Question

If $(7+4\sqrt{3}{)}^n=p+\beta ,$ where $n$ and $p$ are positive integers, and $\beta$ a proper fraction, show that $(1-\beta )(p+\beta )=1$

Answer 1

given that

$(7+4\sqrt{3}{)}^n=p+\beta$ ...........................(1)

where $\beta$ is a proper fraction $0<\beta <1$

$(7-4\sqrt{3}{)}^n={\beta }^{\prime }$

where $\beta$ is a proper fraction

$0<{\beta }^{\prime }<1$

sum of $\beta$ & ${\beta }^{\prime }$ must be 1.

$\beta +{\beta }^{\prime }=1$

$(7-4\sqrt{3}{)}^n={\beta }^{\prime }=1-\beta$ . .................(2)

multiplying eqn (1) & (2)

$(7+4\sqrt{3}{)}^n\times [(7-4\sqrt{3}{)}^n]=(p+\beta ).(1-\beta )$

$\left[\right(49-48{)}^n]=(p+\beta ).(1-\beta )$

$1=(p+\beta ).(1-\beta )$ n is positive integer

Hence proved