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Fermat's little theorem says that for a prime $p$ and $p\nmid a$, $a^{p-1}\equiv 1(\mathrm{mod}p)$; equivalently $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$. The Wilson theorem says $(p-1)!\equiv -1(\mathrm{mod}p)$ 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 givesso $(p-1)/2$ is even and $p\equiv 1(\mathrm{mod}4)$. Conversely, put $h=(p-1)/2$. Pairing $j$ with $p-j$ in the factorial gives $(p-1)!\equiv (-1{)}^h(h!{)}^2$. For $p\equiv 1(\mathrm{mod}4)$, $h$ is even, so the Wilson theorem yields $(h!{)}^2\equiv -1$. Thusand in the latter case $x=h!$ supplies a solution.

For the multiplicative order assertion, divide $k$ by $d$: $k=qd+r$, $0\le r<d$. Since $x^d\equiv 1$ and $x^k\equiv 1$, it follows that $x^r\equiv 1$. Minimality of the positive order excludes $0<r<d$, hence $r=0$ andIn 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 $n\ge 1$. If a prime $p$ divides $F_n=2^{2^n}+1$, it is odd and $2^{2^n}\equiv -1(\mathrm{mod}p)$. Squaring gives $2^{2^{n+1}}\equiv 1$, so the order $d$ divides $2^{n+1}$. It does not divide $2^n$, since $-1\ne 1$ modulo an odd prime. Every divisor of $2^{n+1}$ is a power of two, so the only possibility isIf 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 $n\ge 1$ every such prime is $1$ modulo $4$. No prime $3$ modulo $4$ can occur.

A prime $1$ modulo $4$ need not occur: take $p=13$. It is prime, $2^6=64\equiv -1(\mathrm{mod}13)$, and $2^4\equiv 3\ne 1$. 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.