Question

If I is the integral part and f is the fractional part of (2 + √3)ⁿ then prove that (I + f)(I - f) = 1. Also prove that I is an odd integer.

Answer 1

Solution -: We will consider that
I + f where I is an integer and 0, f < 1. We have to show that I is odd and that (I + f) (I - f) = 1
1st we will solve this expression by multiplication we get

(2 + √3)ⁿ (2 - √3)ⁿ = (4 - 3)ⁿ = 1 by using ( a-b)(a+b) = a^2 - b^2
or
(2 + √3)ⁿ(2 - √3)ⁿ =1. It is thus required to prove that (2 - √3)ⁿ = I-f
We know that by theorem

(2 + √3)ⁿ + (2 + √3)ⁿ = [2ⁿ - C₁.2^n-1.√3 + C₂2^n-2.(√3)² - ...]

+ [2ⁿ - C₁.2^n-1.√3 + C₂2^n-2.(√3)² - ...]

So we will get an even term

2[2ⁿ - C₂.2^n-2.3 + C₄2^n-4.3² - ...]
Now it is true that
0 < (2 - √3) < 1
Apply power n both sides we get

0 < (2 - √3)ⁿ < 1
So w e can say that

If (2 - √3)ⁿ = f' then I + f + f'
Will be even .
Again
0 < f < 1 and 0 < f' < 1 we get
f + f' = integer
So it will be
f + f' = 1 (·.· 0 < f + f' < 2) as above,
It means
I is odd and f' = I - f which will give

(I + f)(I - f) = 1 which is odd and it is true .