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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Setting (p. 1). A prime p>5p>5 is a socialist prime (Trudgian's term, which the paper adopts) when the residues of 2!,3!,…,(p−1)!2!,3!,\ldots,(p-1)! modulo pp are all distinct. Kurepa's left factorial is !n=0!+1!+⋯+(n−1)!!n=0!+1!+\cdots+(n-1)!.

Let pp be a socialist prime, and let rr be the one nonzero residue modulo pp that is not among 2!,…,(p−1)!2!,\ldots,(p-1)! (there are p−2p-2 distinct values among p−1p-1 nonzero residues). The paper derives in Section 2 (pp. 2--3):

  • (2.4) (p. 2). p≡1(mod4)p\equiv1\pmod 4 and ((p−12)!)2≡−1(modp)\bigl(\bigl(\tfrac{p-1}{2}\bigr)!\bigr)^2\equiv-1\pmod p.
  • (2.5) (p. 2). r≡−(p−12)!(modp)r\equiv-\bigl(\tfrac{p-1}{2}\bigr)!\pmod p, and consequently (p2−1)/8(p^2-1)/8 is odd, so p≡5(mod8)p\equiv5\pmod 8.
  • (2.6) (p. 2). (p−12)!≡  !p−2(modp)\bigl(\tfrac{p-1}{2}\bigr)!\equiv\;!p-2\pmod p, and hence the necessary condition
(!p−2)2≡−1(modp).(!p-2)^2\equiv-1\pmod p.

The value of the missing residue and p≡5(mod8)p\equiv5\pmod 8 were already proved by Rokowska and Schinzel (1960), as the paper recalls on p. 1 with (1.1); the paper rederives them here. Condition (2.6), linking socialist primes to Kurepa's left factorial, is the paper's new condition.

Proof pointer

P. 2. Wilson's theorem gives (p−1)!≡−1(p-1)!\equiv-1 and (p−2)!≡1(p-2)!\equiv1, and the reflection (p−k)! (k−1)!≡(−1)k(modp)(p-k)!\,(k-1)!\equiv(-1)^k\pmod p for 1≤k≤p1\le k\le p gives ((p−12)!)2≡(−1)(p+1)/2\bigl(\bigl(\tfrac{p-1}{2}\bigr)!\bigr)^2\equiv(-1)^{(p+1)/2}. Distinctness forbids (p−12)!≡±1\bigl(\tfrac{p-1}{2}\bigr)!\equiv\pm1, which forces $p\equiv1\pmod 4$ and (2.4). Writing (p−1)!(p-1)! as rr times the product of all k!k!, 2≤k≤p−12\le k\le p-1, and pairing factorials by the reflection, gives r≡(−1)(p2−1)/8(p−12)!r\equiv(-1)^{(p^2-1)/8}\bigl(\tfrac{p-1}{2}\bigr)!; since rr differs from (p−12)!\bigl(\tfrac{p-1}{2}\bigr)!, (2.5) follows. Finally the residues 2!,…,(p−1)!2!,\ldots,(p-1)! together with rr run over 1,…,p−11,\ldots,p-1, whose sum is 00 mod pp; this gives (p−12)!≡  !p−2\bigl(\tfrac{p-1}{2}\bigr)!\equiv\;!p-2, and (2.4) turns it into (2.6).

Read depth

Claims checked: the definitions and (2.1)--(2.6) were read clause by clause on the arXiv v1 print, pp. 1--2, and the derivation on p. 2 was followed. Nothing here is independently reviewed.

Dependencies

None in the corpus. External input: Wilson's theorem.

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: for p≥5p\ge5 one has 1!≡(p−2)!(modp)1!\equiv(p-2)!\pmod p, so the problem's ApA_p has at most p−2p-2 elements, with equality exactly when the residues of 2!,…,(p−1)!2!,\ldots,(p-1)! are distinct: at p=5p=5, and for p>5p>5 exactly when pp is a socialist prime (an observation of this page, not of the paper). The conditions here constrain only that extreme case; they say nothing about the size of ApA_p in general or about the asymptotic the problem asks for.