Question

If $aandb$ are natural numbers such that $a^2-b^2$ is a prime number, then $a^2-b^2$ is equal to

• A. $a+b$
• B. $a–b$
• C. $ab$
• D. $1$

Answer 1

The correct option is A

$a+b$

Explanation for the correct option:

Finding the value of $\mathbf{(}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{)}\mathbf{}$:
Given $(a^2–b^2)$ is prime.

So $(a^2–b^2)$ is divisible by $1$ or the number itself.

So$a–b=1$

$a+b$ is not equal to $1$, since $a,$and $b$ are natural numbers.

Then $a^2–b^2=a+b$

Hence, option (A) is the correct answer.