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 [.] denotes greatest integer. is divisible by:-
A) \[p + 1\]
B) \[p - 1\]
C) \[p + \sqrt 5 \]
D) \[p\]

Answer 1

Hint: Here we will use the binomial expansion of two terms and the property of greatest integer function i.e. the value of the quantity inside the greatest is less than or equal to that quantity and is rounded off to the nearest integer less than that quantity.
The binomial expansion of general term \[{\left( {a + b} \right)^n}\] is given by:-
\[{\left( {a + b} \right)^n}{ = ^n}{C_0}{\left( a \right)^0}{b^n}{ + ^n}{C_1}{\left( a \right)^1}{b^{n - 1}} + {................^p}{C_p}{\left( a \right)^p}{b^0}\]

Complete step-by-step answer:
It is given that p is a prime number and is greater than 2 this means p is an odd prime number.
Now we know that the binomial expansion of general term \[{\left( {a + b} \right)^n}\]is given by:-
\[{\left( {a + b} \right)^n}{ = ^n}{C_0}{\left( a \right)^0}{b^n}{ + ^n}{C_1}{\left( a \right)^1}{b^{n - 1}} + {................^p}{C_p}{\left( a \right)^p}{b^0}\]
Therefore, the binomial expansion of \[{\left( {\sqrt 5 + 2} \right)^p}\]is given by:-
\[{\left( {\sqrt 5 + 2} \right)^p}{ = ^p}{C_0}{\left( {\sqrt 5 } \right)^0}{2^p}{ + ^p}{C_{^{1}}}{\left( {\sqrt {5} } \right)^1}{2^{p - 1}} + ...{ + ^p}{C_p}{\left( {\sqrt {5} } \right)^p}{2^0}\]………………………. (1)
Also, the binomial expansion of \[{\left( {\sqrt 5 - 2} \right)^p}\]is given by:-
\[{\left( {\sqrt 5 - 2} \right)^p}{ = ^p}{C_0}{\left( {\sqrt 5 } \right)^0}{\left( { - 2} \right)^p}{ + ^p}{C_{^{1}}}{\left( {\sqrt {5} } \right)^1}{\left( { - 2} \right)^{p - 1}} + ...{ + ^p}{C_p}{\left( {\sqrt {5} } \right)^p}{\left( { - 2} \right)^0}\]
Now since p is an odd number therefore,
\[{\left( {\sqrt 5 - 2} \right)^p} = { - ^p}{C_0}{\left( {\sqrt 5 } \right)^0}{\left( 2 \right)^p}{ - ^p}{C_{^{1}}}{\left( {\sqrt {5} } \right)^1}{\left( 2 \right)^{p - 1}} - ...{ - ^p}{C_p}{\left( {\sqrt {5} } \right)^p}{\left( 2 \right)^0}\]………………………. (2)
Subtracting equation 2 from 1 we get:-
\[{\left( {\sqrt 5 + 2} \right)^p} - {\left( {\sqrt 5 - 2} \right)^p} = 2\left\{ {^p{C_0}{{\left( {\sqrt 5 } \right)}^0}{2^p}{ + ^p}{C_{^{1}}}{{\left( {\sqrt {5} } \right)}^1}{2^{p - 1}} + ...{ + ^p}{C_p}{{\left( {\sqrt {5} } \right)}^p}{2^0}} \right\}\]
Now we know that \[\left( {\sqrt 5 - 2} \right)\]always lies between 0 and 1, any power of it will also lie in between 0 and 1 and that is why the greatest integer function of it will always be zero.
Hence, on taking the greatest integer of the above equation we get:-
\[\left[ {{{\left( {\sqrt 5 + 2} \right)}^p} - {{\left( {\sqrt 5 - 2} \right)}^p}} \right] = 2\left[ {^p{C_0}{{\left( {\sqrt 5 } \right)}^0}{2^p}{ + ^p}{C_{^{1}}}{{\left( {\sqrt {5} } \right)}^1}{2^{p - 1}} + ...{ + ^p}{C_p}{{\left( {\sqrt {5} } \right)}^p}{2^0}} \right]\]
Since \[\left[ {{{\left( {\sqrt 5 - 2} \right)}^p}} \right] = 0\]
Hence, putting the value we get:-
\[\left[ {{{\left( {\sqrt 5 + 2} \right)}^p} - 0} \right] = 2\left[ {^p{C_0}{{\left( {\sqrt 5 } \right)}^0}{2^p}{ + ^p}{C_{^{1}}}{{\left( {\sqrt {5} } \right)}^1}{2^{p - 1}} + ...{ + ^p}{C_p}{{\left( {\sqrt {5} } \right)}^p}{2^0}} \right]\]
\[ \Rightarrow \left[ {{{\left( {\sqrt 5 + 2} \right)}^p}} \right] = 2\left[ {{2^p}{ + ^p}{C_{^{1}}}{{\left( {\sqrt {5} } \right)}^1}{2^{p - 1}} + ...{ + ^p}{C_p}{{\left( {\sqrt {5} } \right)}^p}{2^0}} \right]\]
Multiplying 2 inside the bracket we get:-
\[ \Rightarrow \left[ {{{\left( {\sqrt 5 + 2} \right)}^p}} \right] = {2^{p + 1}} + \left[ {2\left( {^p{C_{^{1}}}{{\left( {\sqrt {5} } \right)}^1}{2^p} + ...{ + ^p}{C_p}{{\left( {\sqrt {5} } \right)}^p}{2^0}} \right)} \right]\]
Simplifying it further we get:-
\[\left[ {{{\left( {\sqrt 5 + 2} \right)}^p}} \right] - {2^{p + 1}} = \left[ {2\left( {^p{C_{^{1}}}{{\left( {\sqrt {5} } \right)}^1}{2^p} + ...{ + ^p}{C_p}{{\left( {\sqrt {5} } \right)}^p}{2^0}} \right)} \right]\]
Now since every term on right hand side has p as a factor therefore, it is divisible by p
Therefore,
\[\left[ {{{\left( {\sqrt 5 + 2} \right)}^p}} \right] - {2^{p + 1}}\] is divisible by p

Therefore, option D is correct.

Note: Here the important fact is that the greatest integer function rounds off the value of the quantity inside it to an integer less or equal to it. It rounds off the number to the nearest integer.