If ${\left(7+4\sqrt{3}\right)}^{\mathrm{n}}=\mathrm{I}+\text{⨍ }$ where I is an integer and $0<\text{⨍ <1,}$ then which of the following is true? 

I) I is an even integer

 II)$\left(\mathrm{I}+\text{⨍ }\right)\left(1-\text{⨍ }\right)=1$

$$\begin{gathered} \mathrm{I}+f={(7+4\sqrt{3})}^{\mathrm{n}}, 0<f<1 \\ \mathrm{Let} \mathrm{F}={(7-4\sqrt{3})}^{\mathrm{n}}, 0<\mathrm{F}<1 \\ \mathrm{I}+f+\mathrm{F}={(7+4\sqrt{3})}^{\mathrm{n}}+{(7-4\sqrt{3})}^{\mathrm{n}}=\mathrm{Even} \mathrm{Integer} \end{gathered}$$

$f+\mathrm{F}$ must be van integer.

$0<f+\mathrm{F}<2 \Rightarrow f+\mathrm{F}=1$

$\therefore$ I=Even Integer-1=Odd Integer

$$\begin{gathered} (\mathrm{I}+f) (\mathrm{I}-\mathrm{f})=(\mathrm{I}+f) \left(\mathrm{F}\right)={(7+4\sqrt{3})}^{\mathrm{n}} {\left(7-4\sqrt{3}\right)}^{\mathrm{n}} \\ =1 \end{gathered}$$