= Solution
<Fermat's little theorem> says that for a prime $p$ and $p\nmid a$, $a^{p-1}\equiv1\pmod p$; equivalently $a^p\equiv a\pmod p$ for every <integer> $a$. The <Wilson theorem> says $(p-1)!\equiv-1\pmod p$ for a prime $p$ (and, conversely, this congruence characterizes primes among <integers> greater than one).
For $p=2$, $x=1$ is a <square root of minus one modulo a prime>. If $p$ is odd and $x^2\equiv-1$, <Fermat's little theorem> gives
$$
1\equiv x^{p-1}=(-1)^{(p-1)/2}\pmod p,
$$
so $(p-1)/2$ is even and $p\equiv1\pmod4$. Conversely, put $h=(p-1)/2$. Pairing $j$ with $p-j$ in the <factorial> gives $(p-1)!\equiv(-1)^h(h!)^2$. For $p\equiv1\pmod4$, $h$ is even, so the <Wilson theorem> yields $(h!)^2\equiv-1$. Thus
$$
\boxed{p=2\text{ or }p\equiv1\pmod4,}
$$
and in the latter case $x=h!$ supplies a solution.
For the <multiplicative order> assertion, divide $k$ by $d$: $k=qd+r$, $0\leq r<d$. Since $x^d\equiv1$ and $x^k\equiv1$, it follows that $x^r\equiv1$. Minimality of the positive order excludes $0<r<d$, hence $r=0$ and
$$
\boxed{d\mid k.}
$$
In particular $d\mid p-1$ by <Fermat's little theorem>. Negative $k$, if included, are handled by the same division using the modular inverse of $x$.
A <Fermat number> in the paper has $n\geq1$. If a prime $p$ divides $F_n=2^{2^n}+1$, it is odd and $2^{2^n}\equiv-1\pmod p$. Squaring gives $2^{2^{n+1}}\equiv1$, so the order $d$ divides $2^{n+1}$. It does not divide $2^n$, since $-1\ne1$ modulo an odd prime. Every divisor of $2^{n+1}$ is a power of two, so the only possibility is
$$
\boxed{\operatorname{ord}_p(2)=2^{n+1}.}
$$
If two different <Fermat numbers> shared a <prime factor>, the same element $2$ modulo that prime would have two different orders. This is impossible, proving pairwise <coprimality>. Also $2^{n+1}\mid p-1$, so for $n\geq1$ every such prime is $1$ modulo $4$. No prime $3$ modulo $4$ can occur.
A prime $1$ modulo $4$ need not occur: take \b[$p=13$]. It is prime, $2^6=64\equiv-1\pmod{13}$, and $2^4\equiv3\ne1$. Since the order divides $12$, the first congruence excludes every divisor of $6$ and the second excludes $4$; hence the order is $12$. This is not a power of two, so $13$ is not a <prime divisor of a Fermat number>. The index restriction matters: the conventional extra number $F_0=3$ is outside the paper's positive-$n$ family.
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