If m and n are positive integers and (m-n) is | vedclass

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(A) We know that $m^{2}-n^{2} = (m-n)(m+n)$.Given that $(m-n)$ is an even number,let $(m-n) = 2k$ for some integer $k$.Since $(m+n) = (m-n) + 2n$,and

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$4$

Vedclass · Answer

(A) We know that $m^{2}-n^{2} = (m-n)(m+n)$.Given that $(m-n)$ is an even number,let $(m-n) = 2k$ for some integer $k$.Since $(m+n) = (m-n) + 2n$,and $(m-n)$ is even,$(m+n)$ must also be an even number because the sum of two even numbers is even.Let $(m+n) = 2j$ for some integer $j$.Therefore,$(m^{2}-n^{2}) = (2k)(2j) = 4kj$.This shows that $(m^{2}-n^{2})$ is always divisible by $4$.

If $m$ and $n$ are positive integers and $(m-n)$ is an even number,then $(m^{2}-n^{2})$ will be always divisible by

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