For $a_n=6^{(2^n)}+1$, how can I show that $a_n\mid(a_{n+1}-2)$ ?
Proving divisibility of sequences
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sequences-and-series
elementary-number-theory
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0Yeah, I edited the question thanks – 2017-01-13
1 Answers
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Observe that $$a_{n+1}-2=6^{2^{n+1}}+1-2$$ $$=6^{2^{n}\cdot2}-1$$ $$=\left(6^{2^{n}}\right)^2-1$$ $$=\left(6^{2^{n}}-1\right)\left(6^{2^{n}}+1\right)$$ $$=a_n\left(6^{2^{n}}-1\right)$$ So, $\left(a_{n+1}-2\right)$ is a multiple of $a_n$
$\therefore$ $$a_{n+1}-2=a_n\left({6^2}^n-1\right)$$