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I'm trying to prove $2^{(2^n)}+1$ divides $2^{(2^nb)}+1$ for $n,b\in \mathbb{N}$. I've considered factoring $2^{(2^n)}+1$ into $(2+1)(2^{(2^n)-1}-2^{(2^n)-2}+...+1)$ and likewise $2^{(2^nb)}+1$ into $(2+1)(2^{(2^nb)-1}-2^{(2^nb)-2}+...+1)$

Proofs follow using the same methods as in the dupe, e.g. $$\bmod 2^{2^n}\!+1\!:\ 2^{2^n}\equiv -1 \Rightarrow (2^{2^n})^b\equiv (-1)^b\equiv -1$$ by the Congruence Power Rule. Or, equivalently use the Factor Theorem or other closely related methods mentioned there (and its links).

J. W. Tanner
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Quick
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    This is a dupe of a FAQ ...hold on ... – Bill Dubuque Nov 04 '21 at 00:36
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    It's always true when $b$ is odd because $a+1 \mid a^b+1$ for odd $b$. But that doesn't quite get you where you need to be. – Robert Shore Nov 04 '21 at 00:37
  • I forgot to mention $b$ is odd. I think I might know where to go from there. Thanks. – Quick Nov 04 '21 at 00:38
  • Good, because $5=2^2+1 \nmid 2^4+1=17$. – Robert Shore Nov 04 '21 at 00:39
  • Proofs follow using the same methods as in the dupe, e.g. $$\large \bmod 2^{2^n}!+1!:,\ \color{#c00}{2^{2^n}\equiv -1} \Rightarrow (\color{#c00}{2^{2^n}})^b\equiv (\color{#c00}{-1})^b\equiv -1\qquad$$ by raising the congruence to power $,b,$ via the Congruence Power Rule. Or, equivalently, use the Factor Theorem or other closely related methods mentioned there (and its links). – Bill Dubuque Nov 04 '21 at 00:48

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