If $(8+3\sqrt{7})^n = I + F $ where $ I $ is integer and $F $ is proper fraction, what is $I$?

Question

I am beginner to Binomial Theorem and I want to find out weather $I$ is even or odd in $$(8+3\sqrt{7})^n= I+F$$ if it can be expressed as a sum of an Integer $I$ and a proper fraction $F$

How could I find out ?

Answer 1

let set $a=8+3\sqrt{7}\quad$ and $\quad\bar a=8-3\sqrt{7}$

From $(a-8)^2=63$ you deduce $a,\bar a$ are roots of $x^2-16x+1=0$.

Which would be the characteristic equation of the linear recurrence relation

$\begin{cases} f(0)=2\\f(1)=16\\f(n)=16f(n-1)-f(n-2)\end{cases}\quad$ where $f(n)=a^n+{\bar a}^n$

Now remark that $\bar a\approx 0.0627$ thus $\bar a^n\to 0$ extremely quickly.

So $f(n)\sim a^n$.

To be more precise since $\bar a>0$ and since $f(n)$ is an integer number, we have $\lfloor a^n\rfloor=I=f(n)-1$

Since $f(n)$ is even (by induction we only multiply and subtract even terms), so $I$ is odd.

Answer 2

Notice that $$\begin{align} (8 +3 \sqrt{7})^n + (8 - 3 \sqrt{7})^n &= \sum_{i=0}^n \binom{n}{i} 8 ^i (3 \sqrt{7})^{n-i} + \sum_{i=0}^n \binom{n}{i} (-1)^i 8 ^i (3 \sqrt{7})^{n-i} \\ &= 2 \sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i} 8^{2i} 3^{2(n-i)} 7^{n-i} \qquad \text{(the odd numbered terms cancel out)}\\ &= 2 N \end{align}$$ where $N$ is a positive integer. Since $8 - 3 \sqrt{7} \approx 0.06$, we have $$(8 +3 \sqrt{7})^n + \epsilon = 2N $$ where $0 < \epsilon < 1$.

So the integer part of $(8 +3 \sqrt{7})^n$ is odd.