If ` (7 + 4 sqrt(3))^(n) = s + t ` , where n and s are
positive integers and t is a proper fraction , show that
(1 - t) (s + t) = 1

Explanation

`(7 + 4 sqrt(3))^(n)` can be written as ` (7 + sqrt(48))^(n)`
` therefore s + t = (7 + sqrt(48))^(n) ` …(i)
` 0 lt t lt 1 ` …(ii)
Now , let ` t' = (7- sqrt(48))^(n)` …(iii)
` 0 lt t' lt 1 ` …(iv)
On adding Eqs. (i) and (iii) , we get
` x + t + t' = ( 7 + sqrt(48))^(n) + (7- sqrt(48))^(n)`
` s + 1 = 2p , AA p in ` N = Even integer [ from theorem 2]
` therefore t + t' = 1 " or " 1 - t = t' `
Then , ` (1 - t) (x + t) = t'(s + t) = (7 - sqrt(48))^(n) (7 + sqrt(48))^(n) " " ` [ from Eqs. (i) and (iii) ]
`= ( 49 - 48)^(n) = (1)^(n) = 1`