Question

If $a$ and $b$ are natural numbers such that $a^2-b^2$ is 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 $a+b$

Consider $a^2-b^2=(a+b)(a-b)$ is a prime (given).
$\therefore a^2-b^2$ is either divisible by lor the number itself.
$\therefore a-b=1$ and $a^2-b^2=a+b$ $(\because a+b\ne 1$ as $a,b\in N)$