Suppose that n is a natural number and I, F are respectively the integral part and fractional part of 
${\left(7+4\sqrt{3}\right)}^n.$ Then show that 
i) I is an odd integer 
$ii) \left(I+F\right)\left(I-F\right)=1$

 

Given $I+F={\left(7+4\sqrt{3}\right)}^n$ where n is a natural number and I, F are respectively the integral part and fractional part of ${\left(7+4\sqrt{3}\right)}^n$.

$0<F<1$$............\left(1\right)$

$Let f={\left(7-4\sqrt{3}\right)}^n$

$36<48<49; 7-6>7-4\sqrt{3}>7-7$

$-6>-4\sqrt{3}>-7; 7-6>7-4\sqrt{3}>7-7$

$0<{\left(7-4\sqrt{3}\right)}^n<1$

$0<f<1.............\left(2\right)$

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

$=2\left(n_{C_0} 7^n+n_{C_2} 7^{n-2}{\left(4\sqrt{3}\right)}^2+.............\right)$

$=2 \left[Integers\right]$

$I+F+f=even integers ........\left(3\right)$

Adding $\left(1\right) and \left(2\right)$

$\Rightarrow 0<F<f<2$

$\Rightarrow F+f=1$$(\therefore F+fisan \mathrm{int}eger)$

$f=1-F$

$From \left(3\right)$

$\left(i\right) I+1=even integer$

$I=even integer -1$

$\therefore$I=odd integer

$ii) \left(I+F\right)\left(I-F\right)=1$

$\left(I+F\right)\left(I-F\right)={\left(7^2-{\left(4\sqrt{3}\right)}^2\right)}^n$

$={\left(49-48\right)}^n=1^n=1$