Question

Show that the integral part of $(5+2\sqrt{6}{)}^n$ is odd, if $n$ be a positive integer.

Answer 1

Assuming that $I$ denotes the integral part and $f$ denotes the functional part of $(5+2\sqrt{6}{)}^n$. Then,

$I+f=5^n+C_1{3}^{n-1}\sqrt{6}+C_2{3}^{n-2}\cdot 6+C_3{3}^{n-3}(\sqrt{6}{)}^3+\dots \dots \dots (1)$

Considering $5-2\sqrt{6}<1$ and is also positive, therefore $(5-2\sqrt{6}{)}^n$ is a proper fraction with out an integral part.

$\therefore f^{\prime }=(5-2\sqrt{6}{)}^n=5^n-C_1{3}^{n-1}\sqrt{6}+C_2{3}^{n-2}\cdot 6-C_3{3}^{n-3}(\sqrt{6}{)}^3+\dots \dots \dots \left(2\right)$

Adding (1) and (2), we get

$I+f+f^{\prime }=2(5^n+C_2{3}^{n-2}\cdot 6+\dots \dots )=\text{An even Integer}$

but since $f$ and $f^{\prime }$ are proper integer their sum must be 1 for $I+f+f^{\prime }$ to be an integer.

Therefore $I$ is an odd integer