

of

P;

reduce

\beta^7

using

\beta^3 = -\beta - 1

Let

P(x) = x^3 + x + 1,

so the product is

\prod_{k=0}^{6} P(\omega^k),

where

\omega^0, \ldots, \omega^6

are all seventh roots of unity. Since

x^7 - 1 = \prod_k (x - \omega^k),

writing the factorization

P(x)

= (x - \beta_1)(x - \beta_2)(x - \beta_3)

and swapping the order of the double product gives

\begin{gathered} \prod_{k=0}^{6} P(\omega^k) \\ = \prod_{j=1}^{3} \prod_{k=0}^{6} (\omega^k - \beta_j) \\ = \prod_{j=1}^{3} \bigl(-(\beta_j^7 - 1)\bigr) \\ = \prod_{j=1}^{3} (1 - \beta_j^7). \end{gathered}

For a root

\beta

of

P,

repeatedly using

\beta^3 = -\beta - 1

gives

\beta^4 = -\beta^2 - \beta,

\beta^5 = -\beta^2 + \beta + 1,

\beta^6 = \beta^2 + 2\beta + 1,

and

\beta^7 = 2\beta^2 - 1.

Hence

1 - \beta^7 = 2(1 - \beta)(1 + \beta),

and

\begin{gathered} \prod_{j} (1 - \beta_j^7) \\ = 2^3 \prod_j (1 - \beta_j) \prod_j (1 + \beta_j) \\ = 8 \cdot P(1) \cdot \bigl(-P(-1)\bigr) \\ = 8 \cdot 3 \cdot 1 = 24. \end{gathered}

So the requested product equals

24.

Problem 8 in other years