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Q. If $a$ and $ b$ are positive integers such that $a^2 - b^2$ is a prime number, then $a^2 - b^2$ is

COMEDKCOMEDK 2007Binomial Theorem

Solution:

$a $ and $ b$ are positive integers and $a^2 - b^2$ is a prime number.
Since, $a^2 - b^2 = (a+ b) (a - b)$ product of two numbers.
$\therefore $ Either $a + b = 1$ or $a - b = 1$
Case I : $a + b = 1 \Rightarrow $ Either $a = 0$ or $b = 0$ then $a^2 - b^2 = 1$ or $-1$
which is not a prime number.
$ \therefore $ This case is not possible
Case II : $a - b = 1 \rightarrow a$ and $b$ can be taken anything with $b < a$.
ln this case $a^2 - b^2 = a + b$ is a prime number.
$\therefore \:\:\: a^2 - b^2 = a + b.$