$3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$.
I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
 A: By induction: case $n=0,1$ is obvious, assume the claim for $n \in \mathbb{N}$. Then, $$2^{3^{n+1}} +1 = ((2^{3^{n}}+1)-1)^3 +1 = (2^{3^{n}}+1)^3 -3(2^{3^{n}}+1)^2 +3 (2^{3^{n}}+1)$$ and by the induction hypothesis $ 3^{n+2}$ divides the last two terms. For the first term (call it $a$), induction again gives $3^{n+1}$ divides $a^{1/3}$. Then, $3^{n+2}$ divides $3^{3n+3}$ which divides $a$ so by transitivity we're done.
A: It follows from one of the Lifting The Exponent Lemmas (LTE):

Let $\upsilon_p(a)$ denote the exponent of the largest power of $p$ that divides $a$.
If $p$ odd prime, $n\in\mathbb Z^+$ is odd, $a,b\in\mathbb Z$, $a\equiv -b\not\equiv 0\pmod{p}$, then
$$\upsilon_p\left(a^n+b^n\right)=\upsilon_p(a+b)+\upsilon_p(n)$$

Therefore $\upsilon_3\left(2^{3^k}+1\right)=\upsilon_3(2+1)+\upsilon_3\left(3^k\right)=k+1$. So in fact, I've proved a stronger result: We have $3^{k+1}\mid 2^{3^k}+1$ and $3^{k+2}\nmid 2^{3^k}+1$, for all $k\in\mathbb Z_{\ge 0}$.
A: This actually looks like an induction proof. Let $P(n)$ be the statement : $$3^{n+1} \ \mid \ 2^{3^n} + 1 $$
Then we can verify that it is indeed true for $n = 1$ since $9$ divides itself. So suppoose $P(n)$ is true. Now, 
$$2^{3^{n+1}} + 1 = \left({2^{3^n}}\right)^3 + 1 $$
Notice that $$\left({2^{3^n} + 1}\right)^3 = \left({2^{3^n}}\right)^3 + 3 \cdot \left({2^{3^n}}\right)^2 + 3 \cdot \left({2^{3^n}}\right) + 1 $$
So, 
\begin{align}
\left({2^{3^n}}\right)^3 + 1 & =  \left({2^{3^n} + 1}\right)^3  -  3 \cdot \left({2^{3^n}}\right)^2 - 3 \cdot \left({2^{3^n}}\right) \\
& =  \left({2^{3^n} + 1}\right)^3 - 3 \cdot \left[{\left({2^{3^n } + 1}\right)^2 - 2\left({2^{3^n}}\right) - 1}\right] - 3 \cdot \left({2^{3^n} }\right) \\
& = \left({2^{3^n} + 1}\right)^3  - 3 \cdot \left({2^{3^n } + 1}\right)^2 + 3 \cdot \left({2^{3^n } + 1}\right) \\ 
& = \left({3^{n+1} \cdot k}\right)^3 - 3 \left({3^{n+1} \cdot k}\right)^2 + 3 \left({3^{n+1} \cdot k}\right) \\
& = 3^{n +2} \left({3^{2n+1} \cdot k^3 - k^2 + k}\right)  
\end{align}
which implies $$3^{n+2} \ \mid \ 2^{3^{n+1}} + 1 $$
And so the result is true for every $n$. 
So all the integers satisfy the given condition and that is the answer to the relevant question. 
A: We can easily verify that $n=1$ is a solution.
Assume for some $k$ that,
$3^{k+1}|2^{3^{k}}+1$, hence there exists an integer $m$, such that,
$m(3^{k+1})=2^{3^{k}}+1$.
Then we have, 
$2^{3^{k+1}}+1=(2^{3^{k}})^{3}+1=(2^{3^{k}}+1)[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]$. (*)
Now, $3^{k}$ is clearly odd for all integers $k$, hence,
$(2^{3^{k}})^{2}-2^{3^{k}}+1 \equiv ((-1)^{3^{k}})^{2}-(-1)^{3^{k}}+1 \equiv 1+1+1 \equiv 0\mod 3$.
Thus let,
$(2^{3^{k}})^{2}-2^{3^{k}}+1=3n$ where $n$ is an integer.
Continuing from (*), we are left with,
$m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[3n]=m(3^{k+2})(n)$
And so, 
$3^{k+2}|2^{3^{k+1}}+1$.
Thus our inductive proof is complete.
A: A proof without induction can be given by using Lifting The Exponent Lemma. Using this lemma we also prove that $3^{n+2}$ does not divide $2^{3^n}+1$.
By this lemma highest power of $3$ in $2^{3^n}+1$ is equal to highest of 3 in $3^{n}$ +highest power of 3 in $2+1$ which is equal to $n+1$.
