Question

If n be any positive integer, show that the integral part of $(7+4\sqrt{3}{)}^n$ is an odd number . Also if $(7+4\sqrt{3}{)}^n=I+f$ where i is a +ive integer and f is a proper fraction, show that $(1-f)(I+f)$

Answer 1

$(2+\sqrt{3}{)}^n$ = I + f.
or I + f = $2^n+{}^n{C}_1{2}^{n-1}\sqrt{3}+{}^n{C}_2{2}^{n-2}(\sqrt{3}{)}^3+{}^n{C}_3{2}^{n-3}(\sqrt{3}{)}^3+\cdots$
Now, 0 < 2 - $\sqrt{3}$ < 1. $\therefore 0(2-\sqrt{3}{)}^n<1.$
Let $(2-\sqrt{3}{)}^n$ = f' where 0 < f' < 1.
$\therefore f^{\prime }=2^n-{}^n{C}_1{2}^{n-1}\sqrt{3}+{}^n{C}_2{2}^{n-2}\left(\sqrt{3}{)}^2{}^n{C}_3{2}^{n-3}\right(3\sqrt{3}{)}^3+\cdots$
Adding (1) and (2),
I + f + f' = 2$[2^n+{}^n{C}_2{2}^{n-2}.3+.....]$
or I + f + f' = even integer.
Now 0 < f < 1 and 0 < f' < 1.
$\therefore$ 0 < f + f' < 2.
Hence from (3) we conclude that f + f' is an integer between 0 and 2.
$\therefore$ f + f' = 1 $\therefore$ f' = 1 - f.
From (3) and (4) we get I + 1 = even integer.
$\therefore$ I is an odd integer.
Now I + f = $(2+\sqrt{3}{)}^n$, f' = 1 - f = $(2-\sqrt{3}{)}^n$
$\therefore$ (I + f) (1 - f) = [(2 + $\sqrt{3})$ (2 - $\sqrt{3}){]}^n=(4-3{)}^n=1.$
$\therefore$ (I + f) (1 - f) = 1