Question

If $p=(8+3\sqrt{7}{)}^n$ and $f=p-\left[p\right]$, then
(where [.] denotes the greatest integer function)

• A. $p^{1/n}-(1-f{)}^{1/n}=6\sqrt{7}$
• B. $p(1-f)=4$
• C. $p(1-f)=1$
• D. $p^{1/n}+(1-f{)}^{1/n}=16$

Answer 1

Answer: (A), (C), (D)

$p^{1/n}+(1-f{)}^{1/n}=16$

Given $p=(8+3\sqrt{7}{)}^n$
Assuming $p_1=(8-3\sqrt{7}{)}^n$
Clearly, $0<p_1<1$
Now,
$p+p_1=(8+3\sqrt{7}{)}^n+(8-3\sqrt{7}{)}^n=2[{}^n{C}_0{8}^n+{}^n{C}_2{8}^{n-2}(3\sqrt{7}{)}^2+\dots ]=2k,k\in Z$
Given
$p=\left[p\right]+f\Rightarrow \left[p\right]+f+p_1=2k$
$\Rightarrow f+p_1=2k-\left[p\right]\cdots \left(1\right)$
Which is integer
We know that,
$p_1\in (0,1),f\in (0,1)\Rightarrow 0<p_1+f<2$
Using equation $\left(1\right)$, the only interger possible
$\Rightarrow f+p_1=1\Rightarrow p_1=1-f$

Therefore,
$p(1-f)=p{p}_1=\left[\right(8+3\sqrt{7}{)}^n(8-3\sqrt{7}{)}^n]=(64-9\times 7{)}^n=1$
and
$p^{1/n}-(1-f{)}^{1/n}=p^{1/n}-p_1^{1/n}\Rightarrow (8+3\sqrt{7})-(8-3\sqrt{7})\Rightarrow 6\sqrt{7}$
and
$p^{1/n}+(1-f{)}^{1/n}=p^{1/n}+p_1^{1/n}\Rightarrow (8+3\sqrt{7})+(8-3\sqrt{7})\Rightarrow 16$