Question

If $p$ is a prime number greater than 2, then the difference $\left[ {{{\left( {2 + \sqrt 5 } \right)}^p}} \right] - {2^{p + 1}}$, where $\left[ . \right]$ denotes the greatest integer .is divisible by?
A) ${\text{p}} + 1$
B) ${\text{p}} - 1$
C) ${\text{p}} + \sqrt 5$
D) ${\text{p}}$

Answer 1

We are required to find that the given function will be divisible by what prime number. The function consists of the greatest integer function, hence, to solve the question the concept of greatest integer function is needed. We will manipulate the terms of the function using the Binomial expansion and then use the result to find our divisor.
Formula Used: We will use the Binomial expansion formula, ${\left( {a + b} \right)^n} = {}^n{C_0}{a^0}{b^n} + {}^n{C_1}{a^1}{b^{n - 1}} + ........ + {}^n{C_n}{a^n}{b^0}$ to solve the question.

Complete step by step solution:
We are given that ${\text{p}}$ is a prime number greater than 2. That means that ${\text{p}}$ is definitely an odd number. This is because 2 is the only prime number that is even. And since ${\text{p}}$ is a prime number greater than 2, so it has to be odd.
Let us now find the value of ${\left( {\sqrt 5 + 2} \right)^{\text{p}}}$. This is because ${\left( {2 + \sqrt 5 } \right)^{\text{p}}} = {\left( {\sqrt 5 + 2} \right)^{\text{p}}}$ as addition is commutative. To do this we will use the formula for the binomial expansion.
We will substitute $a = \sqrt 5$, $b = 2$ and $n = {\text{p}}$ in the formula for the Binomial expansion.
Hence, our expansion of ${\left( {\sqrt 5 + 2} \right)^{\text{p}}}$ becomes,
${\left( {\sqrt 5 + 2} \right)^{\text{p}}} = {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0}$……….$\left( 1 \right)$
Let us now find the value of ${\left( {\sqrt 5 + 2} \right)^{\text{p}}}$. To do this we will use the formula for the binomial expansion. Again, we will substitute $a = \sqrt 5$, $b = - 2$ and $n = {\text{p}}$ in the formula for the Binomial expansion.
Let us now find the value of ${\left( {\sqrt 5 - 2} \right)^{\text{p}}}$. To do this we will use the formula for the binomial expansion. Make the following substitutions in the formula for the Binomial expansion.
a = \sqrt 5 \\\ b = - 2 \\\ n = {\text{p}} \\\
Hence, our expansion of ${\left( {\sqrt 5 - 2} \right)^p}$ becomes,
${\left( {\sqrt 5 - 2} \right)^{\text{p}}} = {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{\left( { - 2} \right)^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{\left( { - 2} \right)^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{\left( { - 2} \right)^0}$
$\Rightarrow {\left( {\sqrt 5 - 2} \right)^{\text{p}}} = - {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0}$…………..$\left( 2 \right)$
In the above equation, we can note that the alternate terms are negative and positive. This is because ${\text{p}}$ is an odd number. So, any negative term raised to power ${\text{p}}$ i.e., a negative number raised to a negative power will be negative. Now since ${\text{p}}$ is odd so ${\text{p}} - 1$ will be even as it is its subsequent term. So, any negative number raised to ${\text{p}} - 1$ or a positive power will be positive. Due to this, we get alternate positive and negative terms.
Now, subtracting equation $\left( 2 \right)$ from equation $\left( 1 \right)$, we get,
{\left( {\sqrt 5 + 2} \right)^{\text{p}}} - {\left( {\sqrt 5 - 2} \right)^{\text{p}}} = {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0} - \left\\{ - {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0} \right\\} \\\ \Rightarrow {\left( {\sqrt 5 + 2} \right)^{\text{p}}} - {\left( {\sqrt 5 - 2} \right)^{\text{p}}} = {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0} + {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} - {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ - {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0}
On gathering the like terms together and then solving the expression, we get,
${\left( {\sqrt 5 + 2} \right)^{\text{p}}} - {\left( {\sqrt 5 - 2} \right)^{\text{p}}} = {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} - \\\ {}^{\text{p}}{C_1}{\left( {\sqrt 5 } \right)^1}{2^{{\text{p}} - 1}} + ........ + {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0} - {}^{\text{p}}{C_{\text{p}}}{\left( {\sqrt 5 } \right)^{\text{p}}}{2^0} \\\ {\Rightarrow \left( {\sqrt 5 + 2} \right)^{\text{p}}} - {\left( {\sqrt 5 - 2} \right)^{\text{p}}} = 2{}^{\text{p}}{C_0}{\left( {\sqrt 5 } \right)^0}{2^{\text{p}}} + 2{}^{\text{p}}{C_2}{\left( {\sqrt 5 } \right)^2}{2^{{\text{p}} - 2}} + ........ + 2{}^{\text{p}}{C_{{\text{p}} - 1}}{\left( {\sqrt 5 } \right)^{{\text{p}} - 1}}{2^1}$
On taking out 2 common from all the terms, we get,
\Rightarrow {\left( {\sqrt 5 + 2} \right)^{\text{p}}} - {\left( {\sqrt 5 - 2} \right)^{\text{p}}} = 2\left\\{ {{}^{\text{p}}{C_0}{{\left( {\sqrt 5 } \right)}^0}{2^{\text{p}}} + {}^{\text{p}}{C_2}{{\left( {\sqrt 5 } \right)}^2}{2^{{\text{p}} - 2}} + ........ + {}^{\text{p}}{C_{{\text{p}} - 1}}{{\left( {\sqrt 5 } \right)}^{{\text{p}} - 1}}{2^1}} \right\\}…………$\left( 3 \right)$ Now we know that $\left( {\sqrt 5 - 2} \right)$ will always lie between 0 and 1. This is because,
$\begin{array}{c}\left( {\sqrt 5 - 2} \right) = 2.23 - 2\\\ = 0.23\end{array}$
Thus, $0 < \left( {\sqrt 5 - 2} \right) < 1$.
Hence, any power of $\left( {\sqrt 5 - 2} \right)$ will also lie between 0 and 1.
So, the greatest integer function of $\left( {\sqrt 5 - 2} \right)$ will always be 0.
On substituting 0 for ${\left( {\sqrt 5 - 2} \right)^{\text{p}}}$ in equation $\left( 3 \right)$, we get,
{\left( {\sqrt 5 + 2} \right)^{\text{p}}} = 2\left\\{ {{}^{\text{p}}{C_0}{{\left( {\sqrt 5 } \right)}^0}{2^{\text{p}}} + {}^{\text{p}}{C_2}{{\left( {\sqrt 5 } \right)}^2}{2^{{\text{p}} - 2}} + ........ + {}^{\text{p}}{C_{{\text{p}} - 1}}{{\left( {\sqrt 5 } \right)}^{{\text{p}} - 1}}{2^1}} \right\\}
{\left( {\sqrt 5 + 2} \right)^{\text{p}}} = {2^{{\text{p}} + 1}} + 2\left\\{ {{}^{\text{p}}{C_2}{{\left( {\sqrt 5 } \right)}^2}{2^{{\text{p}} - 2}} + ........ + {}^{\text{p}}{C_{{\text{p}} - 1}}{{\left( {\sqrt 5 } \right)}^{{\text{p}} - 1}}{2^1}} \right\\}……….$\left( 4 \right)$
Now every term on the RHS of the equation $\left( 4 \right)$ is a factor of ${\text{p}}$. Hence, $\left[ {{{\left( {2 + \sqrt 5 } \right)}^{\text{p}}}} \right] - {2^{{\text{p}} + 1}}$ will be divisible by ${\text{p}}$.

Since our answer matches with option (D) out of all the provided choices, hence, option (D) is the correct answer.

Note:
The greatest integer function is a piecewise function which produces the greatest integer less than or equal to the number, operated upon. When the number is an integer, we use that value. If the number is not an integer, we use the next smaller value.
We will consider $\left( {\sqrt 5 - 2} \right)$ for our manipulation rather than $\left( {2 - \sqrt 5 } \right)$. This is because when we consider $\left( {\sqrt 5 - 2} \right)$, then only we get our desired greatest integer, hence, the desired function. If we will consider $\left( {2 - \sqrt 5 } \right)$ instead of $\left( {\sqrt 5 - 2} \right)$, then the range will change to $\left[ { - 1,0} \right]$ from $\left[ {0,1} \right]$. Due to this, the greatest integer for $\left( {2 - \sqrt 5 } \right)$ will become $- 1$. So, we will not get our required function from manipulations.