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For integer $n \ge 3$, is $3^{2^{n-2}} -1 \equiv 0 \pmod {2^n}$

Here's what I am seeing:

$$3^{2} - 1 = 2^3$$ $$3^{4} - 1 = 5\times2^4$$ $$3^{8} - 1 = 205\times2^5$$ $$3^{16} - 1 = 672,605\times2^6$$

I was trying to prove that this is true for all $n \ge 3$ but I am not clear how to do it.

How would one establish the inductive step?

Larry Freeman
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  • Thanks, @JohnOmielan I didn't realize it generalized to all odd integers. Cool. – Larry Freeman May 14 '21 at 18:55
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    You're welcome. Note another duplicate one is How to prove that $a^{2^{n-2}} \equiv 1 \pmod{2^n}$?, but it's more abstract & complicated than what you need, and possibly want, for your question here. – John Omielan May 14 '21 at 18:56
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    Note when determining the divisibility of an expression, a key thing to check on is if the expression can be factored. With your question, note the difference of squares means $3^{2^{k-1}} - 1 = \left(3^{2^{k-2}} - 1\right)\left(3^{2^{k-2}} + 1\right)$. Since $3^{2^{k-2}} + 1$ is even, then $2 \mid 3^{2^{k-2}} + 1$. Thus, if $2^{k} \mid 3^{2^{k-2}} - 1$, then $2^{k+1} \mid 3^{2^{k-1}} - 1$. Note this inductive step also applies if $3$ is replaced by any odd $a$, with the base case due to $a^2 \equiv 1 \pmod{8}$ also always being true. – John Omielan May 15 '21 at 23:13

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