---

Statement

Socialist primes are defined on the page for (2.6); $\bigl(\tfrac{\cdot}{p}\bigr)$ is the Legendre symbol.

Let $p$ be a socialist prime and $H=\{2,3,4,\ldots,p-3\}\setminus\{\tfrac{p-1}{2}\}$. The paper shows (pp. 3--4):

• (3.1) (p. 3). There is a function $f$ on $H$ with $(f(k))!\equiv-k!\pmod p$ for all $k\in H$, and $f$ is an involution of $H$.
• Parity and quadruples (p. 4). $f(k)\equiv k\pmod 2$, and $H$ splits into $(p-5)/4$ quadruples $U_k=\{k,\,f(k),\,p-1-k,\,p-1-f(k)\}$, each with $\prod_{x\in U_k}x\equiv1$ and $\sum_{x\in U_k}x\equiv0\pmod p$ and all members of the same parity.
• Quadratic characters (p. 4). The Legendre symbol $\bigl(\tfrac{x!}{p}\bigr)$ takes the same value for all $x\in U_k$. The print words this as "all members of $U_k$ have the same quadratic residue modulo $p$" (p. 4); the display before it, and the use made of it, concern the factorials $x!$. Consequently $\Bigl(\tfrac{2!\cdot3!\cdots\frac{p-3}{2}!}{p}\Bigr)=1$, and also $\Bigl(\tfrac{2!\cdot4!\cdots((p-5)/2)!}{p}\Bigr)=1= \Bigl(\tfrac{3!\cdot5!\cdots((p-3)/2)!}{p}\Bigr)$.
• Conclusion (p. 4). $\Bigl(\tfrac{\frac{p-1}{4}!}{p}\Bigr)=1$: the residue $\bigl(\tfrac{p-1}{4}\bigr)!$ is a quadratic residue modulo $p$.

Proof pointer

Pp. 3--4. Since the factorial residues are distinct and miss only $-\bigl(\tfrac{p-1}{2}\bigr)!$, each $-k!$ with $k\in H$ is $(f(k))!$ for a unique $f(k)\in H$. Multiplying (3.1) by $(p-1-f(k))!\,(p-1-k)!$ and using the reflection (2.3) relates $(p-1-k)!$ and $(p-1-f(k))!$ with sign $(-1)^{k+f(k)+1}$; distinctness then forces equal parity, and the four indices $k,f(k),p-1-k,p-1-f(k)$ close up into $U_k$. The characters agree because $\bigl(\tfrac{-1}{p}\bigr)=1$ (as $p\equiv1\pmod4$) and $\bigl(\tfrac{x!}{p}\bigr)=\bigl(\tfrac{1/x!}{p}\bigr)$. The last step rewrites the product of factorials as a product of odd numbers, expresses it through $\bigl(\tfrac{p-1}{2}\bigr)!$, $2^{(p-1)/4}$ and $\bigl(\tfrac{p-1}{4}\bigr)!$, and evaluates $\Bigl(\tfrac{((p-1)/2)!}{p}\Bigr)=-1$ and $\Bigl(\tfrac{2^{(p-1)/4}}{p}\Bigr)=-1$ using $p\equiv5\pmod 8$.

Read depth

Claims checked: the statements of Section 3 were read clause by clause on the arXiv v1 print, pp. 3--4, and the argument was followed. Nothing here is independently reviewed.

Dependencies

(2.3)–(2.5), including $p\equiv5\pmod 8$.

Source. V. Andrejić and M. Tatarevic, On distinct residues of factorials, arXiv:1603.04086v1 (2016); published in Publ. Inst. Math. (Beograd) (N.S.) 100(114) (2016), 101--106. Labels and pages here are those of the arXiv v1 print; the edition read is named on the source card.

Bears on

• Problem 478: more necessary structure for the extreme case $\lvert A_p\rvert=p-2$ (socialist primes; see the page for (2.6)); nothing about $\lvert A_p\rvert$ in general.