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Prove for all natural numbers $n$ such that $2^{3^n}+1$ is divisible by $3^{n+1}$, but is not divisible by $3^{n+2}$.

I know how to prove why is it divisible, but I need help with why is it not divisible.

I used a similar proof for why is it divisible:

https://math.stackexchange.com/a/1591199/925421

My working for not divisible: Induction proof

Base case: n = 0 $$3^{0+2}=9$$ and $$2^{3^0}+1=3$$ 9 cannot divide 3, so base case is true.

Assume n=k is true. That means $3^{k+2}$ does not divide $2^{3^k}+1$. Now prove $3^{k+3}$ does not divide $2^{3^{k+1}}+1$. $$(2^{3^k})^3+1=(3^{k+1}*m)^3-3(3^{k+1}*m)^2+3(3^{k+1}*m)$$ $$=3^{3k+3}*m^3-3^{2k+3}*m^2+3^{k+2}*m$$ $$=3^{k+3}(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$$

That's where I need help! I'm so close to finish, I want to show that $$(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$$ is not an integer, so that means $3^{n+3}$ does not divide $2^{3^{n+1}}+1$

To show that $(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$ is not an integer, I have to show m can never be 3 or multiple of 3. But I don't know how to show that. Please help.

amWhy
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James Harden
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1 Answers1

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The linked problem also has an answer involving the Lifting The Exponent (LTE) result. Here is a plain solution using only induction for the proposition:

$$ \color{blue}{ P(n)\text{ : We have $2^{3^n}=-1 + 3^{n+1}(1+3N(n))$ for some integer $N(n)$ .} } $$

Let us inductively show that $P(n)$ is true for all integers $n\ge 0$.

$P(0)$ is true, $2^{3^0}=2^1=2=-1+3=-1+3^{0+1}$ and we take $N(0)=0$.

$P(1)$ is true, $2^{3^1}=2^3=9=-1+3^2=-1+3^{1+1}$ and we take $N(1)=0$.

Fix some $n\ge 1$. Assume that (for this $n$) the proposition $P(n)$ is true. Then, working modulo $3^{(n+1)+2}$ at the places with the $\equiv$ relation symbol: $$ \begin{aligned} 2^{3^{n+1}} &= 2^{3^n\cdot 3} = \left(2^{3^n} \right)^3\\ &\overset{P(n)}{=\!=\!=} \Bigg[\ -1 + 3^{n+1}(1+3N(n))\ \Bigg]^3 \\ &=(-1)^3 +3\cdot 3^{n+1}(1+3N(n)) \\ &\qquad\qquad\qquad -\underbrace{3\cdot 3^{2(n+1)}}_{\equiv 0}(1+3N(n))^2 + \underbrace{3^{3(n+1)}}_{\equiv 0}(1+3N(n))^3 \\ &\equiv -1 + 3^{n+2}(1+3N(n))\qquad\text{ modulo }3^{n+3} \ . \end{aligned} $$ (We have used $n+3\le 1+2(n+1)\le 3(n+1)$, which is valid for $n\ge 1$.)

The claim is now inductively shown.

$\square$

Alternatively, note that $3^n$ is odd, use $$ 1+2^{3^n}=1+(\color{brown}3-1)^{\color{red}{3^n}}\\ =1+(-1) + \binom{\color{red}{3^n}}1\cdot \color{brown}3^1 - \binom{\color{red}{3^n}}2\cdot \color{brown}3^2 + \binom{\color{red}{3^n}}3\cdot \color{brown}3^3 -\dots \pm\binom{\color{red}{3^n}}k\cdot \color{brown}3^k \mp\dots $$ and show (inductively) that the terms corresponding to $k\ge 2$ are each divisible by $3^{n+2}$.

dan_fulea
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  • Ummm, I don't see how does that prove $3^{n+2}$ does not divide $2^{3^n}+1$. I only see that it proves $3^{n+1}$ divides $2^{3^n}+1$. Also, how did you get the equation $2^{3^n}=-1+3^{n+1}(1+3N(n))$. – James Harden May 12 '21 at 13:40
  • The proposition $P(n)$ above shows that we have $2^{3^n}+1=3^{n+1}\ +\ ($ multiple of $3^{n+2}\ )$ . In particular $3^{n+1}$ divides $(2^{3^n}+1)$, and that the rest by division with $3^{n+2}$ is not zero, but $3^{n+1}$. This is the first proof, and it uses simple congruences modulo some (needed) power of $3$. The alternative proof, stated rather as a hint, is also simple... – dan_fulea May 13 '21 at 17:40