Question

If we are given that \[{{p}_{1}}\] and \[{{p}_{2}}\] are two odd prime numbers such that \[{{p}_{1}}>{{p}_{2}}\], then \[p_{1}^{2}-p_{2}^{2}\] is : -
(a) An even number
(b) An odd prime number
(c) An odd number
(d) A prime number

Answer 1

Hint: Apply the algebraic identity: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to simplify the given expression: - \[p_{1}^{2}-p_{2}^{2}\]. Now, check the results when two odd numbers are subtracted and added by taking some examples. If both the subtraction and addition give odd numbers, the given expression will give an odd number. If one of the operations gives an odd number and the other even then the expression will give an even number and if both the operations give even numbers then the expression will give an even number again.

Complete step-by-step solution
Here, we have been provided with two odd prime numbers \[{{p}_{1}}\] and \[{{p}_{2}}\] and we have to check the type of number we will get from the expression: - \[p_{1}^{2}-p_{2}^{2}\].
Using the algebraic identity, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we get,
\[\Rightarrow p_{1}^{2}-p_{2}^{2}=\left( {{p}_{1}}+{{p}_{2}} \right)\left( {{p}_{1}}-{{p}_{2}} \right)\]
Here, \[\left( {{p}_{1}}+{{p}_{2}} \right)\] is the sum of two odd prime numbers and \[\left( {{p}_{1}}-{{p}_{2}} \right)\] is the difference of the same. Let us see the definition of a prime number.
A prime number is a number, which does not have more than 2 factors. The two factors, i.e. the number which can divide them, are 1 and the number itself. For example: - 2, 3, 5, 7, 11, 13, …… and so on. Here, one important thing to note is that except 2 all other prime numbers are odd.
In the above question we have to consider \[{{p}_{1}}\] and \[{{p}_{2}}\] as two odd primes. That means we cannot choose them as 2 because 2 is an even number. So, let us choose \[{{p}_{1}}\] and \[{{p}_{2}}\] as 7 and 5 respectively.
Now, let us check the results of the given expression, \[p_{1}^{2}-p_{2}^{2}\].So, we have,
\[\begin{align}
  & \Rightarrow p_{1}^{2}-p_{2}^{2}=\left( {{p}_{1}}+{{p}_{2}} \right)\left( {{p}_{1}}-{{p}_{2}} \right) \\
 & \Rightarrow p_{1}^{2}-p_{2}^{2}=\left( 7+5 \right)\left( 7-5 \right) \\
 & \Rightarrow p_{1}^{2}-p_{2}^{2}=12\times 2 \\
 & \Rightarrow p_{1}^{2}-p_{2}^{2}=24 \\
\end{align}\]
Here, we have obtained both \[\left( {{p}_{1}}+{{p}_{2}} \right)\] and \[\left( {{p}_{1}}-{{p}_{2}} \right)\] as even numbers therefore the product is also even.
Hence, option (a) is the correct answer.

Note: One may note that we can choose the value of \[{{p}_{1}}\] and \[{{p}_{2}}\] as any odd prime numbers but we have to check the given condition \[{{p}_{1}}>{{p}_{2}}\] carefully, otherwise \[\left( {{p}_{1}}-{{p}_{2}} \right)\] will given a negative number. Remember that both \[{{p}_{1}}\] and \[{{p}_{2}}\] cannot be chosen as 2 because it is an even number. It is important to remember the results of addition or subtraction of even and odd numbers to solve the question without taking examples.