The remainder obtained when (2m + 1)^{2n} (m, | vedclass

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(A) We need to find the remainder when $(2m + 1)^{2n}$ is divided by $8$. Let $x = 2m + 1$. Since $m \in N$,$x$ is an odd integer. Any odd integer can

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$1$

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(A) We need to find the remainder when $(2m + 1)^{2n}$ is divided by $8$. Let $x = 2m + 1$. Since $m \in N$,$x$ is an odd integer. Any odd integer can be written as $2m + 1$. Consider $(2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1$. Since $m(m + 1)$ is the product of two consecutive integers,it is always even. Let $m(m + 1) = 2k$ for some integer $k$. Then $(2m + 1)^2 = 4(2k) + 1 = 8k + 1$. Now,$(2m + 1)^{2n} = ((2m + 1)^2)^n = (8k + 1)^n$. Using the Binomial Theorem,$(8k + 1)^n = \binom{n}{0}(8k)^n + \binom{n}{1}(8k)^{n-1} + \dots + \binom{n}{n-1}(8k) + 1$. All terms except the last one are multiples of $8$. Therefore,$(2m + 1)^{2n} = 8K + 1$ for some integer $K$. Thus,the remainder when divided by $8$ is $1$.

The remainder obtained when $(2m + 1)^{2n}$ $(m, n \in N)$ is divided by $8$ is

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