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The order of 2 modulo p is the minimal solution of $2^t\equiv 1 \pmod{p}$

Euler's theorem guarantees that the congruence has a solution. The challenge is to demonstrate that $k=2^{n+1}$ is the minimal solution where p is a prime divisor of the n-th Fermat number $F_n$

We know that all solutions are multiples of the minimal solution:

If $2^t\equiv 1 $ and $k\nmid t$ then $t=kq+r$ with $0\lt r \lt k$ and

$2^r 2^{kq} \equiv 1\equiv 2^{kq}$ with $gcd(p,2^{kq})=1$ so by the cancellation law,

$2^r \equiv 1$

but since k is minimal, this is a contradiction, so $k|t$

From here I don't know how to proceed.

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Since $p \mid 2^{2^n}+1$, expressing this in congruence form and squaring both sides gives

$$2^{2^n} \equiv -1 \pmod{p} \implies 2^{2^{n+1}} \equiv 1 \pmod{p} \tag{1}\label{eq1A}$$

Thus, the multiplicative order of $2$ modulo $p$, which you're calling $k$, must divide $2^{n+1}$, so $k$ is a power of $2$, say $k = 2^j$. If any $j \lt n + 1$ gives $2^k \equiv 1 \pmod{p}$, then $k$ being any higher power of $2$ would also be congruent to $1$. However, that contradicts $2^{2^n} \equiv -1 \pmod{p}$. This means $j = n + 1,$ so the multiplicative order, i.e., minimal solution, of $2^k \equiv 1 \pmod{p}$ is $k = 2^{n+1}$.

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  • $\begingroup$ I guess since k is a power of two, $2^{2^n}\equiv -1$ and $2^{2^n+1}\equiv 1$, $2^{2^n+1}$ is the smallest power of two and $2^n+1$ must be the smallest solution. Is that right? $\endgroup$ – Anna Naden Aug 29 at 1:51
  • $\begingroup$ @AnnaNaden Yes, that's correct, except the power is $2^{n+1}$, not $2^{n} + 1$ as you wrote, as the latter indicates just multiplying by $2$ instead of squaring (use 2^{n+1} in MathJax). If you have any $m$ where $2^m \equiv 1 \pmod{p}$, then the smallest $k$ where $2^k \equiv 1 \pmod{p}$ must be one where $k \mid m$, as you show yourself in your question. Since $2^{2^{n+1}} \equiv 1 \pmod{p}$, then $k \mid 2^{n+1}$, i.e., is a power of $2$. As I explained in my answer, $k$ can't be any smaller $2^{n+1}$ due to $2^{2^{n}} \equiv -1 \pmod{p}$, so $k = 2^{n + 1}$ is the smallest solution. $\endgroup$ – John Omielan Aug 29 at 2:37

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