Question

If \[a\] and \[b\] are natural numbers such that \[{a^2} - {b^2}\] is a prime number, then find the value of \[{a^2} - {b^2}\].
A. \[a + b\]
B. \[a - b\]
C. \[ab\]
D. 1

Answer 1

Hint First we apply the formula \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\] in the expression \[{a^2} - {b^2}\]. We know that the prime number is divisible by itself or 1. By Using this condition we determine the value of \[{a^2} - {b^2}\].

Formula used
The identity formula is \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\].

Complete step by step solution
 Given that \[{a^2} - {b^2}\] is a prime number.
Apply the formula \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\]
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Since \[{a^2} - {b^2}\] is a prime number so either \[\left( {a + b} \right) = 1\] or \[\left( {a - b} \right) = 1\].
We know that the sum of natural numbers is natural and the sum must be greater than 1. For example : the sum of two 1 is 2 which is greater than 1.
Since it is given that, \[a\] and \[b\] are natural numbers. So it is not possible that the sum of \[a\] and \[b\]is 1.
Thus \[\left( {a + b} \right) \ne 1\].
So, \[\left( {a - b} \right) = 1\].
Putting \[\left( {a - b} \right) = 1\] in the equation \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].
\[{a^2} - {b^2} = \left( {a + b} \right) \cdot 1\]
\[ \Rightarrow {a^2} - {b^2} = \left( {a + b} \right)\]
Hence option A is the correct option.

Note: Some students put \[\left( {a + b} \right) = 1\] . Thus they get the solution \[{a^2} - {b^2} = a - b\] which is incorrect. They skip the condition that \[a\] and \[b\] are natural numbers. It is not possible that the sum of \[a\] and \[b\]. The correct solution is \[{a^2} - {b^2} = a + b\].