Question

The \[{{x}^{n}}-{{y}^{n}}\] is divisible by x – y when n is an even positive integer.
(a) True
(b) False

Answer 1

Hint: First of all consider the statement that \[P\left( n \right)={{x}^{n}}-{{y}^{n}}\] is divisible by (x – y). Now, substitute n – 1 and prove that P(1) is divisible by (x – y). Now, assume the stamen for n = k and using it, prove the statement for n = k + 1. In case, if it is true for P (k + 1), it would be true for all natural numbers. Finally, use this result to answer the question.

Complete step-by-step answer:
Here, we have to tell if \[{{x}^{n}}-{{y}^{n}}\] is divisible by (x – y) when n is even or not. Let us take the statement that \[P\left( n \right)={{x}^{n}}-{{y}^{n}}\] is divisible by (x – y). By substituting n = 1 in P(n), we get, \[P\left( n \right)={{x}^{1}}-{{y}^{1}}=\left( x-y \right)\] which is divisible by (x – y).
Let us assume that the result is true for n = k, that is \[P\left( k \right)={{x}^{k}}-{{y}^{k}}\] is divisible by (x – y). As \[\left( {{x}^{k}}-{{y}^{k}} \right)\] is divisible by (x – y), so we get,
\[\dfrac{\left( {{x}^{k}}-{{y}^{k}} \right)}{\left( x-y \right)}=\text{integer}=c\]
So, we get,
\[\left( {{x}^{k}}-{{y}^{k}} \right)=c\left( x-y \right)\]
\[{{x}^{k}}={{y}^{k}}+c\left( x-y \right)....\left( i \right)\]
Now, let us prove that P (k + 1) is divisible by (x – y) as well.
\[P\left( k+1 \right)={{x}^{k+1}}-{{y}^{k+1}}\]
We know that \[{{a}^{x+y}}={{a}^{x}}.{{a}^{y}}\]. By using this, we get,
\[P\left( k+1 \right)={{x}^{k}}.x-{{y}^{k}}.y\]
By substituting the value of \[{{x}^{k}}\] from equation (i), we get,
\[P\left( k+1 \right)=\left( {{y}^{k}}+c\left( x-y \right) \right)x-{{y}^{k}}.y\]
\[P\left( k+1 \right)=x{{y}^{k}}+cx\left( x-y \right)-{{y}^{k}}.y\]
\[P\left( k+1 \right)={{y}^{k}}\left( x-y \right)+cx\left( x-y \right)\]
By taking out (x – y) common, we get,
\[P\left( k+1 \right)=\left( x-y \right)\left( {{y}^{k}}+cx \right)\]
As (x – y) is a factor of P (k + 1), so P (k + 1) is also divisible by (x – y).
Hence, by the principle of Mathematical Induction, we can say that \[P\left( n \right)={{x}^{n}}-{{y}^{n}}\] is divisible by (x – y) for all natural numbers.
As we have proved that \[{{x}^{n}}-{{y}^{n}}\] is divisible by (x – y) for all natural numbers. So, \[{{x}^{n}}-{{y}^{n}}\] would also be divisible by (x – y) when n is an even positive integer.
So, the given statement is true. Hence, option (a) is the right answer

Note: In this question, students must note that if it would have been written in the question that \[{{x}^{n}}-{{y}^{n}}\] is divisible by (x – y) “only” where n is an even positive integer, then that statement would be false as it is true for all natural numbers and not just for even numbers but the ‘only’’ word is not there in the statement. So, the given statement is true because natural numbers consist of odd and even positive integers.