Question

Shew that the integral part of $(8+3\sqrt{7}{)}^n$ is odd, if $n$ be a positive integer.

Answer 1

suppose for $(8+3\sqrt{7}{)}^n$

I = integral part

f = fractional part

$(8+3\sqrt{7}{)}^n=I+f$

$0<f<1$

$8-3\sqrt{7}<1$

$\therefore (8-3\sqrt{7}{)}^n=f^{\prime }$ a proper fraction

adding both equation

$(8+3\sqrt{7}{)}^n+(8-3\sqrt{7}{)}^n=I+f+f^{\prime }$

$[C_0{8}^n.(3\sqrt{7}{)}^0+C_1{8}^{n-1}.(3\sqrt{7}{)}^1+.......C_n{8}^0.(3\sqrt{7}{)}^n]+[C_0{8}^n.(3\sqrt{7}{)}^0-C_1{8}^{n-1}.(3\sqrt{7}{)}^1+.......C_n{8}^0.\left(3\sqrt{7}{)}^n\right]$

$=I+f+f^{\prime }$

$f+f^{\prime }=1$

let us assume n = even

$2.[C_0{8}^n.(3\sqrt{7}{)}^0+C_1{8}^{n-2}.(3\sqrt{7}{)}^2+.......C_n{8}^0.(3\sqrt{7}{)}^n]=I+1$

$2.[C_0{8}^n.(3\sqrt{7}{)}^0+C_1{8}^{n-2}.(3\sqrt{7}{)}^2+.......C_n{8}^0.(3\sqrt{7}{)}^n]=even$

$\therefore I=even-1=odd$