Proving that $3^{n+2}$ does not divide $2^{3^n} +1$ for any positive integer $n$

Question

Proving that $3^{n+2}$ does not divide $2^{3^n} +1$ for any positive integer $n$

I am thinking of applying inductionon $n$ but am not being successful.

Answer 1

By induction. Let $S_n=2^{3^n}+1$ We claim that $v_3(S_n)={n+1}$. Easily seen for small $n$.

Notation: here, as is fairly standard, we are writing $v_p(n)$ for prime $p$ and natural number $n$ to indicate the maximal power of $p$ which divides $n$. Thus, with $n=1$ we have $$S_1=2^3+1=9=3^2\implies v_3(S_1)=2=1+1$$ as claimed.

Similarly,$$S_2=2^9+1=513=3^3\times 19 \implies v_3(S_2)=3=2+1$$ as claimed.

And again, $$S_3=2^{27}+1=134217729=3^4\times 19\times 87211\implies v_3(S_3)=4=3+1$$ as claimed.

Proof: Suppose it is true for $n-1$. That is, assume that $$v_3(S_{n-1})=(n-1)+1=n$$ Then we write $$S_n=(2^{3^{n-1}})^3+1=\left(S_{n-1}-1\right)^3+1=S_{n-1}^3-3S_{n-1}^2+3S_{n-1}$$ It is clear that $3^{n+2}$ divides each of the first two terms but only $3^{n+1}$ divides the third, so we are done.

Answer 2

Suppose $\left.3^{n+1}\,\middle|\,2^{3^n}+1\right.$ and $\left.3^{n+2}\not\,\middle|\,\,2^{3^n}+1\right.$; that is $2^{3^n}+1$ has exactly $n+1$ factors of $3$. Then $$ \begin{align} \left(2^{3^n}+1\right)^3 &=2^{3^{n+1}}+3\cdot2^{2\cdot3^n}+3\cdot2^{3^n}+1\\ &=2^{3^{n+1}}+3\cdot2^{2\cdot3^n}\left(2^{2\cdot3^n}+1\right)+1 \end{align} $$ Therefore, $$ 2^{3^{n+1}}+1=\overbrace{\left(2^{3^n}+1\right)^3}^{\substack{\text{exactly }3n+3\\\text{ factors of }3}}-\overbrace{3\cdot2^{2\cdot3^n}\left(2^{2\cdot3^n}+1\right)\vphantom{\left(2^{3^n}\right)^3}}^{\substack{\text{exactly }n+2\\\text{ factors of }3}} $$ has exactly $n+2$ factors of $3$. Since this holds for $n=0$, for all $n\ge0$, $$ \left.3^{n+1}\,\middle|\,2^{3^n}+1\right.\quad\text{and}\quad\bbox[5px,border:2px solid #C0A000]{\left.3^{n+2}\not\,\middle|\,\,2^{3^n}+1\right.} $$

Answer 3

Another way of looking at @lulu’s proof:

For odd $p$, look at $\Bbb Q_p$, and a nonzero root $\zeta$ of $X^p-X$, that is a $(p-1)$-th root of unity. Let $u$ be a unit of $\Bbb Z_p$, i.e. $v_p(u)=0$, and $n\ge1$. Then $(\zeta+p^nu)^p=\zeta+p^{n+1}u'$ for some unit $u'\in\Bbb Z_p$. For, $(\zeta+p^nu)^p=\zeta+p\zeta^{p-1}p^nu+(\text{middle terms})+p^{\,pn}u^p$, where the “middle terms” all have a factor $p$ from the binomial coefficient, and at least $p^{2n}$ from powers of $p^nu$.

For the application, $p=3$, $\zeta=-1$, and we see that $v_3(2^{3^n}-(-1))=n+1$.

Answer 4

For a short analysis let's introduce the notation of curly braces $ \{a,p\} =k $ meaning that $a$ contains the factor $p$ to the $k$'th power, and let's as well introduce the "Iverson brackets" $[a:b]=k$ meaning that $k=1$ if $b$ divides $a$ and $k=0$ if $b$ does not divide $a$. (See here for a detailed essay on the motivation of this notation and its application)

---

In general, for some integer $m$ we have the primefactor $3$ in the expression $2^m-1$ according to the following formula: $$ \{ 2^m -1, 3\} = [m:2](1+\{m,3\}) \tag 1$$

Now we have the well known composition $ 2^m+1 = (2^{2m}-1) / (2^m-1)$ and thus can derive for some integer $m$: $$\begin{array}{} \{ 2^m +1, 3\} &= [2m:2](1 + \{2m,3\})-[m:2](1 + \{m,3\}) \\ &=1 \cdot (1+\{m,3\})-[m:2] \cdot (1+\{m,3\})\\ &=(1-[m:2])(1+\{m,3\}) \end{array} \tag 2$$ Now we assume $m=3^n$ and get $$\begin{array}{} \{ 2^{3^n} +1, 3\} &= (1-[3^n:2])(1+\{3^n,3\}) \\ &= (1-0)(1+ n) \\ &= 1+n \end{array} \tag 3$$ Thus we know, that the primefactor $3$ occurs to the ($1+n$'th) power (meaning $(2^{3^n}+1)=3^{1+n} \cdot x$ where $ 3 \not | x$) and not to the ($2+n$'th) power