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If $x=(9+4\sqrt{5})^{48}=[x]+f$, where $[x]$ is defined as integral part of $x$

and $f$ is a fraction, then $x(1-f)$ equals .

$\color{green}{a.)\ 1} \\ b.)\ \text{less than}\ 1 \\ c.)\ \text{more than}\ 1 \\ d.)\ \text{between}\ 1 \text{and }\ 2 \\ e.)\ \text{none of these}\ \\ $

This question looks scary from the get go. I tried to go with pattern

$(9+4\sqrt{5})^{1}\approx 17.94=17+0.94 \implies 17(1-0.94)=1.02\\ (9+4\sqrt{5})^{2}\approx 321.99=321+0.99 \implies 321(1-0.99)=3.21\\ $

i don't know if i interpreted the question correctly .

I look for a short and simple way.

I have studied maths upto $12$th grade. Thanks.

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    $\begingroup$ HInt: The whole idea is to notice that $\big(9-4\sqrt5\big)\big(9+4\sqrt5\big)=1$. $\endgroup$ – Lucian Jun 28 '15 at 18:40
  • $\begingroup$ i didnt understand ur hint.but it looks useful $\endgroup$ – R K Jun 28 '15 at 18:42
  • $\begingroup$ $$ x_n = \left( 9 + \sqrt {80}\right)^n + \left( \frac{1}{ 9 + \sqrt {80}}\right)^n $$ is always an integer, also obeys a degree two linear recurrence of roughly Fibonacci type, $x_{n+2} = 18 x_{n+1} - x_n$ $\endgroup$ – Will Jagy Jun 28 '15 at 18:54
  • $\begingroup$ note $$ \left( \frac{1}{ 9 + \sqrt {80}}\right) = 9 - \sqrt {80} $$ and $\sqrt {80} = 4 \sqrt 5$ $\endgroup$ – Will Jagy Jun 28 '15 at 19:02
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Consider $$P=(9+4\sqrt 5)^{48}+(9-4\sqrt 5)^{48}.$$

Note that $P$ is an integer.

Now we have $0\lt 9-4\sqrt 5\lt 1$. Hence we have $$0\lt (9-4\sqrt 5)^{48}\lt 1.$$ Hence, we have $$x=(9+4\sqrt 5)^{48}=P-1+1-(9-4\sqrt 5)^{48}.$$ This implies that $f=1-(9-4\sqrt 5)^{48}$.

Thus, we have $$x(1-f)=(9+4\sqrt 5)^{48}(9-4\sqrt 5)^{48}=1.$$

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  • $\begingroup$ I deleted my answer once because I thought $(9+4\sqrt 5)^{48}-(9-4\sqrt 5)^{48}$ is an integer... and I saw Will Jagy's comment...so CW:) $\endgroup$ – mathlove Jun 28 '15 at 19:14
  • $\begingroup$ Meanwhile, the numbers $2,18,322,5778$ are the first entries in these imprimitive $x^2 - 5 y^2 = 4,$ given a solution $(x,y)$ a new one comes from $(9x+20y, 4x+9y).$ So, $(2,0),$ $(18,8),$ $(322,144),$ $(5778, 2584),$ $\endgroup$ – Will Jagy Jun 28 '15 at 19:29
  • $\begingroup$ dividing by $2$ gives $1,9,161, 2889,$ and all positive $x$ in $x^2-5y^2 = 1, $ but that would throw off the original problem. $\endgroup$ – Will Jagy Jun 28 '15 at 19:43

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