A probability problem by harsh poddar

This is quite a popular problem.

By binomial expansion we have:

$(7+ 4\sqrt3)^n = \displaystyle \sum_{r=0}^{n} {n\choose r} (7) ^{n-r} (4\sqrt3) ^{r} = p+q$

for $p \in I, q \in (0,1)$

and

$(7 - 4\sqrt3)^n = \displaystyle \sum_{r=0}^{n} {n\choose r} (7)^{n-r} (-4\sqrt3)^r = t$

Now,

$0<7-4\sqrt3<1$

$\displaystyle \Rightarrow 0< (7-4\sqrt3)^n<1$

$\displaystyle \Rightarrow t \in (0,1)$

Adding the two expansions, we get:

$\big( 7+4\sqrt3\big)^n + \big(7-4\sqrt3\big)^n = 2\big( {n\choose 0} (7)^n\big) + (0) + 2\big( {n\choose2} (7)^{n-2} (4\sqrt3)^2\big) + (0) + \ldots$

All the terms where $r$ is odd get cancelled and only even powers of $(4\sqrt3)$ remain, i.e., the RHS of the equation is an integer.

$\displaystyle \Rightarrow p + q + t = integer$

For $q, t \in (0,1) ; 0<q+t<2$

And since the RHS and $p$ are integers, $q+t$ is also an integer. We conclude $q + t = 1$ since $0< q + t < 2$ .

$\displaystyle \Rightarrow (1-q) = t = (7-4\sqrt3)^n$

$\displaystyle \Rightarrow (1-q)(p+q) =\{ (7-4\sqrt3)(7+4\sqrt3) \} ^{n} =(49 - 16(3))^n = \boxed{1}$

Note : Some problems might even ask to comment whether $p$ is divisible by 2 or 4 or any multiple of 2, also what is the remainder when $p$ is divided by 2 and other similar questions. It is now trivial to prove that $p$ is odd. I'll leave the proof as an exercise.