Let $a$ be a positive number, and we show decimal part of the $a$ with $\left\{a\right\}$.For a positive number $x$ with $\sqrt 2< x <\sqrt 3$ such that, $\left\{\frac{1}{x}\right\}$=$\left\{x^2\right\}$.Find value of the $$x(x^7-21)$$
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Tags: number theory, number theory unsolved
joa51736
31.01.2016 14:53
$\sqrt{2}<x<\sqrt{3}$ so $\{\frac{1}{x}\}=\frac{1}{x}$ and $\{x^2\}=x^2-2$. Hence $\frac{1}{x}=x^2-2$. So $x^3-2x-1=0=(x+1)(x^2-x-1)$. We deduce that $x=\frac{1+\sqrt{5}}{2}$. Hence $x(x^7-21)=13$. To calculate $x^8-21x$ you can use $x^2=x+1$.
Emirhan
31.01.2016 14:58
joa51736 wrote: $\sqrt{2}<x<\sqrt{3}$ so $\{\frac{1}{x}\}=\frac{1}{x}$ and $\{x^2\}=x^2-2$. Hence $\frac{1}{x}=x^2-2$. So $x^3-2x-1=0=(x+1)(x^2-x-1)$. We deduce that $x=\frac{1+\sqrt{5}}{2}$. Hence $x(x^7-21)=13$. Why $\{\frac{1}{x}\}=\frac{1}{x}$ and $\{x^2\}=x^2-2$ ?
joa51736
31.01.2016 15:00
Because $0<\frac{1}{x}<1$ and $2<x^2<3$.
Emirhan
31.01.2016 15:02
I understand now. Thanks joa51736.