text: # Thread: not divide
  
  1. ## not divide
  
  Hello, need your help thanks
  
  Prove that if $\displaystyle n\in\mathbb{N}^{*}$ then $\displaystyle 6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor$
  
  2. Originally Posted by maria18
  Prove that if $\displaystyle n\in\mathbb{N}^{*}$ then $\displaystyle 6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor$
  If $\displaystyle x^3 = 28$ then $\displaystyle (x-3)(x^2+3x+9) = x^3-27 = 28-27=1$. Therefore $\displaystyle (x-3)^{-1} = 9+3x+x^2$, and
  $\displaystyle (x-3)^{-n} = (9+3x+x^2)^n = 9^n + a_1x + a_2x^2 + \ldots + a_{2n}x^{2n}$, a polynomial in x of degree 2n with integer coefficients.
  
  Let $\displaystyle \omega = e^{2\pi i/3} = -\tfrac12 + \tfrac{\sqrt3}2i$, a complex cube root of unity. Note that $\displaystyle 1+\omega+\omega^2 = 0$. Let $\displaystyle X = \sqrt[3]{28}$. Then the complex cube roots of 28 are $\displaystyle X,\ \omega X$ and $\displaystyle \omega^2X$. Therefore
  
  $\displaystyle (X-3)^{-n} + (\omega X-3)^{-n} + (\omega^2X-3)^{-n}$
  . . . . .\displaystyle \begin{aligned}= 3*9^n + {}& (1+\omega+\omega^2)a_1X + (1+\omega^2+\omega)a_2X^2 + 3a_3X^3 \\\quad&+ (1+\omega+\omega^2)a_4X^4 + (1+\omega^2+\omega)a_5X^5 + 3a_6X^6 + \ldots\\ = 3*9^n + {}&3*28(a_3 + a_6X^3 + \ldots).\end{aligned}
  
  This is a (real) integer of the form $\displaystyle 3*9^n + 84k$ for some integer k, and is therefore an odd multiple of 3.
  
  Next, notice that $\displaystyle \omega X-3 = -(3+\tfrac12X) + \tfrac{\sqrt3}2Xi$, and $\displaystyle X > 3$. Hence $\displaystyle |\omega X-3|>3\sqrt3$ and so $\displaystyle |(\omega X-3)^{-n}|<3^{-3n/2}$, and similarly $\displaystyle |(\omega^2 X-3)^{-n}|<3^{-3n/2}$. So both these quantities are very small (certainly less than 1/2). It follows from the previous paragraph that $\displaystyle (X-3)^{-n}$ is within distance less than 1 from an odd multiple of 3. So its integer part cannot be a multiple of 6.
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}