If ` x= (8 + 3 sqrt(7))^(n)` , where n is a natural
number, power that the integral part of x is an odd
integer and also show that ` x - x^(2) + x[x] = 1 , ` where [.]
denotes the greatest integer function .

Explanation

` (8 + 3 sqrt(7))^(n) ` can be written as ` (8 + sqrt(63))^(n)`
` therefore x = [x] + f `
or ` [x] + f = (8 + sqrt(63 ))^(n)` …(i)
` 0 le f lt 1 ` …(ii)
Now , let ` f' = (8 - sqrt(63))^(n)` …(iii)
` 0 lt f' lt 1 ` …(iv)
On adding Eqs. (i) and (iii) , we get
` [x] + 1 = 2p , AA p in ` N = Even integer [ from theorem 2]
` therefore [x] = 2p - 1 ` = Odd integer
i.e., Integral part of x = Odd integer
`because f + f' = 1 rArr 1 - f f' ` ...(iv)
LHS `= x - x^(2) + x[x] = x - x (x - [x]) = x - xf `
` [ because x = [x] + f]`
` = x (1 - f) - x f' ` [from Eq.s (v)]
` = (8 + sqrt(63))^(n) (8 - sqrt(63))^(n)" " ` [ from Eqs (i) and (iii) ]
` = (8 + sqrt(63))^(n) = (1)^(n) = 1 = RHS ` .