If $(2+\sqrt{3})^n$ has the integral part $I$ and the decimated part $f$ then prove that
$(I+f) (I-f) =1$
I did the following
$(2+\sqrt{3})^n=I+f$
$(2-\sqrt{3})^n=f'$
Thus adding both
We get
$(2+\sqrt{3})^n+(2-\sqrt{3})^n=I+f+f'$
Since the left part is integer by
binomial expansion therefore
$f+f'=1 $- - - - - - - eqn A
If we multiply $(2+\sqrt{3})^n$ and
$(2-\sqrt{3})^n$ then we get $1$
Hence $(I+f) (f') =1$
Thus we get from eqn A we get
$(I+f) (1-f)=1$
But we have to prove that $(I+f) (I-f) =1$
How to do so? Is the question
wrong or I am missing something?