Let $F_n = 2^{2^n}+1$ be the $n$th Fermat number for $n \geq 2$ and $p$ a prime factor of $F_n$.
How can I show that that $2^{(p-1)/2} \equiv 1 \:\mathrm{mod}\:p$?
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did you mean $2^{\color{red}2^n}$? – J. W. Tanner May 25 '20 at 21:40
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2did you mean $2^{2^n} + 1$? The only prime that can divide $2^{2^n}$ or $2^{n^n}$ is $2$. – failedentertainment May 25 '20 at 21:41
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Yes, this is what I was trying to understand, thank you – Monya Feldman May 25 '20 at 22:31
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Cf. this question – J. W. Tanner May 25 '20 at 23:41
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You can look to Legendre's symbol you may state this theorem this way as well:
$$\left(\frac{a}{p}\right) \equiv a^{(p-1)/2}\ \pmod{ p}\;\;\text{ and } \left(\frac{a}{p}\right) \in \{-1,0,1\}$$.
where Legendre's symbol is defined as:
$$\left(\frac{a}{p}\right) = \begin{cases}\;\;\,1 \text{ if } a \text{ is a quadratic residue modulo}\ p\text{ and } a \not\equiv 0\pmod{p} \\-1 \text{ if } a \text{ is a quadratic non-residue modulo}\ p\\\;\;\,0 \text{ if } a \equiv 0 \pmod{p}. \end{cases}$$ And you may check here for more informations about the proof
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