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Question Number 183075 by Ml last updated on 19/Dec/22

Answered by BaliramKumar last updated on 19/Dec/22

$$letn=2k$$$$n^2={\left(2k\right)}^2=4k^2=2\left(2k^2\right)=2\left(m\right)even$$

Answered by Rasheed.Sindhi last updated on 20/Dec/22

$$n\equiv 0\left(mod2\right)$$$${\left(n\right)}^2\equiv {\left(0\right)}^2\left(mod2\right)$$$$n^2\equiv 0\left(mod2\right)$$$$\therefore n\in \mathbb{E}\Rightarrow n^2\in \mathbb{E}$$

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