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Statement

Setting (p. 3). The generalized left factorial is $!^kn=(0!)^k+(1!)^k+\cdots+((n-1)!)^k$, so that $!^1n=\,!n$. Socialist primes are defined on the page for (2.6).

Condition (2.7) (p. 3). If $p$ is a socialist prime, then

$$\begin{aligned} (!^kp-2)^2+1&\equiv0\pmod p &&\text{if $k$ is odd,}\\ !^kp&\equiv1\pmod p &&\text{if $k=4t$,}\\ !^kp&\equiv3\pmod p &&\text{if $k=4t+2$.} \end{aligned}$$

Range of $k$. The display as printed names no range for $k$. The derivation on p. 3 passes through the congruence $!^kp\equiv2-\bigl(-\bigl(\tfrac{p-1}{2}\bigr)!\bigr)^k\pmod p$, which rests on $1^k+2^k+\cdots+(p-1)^k\equiv0\pmod p$, proved there for $1\le k\le p-2$; so (2.7) is established for $1\le k\le p-2$. For $k=1$ it is (2.6).

Proof pointer

P. 3. Since $2!,\ldots,(p-1)!$ together with the missing residue $-\bigl(\tfrac{p-1}{2}\bigr)!$ of (2.5) run over the nonzero residues, the sum of their $k$-th powers is the power sum $\sum_{m=1}^{p-1}m^k$, which vanishes mod $p$ for $1\le k\le p-2$ (shown by telescoping $(m+1)^{n+1}-m^{n+1}$). Adding $(0!)^k+(1!)^k=2$ gives the intermediate congruence above, and (2.4), $\bigl(\bigl(\tfrac{p-1}{2}\bigr)!\bigr)^2 \equiv-1$, evaluates the power in each residue class of $k$ mod 4.

Read depth

Claims checked: the definition and (2.7) were read clause by clause on the arXiv v1 print, p. 3, and the derivation was followed. Nothing here is independently reviewed.

Dependencies

(2.4) and (2.5).

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: further necessary conditions 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.