Question

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

Answer 1

$a^2-b^2=(a-b)(a+b)$
Since, and b are odd primes.
So let $a=2k+1$
$b=2k^{\prime }+1$
$\therefore a-b=2k+1-2k^{\prime }-1$
$=2k-2k^{\prime }$
$=2(k-k^{\prime })$ is even
$a+b=2k+1+2k^{\prime }+1$
$=2k+2k^{\prime }+2$
$=2(k+k^{\prime }+1)$ is even
$\therefore$ Neither (a-b) nor (a+b) is equal to 1.
$\therefore$ Neither of the two divisors $(a-b)and(a+b)of(a^2-b^2)$ is equal to 1.
$\therefore (a^2-b^2)$ is composite.
since, out of the two divisors of a prime number p, one must be equal to 1