Question:
If $p_{1}$ and $p_{2}$ are two odd prime numbers such that $p_{1}>p_{2}$, then $p_{1}^{2}-p_{2}^{2}$ is
(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number
Solution:
Let the two odd prime numbers $p_{1}$ and $p_{2}$ be 5 and 3 .
Then,
$p_{1}^{2}=5^{2}$
$=25$
And
$p_{2}^{2}=3^{2}$
= 9
Thus,
p_{1}{ }^{2}-p_{2}{ }^{2}=25-9
= 16
16 is even number.
Take another example, with $p_{1}$ and $p_{2}$ be 11 and 7 .
Then,
$p_{1}^{2}=11^{2}$
= 121
And
$p_{2}{ }^{2}=7^{2}$
= 496
Thus,
$p_{1}{ }^{2}-p_{2}{ }^{2}=121-49$
=72
72 is even number.
Thus, we can say that $p_{1}{ }^{2}-p_{2}{ }^{2}$ is even number
In general the square of odd prime number is odd. Hence the difference of square of two prime numbers is odd
Hence the correct choice is (a).