Question

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?

Answer 1

$(2+\sqrt{5}{)}^{96}=I+f$
Let $f_1=(\sqrt{5}-2{)}^{96}$
$I+f+f_1=(\sqrt{5}+2{)}^{96}+(\sqrt{5}-2{)}^{96}$
=2x[]
=even number.
Now,, 0 < f<1
$0<f_1<1$
$\Rightarrow 0<f+f_1<2---\left(1\right)$
Now I+f+f_1 =even number
$\Rightarrow f+f_1\in integer---\left(2\right)$
From (1)and (2), $f+f_1=1$
$\Rightarrow f_1=I-f\therefore (I+f)(I-f)=(2+\sqrt{5}{)}^{96}(\sqrt{5}-2{)}^{96}=1\therefore x(1-f)=1$