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**Hint:**In this question, we need to show that $\left( {{a}^{2}}-{{b}^{2}} \right)$ is composite if a and b are odd prime. For this we just need to show that $\left( {{a}^{2}}-{{b}^{2}} \right)$ has factors which are not equal to 1 or itself. We will first write a and b in the form of odd numbers which are prime. Then we will use them in $\left( {{a}^{2}}-{{b}^{2}} \right)$ to prove that it has factors other than 1. We will use the property that ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$.

**Complete step-by-step solution:**

Here a and b are two odd prime and we need to prove $\left( {{a}^{2}}-{{b}^{2}} \right)$ as composite. Thus we need to prove that $\left( {{a}^{2}}-{{b}^{2}} \right)$ has factors other than 1 or itself. As we know from the algebraic property that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$. So (a+b) and (a-b) are the factors of $\left( {{a}^{2}}-{{b}^{2}} \right)$. But we need to prove that these are not equal to 1. For this let us use general notation to write any odd prime.

a is an odd prime, so it can be written as a = 2k+1. Similarly, b is also an odd prime. So it can also be written as b = 2k'+1 where k, k' are integers.

Now (a-b) will become,

$\left( a-b \right)=2k+1-\left( 2k'+1 \right)\Rightarrow 2k+1-2k'-1\Rightarrow 2k-2k'$.

Taking 2 common from both the terms we get, $\left( a-b \right)=2k-2k'$ which is an even number, so it cannot be equal to 1. (1 is not an even number).

Now let us find the value of (a+b) we get,

$\left( a+b \right)=2k+1+2k'+1$.

Simplifying it we get, $\left( a+b \right)=2k+2k'+2$.

Taking 2 common from all the terms on the right side we get $\left( a+b \right)=2\left( k+k'+1 \right)$ which is an even number. Hence it cannot be equal to 1.

Therefore, none of the factors of $\left( {{a}^{2}}-{{b}^{2}} \right)$ are equal to 1 and as we can say they are neither equal to $\left( {{a}^{2}}-{{b}^{2}} \right)$. So $\left( {{a}^{2}}-{{b}^{2}} \right)$ has two factors. Therefore it cannot be prime.

**Hence $\left( {{a}^{2}}-{{b}^{2}} \right)$ is composite even if a and b are both odd primes.**

**Note:**Students should know the mathematical way to represent an odd integer i.e. 2k+1 where k is an integer. If a number can be divided by 2 then it is an even number. Here both (a+b) and (a-b) have 2 as their factor, so they can be divided by 2 easily and hence they are even numbers.

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