Prove that $3^{2 n} + 7$ is a multiple of $8$ .

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Solution

Let $p = 3^{2 n} + 7$
$p = {(3^{2})}^{n} + 7 9^{n} - 1 + 8 p = 9^{n} - 1^{n} + 8 a^{n} - b^{n} = (a - b) (a^{n - 1} + a^{n - 2} b + a^{n - 3} b^{2} . . . . . . . . . . . b^{n - 1}) \Rightarrow p = (9 - 1) (9^{n - 1} + 9^{n - 2} .1 + 9^{n - 3} 1^{2} . . . . . . . . . . . 1^{n - 1}) + 8 \Rightarrow p = 8 (9^{n - 1} + 9^{n - 2} .1 + 9^{n - 3} 1^{2} . . . . . . . . . . . 1^{n - 1}) + 8 \Rightarrow p = 8 (9^{n - 1} + 9^{n - 2} .1 + 9^{n - 3} 1^{2} . . . . . . . . . . . 1^{n - 1} + 1)$
which is mutiple of $8$
Hence proved.

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