Question

If $(8+3\sqrt{7}{)}^n=P+F$, where $P$ is an integer and $0<F<1$, then

• A. $P$ is an odd integer
• B. $P$ is an even integer
• C. $F.(P+F)=1$
• D. $(1-F)(P+F)=1$

Answer 1

Answer: (A), (D)

The correct options are
A $(1-F)(P+F)=1$
D $P$ is an odd integer

Consider the expansion of $(8+3\sqrt{7}{)}^n+(8-3\sqrt{7}{)}^n$.
The alternate terms will cancel from both the expansions.
Hence,
$(8+3\sqrt{7}{)}^n+(8-3\sqrt{7}{)}^n=2\left[8^n{+}^n{C}_2\right(8{)}^{n-2}(3\sqrt{7}{)}^2+...]$ (The last term will depend on whether $n$ is odd or even).
Hence, $(8+3\sqrt{7}{)}^n+(8-3\sqrt{7}{)}^n$ is even.
We have,
$0<8-3\sqrt{7}<1$.
Hence,
$0<(8-3\sqrt{7}{)}^n<1$.
$(8+3\sqrt{7}{)}^n=P+F$, where $0<F<1$ and $P$ is an integer.
Hence, $P$ must be odd.
Hence,
$(8-3\sqrt{7}{)}^n=1-F$.
We have,
$(8+3\sqrt{7})\times (8-3\sqrt{7})=1$.
Hence, $(P+F)\times (1-F)=1$.
Hence, options A and D are the correct answers.