If $p = (8+3\sqrt7)^m$ and $f= p - \lfloor p\rfloor$ where $\lfloor p \rfloor $ denotes the greatest integer function, then the value of $p(1-f)$ is equal to:
a)$ 1$
b)$2$
c) $2^{n}$
d) $2^{2n}$
I have written the binomial expansion of $8+3\sqrt7$ but that doesn't lead me anywhere. How do I attempt this problem? Any hints?
Given that $p = (8+3\sqrt 7)^m$ and $f = p - \lfloor p\rfloor$. Let $q = (8-3\sqrt{7})^m$. Now $pq = 1$ as $(8+3\sqrt 7)^m(8-3\sqrt7)^m = 1^m=1$. So $0<q < 1$.
Now using binomial theorem $p+q \in \Bbb Z$ as all square root terms get cancelled.
From here, $p+q = \lfloor p\rfloor + f + q$. Now since $f\in [0,1)$ and $q \in (0,1)$, their sum belongs in $(0,2)$. And since $p+q \in \Bbb Z$ so only possible value for $f+q$ is $1$ or $q = 1-f$.
Thus $p(1-f) = pq = 1.$