Let $F\_n=2^{{2^n}}+1$, $n\geq 1$ ( Fermat numbers) and $p>2$ a prime number sucht that $p|F\_n$


I want to show if true that :


$p$ is Wieferich prime number $\Longleftrightarrow $ $p^2|F\_n$


the sense $\Longleftarrow$


If $p>2$ is a prime number and divides $F\_n$, then $p=1+k2^{n+1}$ for some $k\in \mathbb N$. If we assume that $p^2$ divides $F\_n$, then $2^{2^n}\equiv -1 \mod p^2$, and with this integer $k$ ($k=\frac{p-1}{2^{n+1}}$), we have $2^{k2^{n+1}}=(2^{2^n})^{2k}\equiv 1 \mod p^2$. Thus, $2^{p-1}\equiv 1 \mod p^2$, which implies that $p^2$ divides $2^{p-1}-1$. thus $p$ is Wieferich prime number .


If the sense $\Longrightarrow$ is true, can you give me help


**Addition**


My research goal is to try to prove;
If prime $p>3,\quad p|F\_n$ then $p^2\not| F\_n$


The difficulty is that I have no way to prove this conjecture and it's also difficult to provide a counterexample because if $p| F\_n$ and $p^2| F\_n$, as above, then p is a Wieferich prime, but only two numbers of this type are known at the moment.



A prime divisor $p$ of Fermat number $F_n$ is a Wieferich prime if and only if $p^2$ divides $F_n$
For the $\Longrightarrow$ sense, note we have, for some positive integer $m$, that


$$p \mid 2^{2^n} + 1 \;\;\to\;\; p \mid 2^{2^{n+1}} - 1 \;\;\to\;\; mp + 1 = 2^{2^{n+1}} \tag{1}\label{eq1A}$$


From your definition of


$$p = 1 + k2^{n+1} \;\;\to\;\; p - 1 = k2^{n+1} \tag{2}\label{eq2A}$$


we then get


$$\begin{equation}\begin{aligned}
2^{p-1} & = (2^{2^{n+1}})^k \\
& = (mp + 1)^k \\
& = (mp)^{k} + k(mp)^{k-1} + \ldots + \frac{k(k-1)}{2}(mp)^2 + k(mp) + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$


Thus,
$$2^{p-1} - 1 \equiv k(mp) \pmod{p^2} \tag{4}\label{eq4A}$$


Using the definition of [Wieferich prime](https://en.wikipedia.org/wiki/Wieferich_prime), we then have


$$p^2 \mid 2^{p-1} - 1 \;\;\to\;\; p \mid km \tag{5}\label{eq5A}$$


Since \eqref{eq2A} shows $p \nmid k$, then $p \mid m$. Thus, \eqref{eq1A} gives that


$$p^2 \mid 2^{2^{n+1}} - 1 = (2^{2^n} + 1)(2^{2^n} - 1) \tag{6}\label{eq6A}$$


Since $p \mid 2^{2^n} + 1$ and $\gcd(2^{2^n} + 1, 2^{2^n} - 1) = 1$, we get


$$p^2 \mid 2^{2^n} + 1 \;\;\to\;\; p^2 \mid F\_n \tag{7}\label{eq7A}$$




---


Alternatively, using the first $2$ parts of \eqref{eq1A} and \eqref{eq2A}, plus that $p \nmid k$, then the [Lifting-the-exponent lemma](https://en.wikipedia.org/wiki/Lifting-the-exponent_lemma) gives that


$$\nu\_p(2^{p-1}-1) = \nu\_p\left((2^{2^{n+1}})^k-1\right) = \nu\_p(2^{2^{n+1}}-1) + \nu\_p(k) = \nu\_p(2^{2^{n+1}}-1) \tag{8}\label{eq8A}$$


Then use the first part of \eqref{eq5A} and continue from \eqref{eq6A} to get to the same conclusion.


