If n be a positive integer, then prove by binomial theorem that the integral part of $(7+4\sqrt3)^n$ is an odd number. I wanted to know, how can i prove this?
If $n$ be a positive integer, then prove by binomial theorem that the integral part of $(7+4\sqrt3)^n$ is an odd number.

I'm working with the hypothesis that "integral part" means "floor".
$$\begin{align}
(7+4\sqrt{3})^n + (7 - 4\sqrt{3})^n &= \sum_{k=0}^n \binom{n}{k}7^{n-k}4^k\sqrt{3}^k + \sum_{k=0}^n \binom{n}{k}7^{n-k}(-1)^k4^k\sqrt{3}^k\\
&= 2\sum_{m=0}^{\lfloor n/2\rfloor} \binom{n}{2m}7^{n-2m}4^{2m}3^m
\end{align}$$
is even. $0 < 7 - 4\sqrt{3} < 1$. Thus the integral part of $(7+4\sqrt{3})^n$ is
$(7+4\sqrt{3})^n + (7-4\sqrt{3})^n - 1$, an odd number.
