Question

Sea $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ donde $i = \sqrt{-1}.$ Halla el valor del producto $\prod_{k=0}^{6} \left(\omega^{3k} + \omega^k + 1\right).$

Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product $\prod_{k=0}^{6} \left(\omega^{3k} + \omega^k + 1\right).$

Respuesta

Solución:

Sea $P(x) = x^3 + x + 1,$ así que el producto es $\prod_{k=0}^{6} P(\omega^k),$ donde $\omega^0, \ldots, \omega^6$ son todas las raíces séptimas de la unidad. Como $x^7 - 1 = \prod_k (x - \omega^k),$ escribir $P(x)$ $= (x - \beta_1)(x - \beta_2)(x - \beta_3)$ e intercambiar el orden del producto doble da $\begin{gathered} \prod_{k=0}^{6} P(\omega^k) \\ = \prod_{j=1}^{3} \prod_{k=0}^{6} (\omega^k - \beta_j) \\ = \prod_{j=1}^{3} \bigl(-(\beta_j^7 - 1)\bigr) \\ = \prod_{j=1}^{3} (1 - \beta_j^7). \end{gathered}$

Para una raíz $\beta$ de $P,$ usar repetidamente $\beta^3 = -\beta - 1$ da $\beta^4 = -\beta^2 - \beta,$ $\beta^5 = -\beta^2 + \beta + 1,$ $\beta^6 = \beta^2 + 2\beta + 1,$ y $\beta^7 = 2\beta^2 - 1.$ Por lo tanto $1 - \beta^7 = 2(1 - \beta)(1 + \beta),$ y $\begin{gathered} \prod_{j} (1 - \beta_j^7) \\ = 2^3 \prod_j (1 - \beta_j) \prod_j (1 + \beta_j) \\ = 8 \cdot P(1) \cdot \bigl(-P(-1)\bigr) \\ = 8 \cdot 3 \cdot 1 = 24. \end{gathered}$

Así que el producto pedido es igual a $24.$

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 $P(x)$ $= (x - \beta_1)(x - \beta_2)(x - \beta_3)$ 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 $24.$

2023 AIME II Problema 8

Solución:

El Problema 8 en otros años