Prove that there are infinitely many $n\in\mathbb{N^*}$ such that $n|2^{2n-1}+1$ Prove that there are infinitely many $n\in\mathbb{N^*}$ such that $n|2^{2n-1}+1$
Hi! I think that I solved this problem but my proof is pretty complicated and I would love if someone could comment under my solution and say if it is correct or not. I would like as many comments as possible from as many people as possible because I want to be 100% sure that it is correct. Thank you!
 A: We will begin with $n=3$ which is obviously a solution because $3\mid 33$.
Now assume that the number of solutions to the problem is finite and let $n$ be the biggest such solution. We know that $n\geq3$.
From the Zsigmondy Theorem we know that we can choose a prime $p$ such that $p$ is a primitive divisor of $2^{2n-1}+1$ because $2n-1\geq5$ so we aren't in the exceptional case in which the theorem isn't true.
$p\mid 2^{2n-1}+1 \Leftrightarrow$
$2^{2n-1} \equiv -1 \pmod p$
$\Rightarrow 2^{4n-2}\equiv 1 \pmod p$
Now, let $m$ be the order of $2$ modulo $p$. We know that $m \mid 4n-2$, but $m\nmid2n-1$ so $m=2d$, where $d$ is a divisor of $2n-1$.
By the definition of the order, $p\mid 2^m-1$ so $p\mid 2^{2d}-1$ so $p \mid(2^d-1)\cdot(2^d+1)$.
$p$ can't divide $2^d-1$ because $d<m$, so $p\mid 2^d+1$, but because $p$ was a primitive divisor of $2^{2n-1}+1$, we have that $d\geq2n-1$ so $d=2n-1$.
So the order of $2$ modulo $p$ is $4n-2$. From Fermat's little theorem, $p\mid 2^{p-1}-1$, so $4n-2\mid p-1$. This means that there is a $k\in \mathbb{N}^*$ such that $p-1=k\cdot(4n-2) \Leftrightarrow p=k\cdot(4n-2)+1$. This guarantees that $p>n$ so $\gcd(p,n)=1$ so because $p\mid 2^{2n-1}+1$ and $n\mid 2^{2n-1}+1$ we have that $pn\mid 2^{2n-1}+1$.
Because $X+1|X^a+1$ for any odd $a\in \mathbb{N}^*$, we have that $$2^{2n-1}+1\mid {(2^{2n-1})}^{4nk+1}+1 \Leftrightarrow 2^{2n-1}+1\mid 2^{(2n-1)\cdot(4nk+1)}+1 \Leftrightarrow 2^{2n-1}+1\mid 2^{8 n^2k-4nk+2n-1}+1 \Leftrightarrow 2^{2n-1}+1\mid 2^{2np-1}+1$$ $np\mid 2^{2n-1}+1\mid 2^{2np-1}+1$ so $np$ is another solution to the problem that is obviously bigger than $n$ so our assumption was wrong, so there are infinitely many such $n$.
$\square$
A: COMMENT (This is a pertinent comment but not an answer).
If $2^{2n-1}+1=qn$ then both $q$ and $n$ should be odd so we have
$$2*16^n+1=q(2n+1)\Rightarrow 1\equiv q(2n+1)\pmod{16}$$ The modular equation
$$q(2n+1)\equiv1\pmod{16}$$ have solution for all $q=1,3,5,\cdots,13,15$ modulo $16$ and these solutions are
$$1\text { con } 0 \text { and } 8\\3\text { con } 5 \text { and } 13\\5\text { con } 6 \text { and } 14\\7\text { con } 3 \text { and } 11\\9\text { con } 4 \text { and } 12\\11\text { con } 1 \text { and } 9\\13\text { con } 2 \text { and } 10\\15\text { con } 7 \text { and } 15$$ Some if not all of these are necessary conditions that could maybe help the proof by somebody. I will try myself to end.
