Let a and b be positive integers and b not a perfect square, then for every positive integer n, the number $(\mathrm{a}+\sqrt{\mathrm{b}}{)}^{\mathrm{n}}$ is irrational. Also,

$\begin{array}{rl} & (\mathrm{a}+\sqrt{\mathrm{b}}{)}^{\mathrm{n}}+(\mathrm{a}-\sqrt{\mathrm{b}}{)}^{\mathrm{n}}\\ & =2{[}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}{+}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}\left(\sqrt{\mathrm{b}}{)}^2{+}^{\mathrm{n}}{\mathrm{C}}_4{\mathrm{a}}^{\mathrm{n}-4}\right(\sqrt{\mathrm{b}}{)}^4+\dots .]\end{array}$

Clearly, R.H.S. is an even integer say E.

Let $(\mathrm{a}+\sqrt{\mathrm{b}}{)}^{\mathrm{n}}=\mathrm{I}+\mathrm{F}$ where I is the integral part and 
F is the fractional part of $(\mathrm{a}+\sqrt{\mathrm{b}}{)}^{\mathrm{n}}\text{i.e.,}0<\mathrm{F}<1$

Let $(\mathrm{a}-\sqrt{\mathrm{b}}{)}^{\mathrm{n}}={\mathrm{F}}^{\mathrm{\prime }},0<{\mathrm{F}}^{\mathrm{\prime }}<1$

Then, $\mathrm{I}+\mathrm{F}+{\mathrm{F}}^{\mathrm{\prime }}=\mathrm{E}$

As, I and E are integers, F + F′ , must be an integer.

But, $0<\mathrm{F}+{\mathrm{F}}^{\mathrm{\prime }}<2\Rightarrow \mathrm{F}+{\mathrm{F}}^{\mathrm{\prime }}=1\Rightarrow \mathrm{I}=\mathrm{E}-1$

Thus, integral part of $(\mathrm{a}+\sqrt{\mathrm{b}}{)}^{\mathrm{n}}$ is The fractional part of $(\mathrm{a}+\sqrt{\mathrm{b}}{)}^{\mathrm{n}} \text{is }1-(\mathrm{a}-\sqrt{\mathrm{b}}{)}^{\mathrm{n}}$