Show that the integer next above `(sqrt(3)+1)^(2m)` contains `2^(m+1)` as a factor.

Explanation

Let `(sqrt(3)+1)^(2m) = I + f`, where I is the integral and f the fractional part.
`:. 0 lt f lt 1`
Also, let `f' = (sqrt(3) - 1)^(2m)`. Now
`I + f + f' = (sqrt(3) + 1)^(2m) + (sqrt(3) - 1)^(2m)`
`= (4+2sqrt(3))m + (4-2sqrt(3))m`
`= 2^(m+1)[2^(m)+.^(m)C_(2)2^(m-2)(sqrt(3))^(2)+"....."]`
or `I + f + f' =` even integer
`rArr f + f'` is an integer
`:' 0 lt f lt 1` and `0 lt f' lt 1`
`:. 0 lt f + f' lt 2` ltbr gt Thus, `f + f'` is an integer between 0 and 2. ltbr `:. f + f' = 1`
Hence, from `(1)`, we conclude that `I + f + f'` is an integer next above `(sqrt(3)+1)^(2m)` and it contains `2^(m+1)` as a factor.