Question

If $n$ is positive integer, then prove that the integral part of ${(7+4\sqrt{3})}^n$ is an odd number.

Answer 1

Let ${(7+4\sqrt{3})}^n=I+f$ ..... $\left(i\right)$
where $I$ & $f$ are its integral and fractional parts respectively.
It means $0<f<1$
Now, $0<7-4\sqrt{3}<1$
$0<{(7-4\sqrt{3})}^n<1$
Let ${(7-4\sqrt{3})}^n=f^{\prime }$ ...... $\left(ii\right)$
$\Rightarrow$ $0<f^{\prime }<1$
Adding (i) and (ii)
$I+f+f^{\prime }={(7+4\sqrt{3})}^n+{(7-4\sqrt{3})}^n$
$=2[{}^n{\text{C}}_0{7}^n{+}^n{\text{C}}_2{7}^{n-2}{\left(4\sqrt{3}\right)}^2+...]$
$I+f+f^{\prime }=$ even integer $\Rightarrow$ ($f+f^{\prime }$ must be an integer)
$0<f+f^{\prime }<2\Rightarrow f+f^{\prime }=1$
$\Rightarrow I+1=$ even integer
Therefore $I$ is an odd integer.