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Standard VI

Mathematics

Prime Factors

If n1 and ...

Question

If $n_{1}$ and $n_{2}$ are two odd prime numbers such that $n_{1} > n_{2}$ , then write the nature of $n_{1}^{2} - n_{2}^{2}$ .

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Solution

$n_{1}^{2} - n_{2}^{2} = (n_{1} + n_{2}) (n_{1} - n_{2})$
Adding of two odd numbers (for e.g. $1 + 3 = 4)$ is even & substitution $(e . g . 3 - 1 = 2)$ is also even
And $e v e n \times e v e n (e . g . 2 \times = 4)$ is even
$\Rightarrow n_{1}^{2} - n_{2}^{2} = e v e n$

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If n1 and n2 are two odd prime numbers such that n1>n2, then write the nature of n21−n22.

n21−n22=(n1+n2)(n1−n2)Adding of two odd numbers (for e.g. 1+3=4) is even & substitution (e.g.3−1=2) is also even And even×even(e.g.2×=4) is even⇒n21−n22=even

If $n_{1}$ and $n_{2}$ are two odd prime numbers such that $n_{1} > n_{2}$ , then write the nature of $n_{1}^{2} - n_{2}^{2}$ .

$n_{1}^{2} - n_{2}^{2} = (n_{1} + n_{2}) (n_{1} - n_{2})$
Adding of two odd numbers (for e.g. $1 + 3 = 4)$ is even & substitution $(e . g . 3 - 1 = 2)$ is also even
And $e v e n \times e v e n (e . g . 2 \times = 4)$ is even
$\Rightarrow n_{1}^{2} - n_{2}^{2} = e v e n$