Question

Show that the integer next above $(\sqrt{3}+1{)}^{2m}$ contains $2^{m+1}$ as a factor

Answer 1

Let $(\sqrt{3}+1{)}^{2m}$ = I + f
and $(\sqrt{3}-1{)}^{2m}$ = f' where 0 < f' < 1
and 0 < f' < 1 and I is an integer.
Thus I + f + f' = $(\sqrt{3}+1{)}^{2m}+(\sqrt{3}-1{)}^{2m}$
= $(4+2\sqrt{3}{)}^m+(4-2\sqrt{3}{)}^m$
= $2^m\left[\right(2+\sqrt{3}{)}^m+(2-\sqrt{3}{)}^m]$
= $2^{m+1}[2^m+{}^m{C}_2{.2}^{m-2}.(\sqrt{3}{)}^2+.....]$
Now as in part (a) f + f' = 1 and so I + f + f' is an integer next above $(\sqrt{3}+1{)}^{2m}$ which by (1) contains $2^{m+1}$ as a factor.