Question

If $n$ be a positive integer and $(7+4\sqrt{3}{)}^n=p+\beta$, where $p$ is a positive integer and $\beta$ is a proper fraction, then

• A. $(1-\beta )(p+\beta )=2$
• B. $(1-\beta {)}^{1/n}+(p+\beta {)}^{1/n}=14$
• C. $(1-\beta )(p+\beta )=1$
• D. $(1-\beta {)}^{1/n}+(p+\beta {)}^{1/n}=8\sqrt{3}$

Answer 1

Answer: (B), (C)

$(1-\beta )(p+\beta )=1$

Given
$p+\beta =(7+4\sqrt{3}{)}^n$
$0\le \beta <1$
Assuming
$f=(7-4\sqrt{3}{)}^n$
Clearly,
$0<f<1$
Now,
$p+\beta +f=(7-4\sqrt{3}{)}^n+(7+4\sqrt{3}{)}^n=2[7^n{+}^n{C}_2\cdot 7^{n-2}\cdot (4\sqrt{3}{)}^2+\cdots ]=2k,k\in N\Rightarrow \beta +f=2k-p\Rightarrow \beta +f=\text{Integer}$
As $0<\beta +f<2$, so
$\Rightarrow \beta +f=1$

So,
$(p+\beta )(1-\beta )=(p+\beta )\left(f\right)=(7+4\sqrt{3}{)}^n\times (7-4\sqrt{3}{)}^n=(49-16\times 3{)}^n\therefore (p+\beta \left)\right(1-\beta )=1$

$(1-\beta {)}^{1/n}+(p+\beta {)}^{1/n}=f^{1/n}+(p+\beta {)}^{1/n}=(7-4\sqrt{3})+(7+4\sqrt{3})=14$