Question

If a and b are any two odd primes, show that $(a^2-b^2)$ is composite.

Answer 1

We know, $a^2-b^2=(a-b)(a+b)$

Since, both $a$ and $b$ are odd prime numbers, both $a+b$ and $a-b$ will be even numbers greater than or equal to $2$

So, $(a-b)(a+b)$ is a multiple of $4$

All multiples of $4$ are composite numbers, so $a^2-b^2$ is a composite number