We can solve this by Euclid's division lemma.
Let a be any integer and b equal to 2.
Therefore, by Euclid's division lemma remainder will be 0 or 1.
a=2q +0 (for r=0)
a=2q here q is some natural no.
It is divisible by 2. Therefore it is even no.
a= 2q +1 (for r =1)
It is not divisible by 2. Therefore it is addressed no.
Okay a can also be written as 2q-1.
Product of two odd interfere= (2q+1)(2q-1)
As it is not divisible by 2. It will be a odd no.