Question

If a and b are two odd primes, show that \[\left( {{a}^{2}}-{{b}^{2}} \right)\] is composite.

Answer 1

Hint: We know that the difference between two odd numbers is equal to an even number. We also know that the sum of two odd numbers are equal to an even number. We know that \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]. We know that the product of two even numbers is an even number. By using this concept, we can solve this problem.

Complete step-by-step answer:
We know that \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]. So, let us assume
\[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right).....(1)\]
From the question, it is clear that a and b are two odd primes.
We know that the difference between two odd numbers is equal to an even number.
So, we can say that \[a-b\] is an even number.
We also know that the sum of two odd numbers are equal to an even number.
So, we can say that \[a+b\] is an even number.
We know that the product of two even numbers is an even number.
So, we can say that the product of \[a-b\] and \[a+b\]is also an even number.
From equation (1), we can say that \[{{a}^{2}}-{{b}^{2}}\] is an even number.
We can also say as both \[a-b\] and \[a+b\]are even numbers, then the product of \[a-b\] and \[a+b\]is a multiple of 2.
We know that a prime number is a number whose factors are 1 and itself.
So, we can say that \[{{a}^{2}}-{{b}^{2}}\] is a composite number.
Hence, we can say that if a and b are two odd primes, show that \[\left( {{a}^{2}}-{{b}^{2}} \right)\] is composite.

Note: Students may have a misconception that the difference between two odd numbers is an equal odd number and the sum of two odd numbers are equal to an odd number. So, it is clear that if this misconception is followed, then we can say that we cannot get a wrong answer. So, if this misconception is followed, then we cannot get the correct answer.