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I would like some hints on what to use to prove that: $$3^{n+3} \mid 2^{3^n}-3^{n+1}+1$$ for $n$ integer, $n \ge 0$. I tried it for $n \le 12$ and it seems to hold. Also, is it possible to extend the relation to something like: $$a^{n+k} \mid b^{a^n}-a^{n+j}+1$$ with $a$, $b$, $k$, $j$ integers?

Fabius Wiesner
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    Maybe express $2=3-1$ and use binomial expansion. – Julian Mejia May 08 '19 at 12:22
  • https://math.stackexchange.com/questions/207071/how-to-determine-highest-power-of-2-in-31024-1/207103#207103 – lab bhattacharjee May 08 '19 at 12:33
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    So, I tried to use my previous hint and it got messy, so I just used induction and it worked. Start with $2^{3^{n}}\equiv 3^{n+1}-1\mod 3^{n+3}$, i.e. there is some integer $k$ such that $2^{3^{n}}=3^{n+1}-1+3^{n+3}k$. By taking cubes in both sides prove that $2^{3^{n+1}}\equiv 3^{n+2}-1$. – Julian Mejia May 08 '19 at 12:45

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Hint:

$$2^{3^{n+1}}+1=(2^{3^n})^3+1=(2^{3^n}+1)(2^{2\cdot3^n}-2^{3^n}+1)$$

Note that the second factor is a multiple of $3$.

ajotatxe
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