Chen: On integers of the forms kʳ−2ⁿ and kʳ2ⁿ+1

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corollary_1: States that for each positive odd integer r there are infinitely many primes p with p^r - 2^n having at least two distinct prime factors for every positive n, and infinitely many with p^r 2^n + 1 doing so.

corollary_2: States that for each positive odd integer r not divisible by 3 there are infinitely many primes p with p^(2r) - 2^n having at least two distinct prime factors for every positive n, and infinitely many with p^(2r) 2^n + 1 doing so.

theorem_1: States that for each positive odd integer r, the positive odd k for which k^r - 2^n has at least two distinct prime factors for every positive n contain an infinite arithmetic progression, and likewise for k^r 2^n + 1.

theorem_2: States that for each positive odd integer r not divisible by 3, the positive odd k for which k^(2r) - 2^n has at least two distinct prime factors for every positive n contain an infinite arithmetic progression, and likewise for k^(2r) 2^n + 1.

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The copy read for this card is the full published article, whose PDF prints "© 2002 Elsevier Science (USA). All rights reserved." on its first page, every other right reserved.

Yong-Gao Chen, "On integers of the forms kʳ−2ⁿ and kʳ2ⁿ+1," Journal of Number Theory, 98(2), 310-319, 2003. https://doi.org/10.1016/s0022-314x(02)00051-3

Overview

Chen studies uniform compositeness, strengthened to the presence of at least two distinct prime divisors, for the nonlinear sequences $k^r-2^n$ and $k^r2^n+1$ with $n\ge 1$. The motivating questions ask whether infinitely many odd $k$, and infinitely many prime $k$, make $k^2-2^n$ composite for every $n$ (Introduction, p. 311).

For every fixed positive odd $r$, Theorem 1(i)–(ii) states that each of the two sets of odd $k$ for which, respectively, $k^r-2^n$ or $k^r2^n+1$ has at least two distinct prime factors for every positive $n$ contains an infinite arithmetic progression (p. 311). Corollary 1 gives infinitely many prime bases $p$ with the corresponding properties (p. 311; proof on p. 315), by applying Dirichlet's theorem to the constructed progression.

The proof uses the covering

$$\mathbb Z=\bigcup_{i=1}^7(a_ir\bmod n_i),\qquad (n_1,\ldots,n_7)=(2,4,8,16,32,64,64),$$

where $a_ir$ represents $0,1,3,7,15,31,63$ modulo the indicated moduli; see equation (4), p. 313. The associated primes are

$$p_i\in\{3,5,17,257,65537,641,6700417\},$$

with $\operatorname{ord}_{p_i}(2)=n_i$, and primes $q_i\mid 2^{p_i}-1$ (p. 314). Chinese-remainder conditions (5)–(6) force $p_i\mid M^r-2^n$ whenever $n$ lies in the $i$th covering class. Equation (8) writes this difference as $p_i^{\alpha_i}K_i$. Lemma 1, equations (1)–(3), supplies a uniform lower bound ensuring $|K_i|>1$ when $\alpha_i=1$ (pp. 312–313). When $\alpha_i\ge2$, Lemma 2 and Corollary 3 lift multiplicative orders through powers of $p_i$ and force the distinct prime $q_i$ to divide $K_i$ (pp. 313–314). This proves Theorem 1(i). For the plus sign, the congruences $2^{a_i}M+1\equiv0\pmod{p_i^2q_i}$ together with $M\equiv1+2^m\pmod{2^{m+1}}$ give the analogous construction; the paper records this argument only as being similar to part (i) (Theorem 1(ii), p. 315).

Lemma 3 gives a 28-class covering with moduli dividing $2^8 3^4$; its proof is a finite verification over $0\le n<2^8 3^4$ (p. 315). The proof of Theorem 2 then assigns primitive prime divisors $p_i$ of $2^{m_i}-1$, primitive prime divisors $q_i$ of $2^{p_i^5}-1$, and square roots of $2$ modulo suitable powers of these primes; congruences (10)–(14) produce the required bases (p. 316). Equation (15), the lower bound from Lemma 1, and the lifting argument following equation (16) again produce a second distinct divisor (p. 317). For the plus sign the paper displays three congruences and says the proof is similar (p. 318); read literally, those congruences give $M^{2r}2^n+1\equiv2\pmod{p_i}$ on the $i$th class, and the prime $p_2=7$ divides no number $k^{2r}2^n+1$, so the printed construction does not, as written, prove Theorem 2(ii) (see the Theorem 2 page below).

Theorem 2 (p. 311) and Corollary 2 (p. 312) are both stated for odd $r$ with $3\nmid r$, in agreement with the abstract's range $(r,12)\le3$ (p. 310) for the even exponent $2r$ and with the proof's use of $(6,r)=1$ (p. 316).

Section 3 formulates, but does not prove, Conjectures 1 and 2 for arbitrary positive $r$, and specifically identifies $r=4,6$ as interesting unresolved cases (p. 318). The paper concerns positive exponents $n$, constructs sufficient arithmetic progressions, and does not classify all bases having either uniform factorization property.

Relation to E1113

This source bears on Problem 1113.

In E1113, write the Sierpiński coefficient as $m$, so the relevant sequence is $m2^n+1$. Chen's Theorem 1(ii) translates as follows: for every fixed odd $r$, there is an infinite arithmetic progression of odd bases $M$ such that

$$m=M^r\quad\Longrightarrow\quad m2^n+1=M^r2^n+1$$

has at least two distinct prime factors for every $n\ge1$ (Theorem 1(ii), p. 311; construction on p. 315). At $n=0$ the value $M^r+1$ is even and larger than $2$ for $M\ge3$, so these $m$ are Sierpiński numbers in the problem's sense. Corollary 1(ii) permits infinitely many such $M$ to be prime (pp. 311, 315), so the paper produces infinitely many Sierpiński coefficients that are odd prime powers $m=p^r$.

Crucially, all these examples come with a finite covering set. For the seven-class construction, if $n\equiv a_ir\pmod{n_i}$, then the prescribed congruence $2^{a_i}M+1\equiv0\pmod{p_i^2q_i}$ implies

$$p_i\mid M^r2^n+1.$$

Equation (4) says that some such class contains every integer $n$ (p. 313). Hence

$$\{3,5,17,257,65537,641,6700417\}$$

is a finite prime cover for every coefficient $m=M^r$ obtained from this construction. The auxiliary primes $q_i$ are used to guarantee a second distinct prime factor when the covering prime occurs with high multiplicity; they are not needed merely to establish the finite cover. This cover is a deduction from the printed congruences, not a statement of the paper. The 28-class construction for even powers uses finitely many covering primes $p_i$ for the minus sign (Lemma 3 and equations (10)–(16), pp. 315–317); for the plus sign the printed conditions do not, as written, give a cover (p. 318), as noted above.

Accordingly, the paper supplies a template for constructing Sierpiński numbers with finite covers, and its examples with two distinct prime divisors in every term all come from such covers. It does not construct a Sierpiński number without a finite covering set, nor does it prove that every Sierpiński number has one. In particular, its proved exponent ranges do not cover fourth powers: Section 3 explicitly leaves $r=4$ among the interesting cases (p. 318). Thus it does not settle E1113, and it proves no result for the fourth-power family $k^42^n+1$.

Results

Page numbers are those of the journal print (pp. 310–319).

• Theorem 1 (p. 311): for odd $r$, the odd $k$ with every $k^r-2^n$, or every $k^r2^n+1$, having at least two distinct prime factors ($n\ge1$) contain an infinite arithmetic progression.
• Corollary 1 (p. 311; proof p. 315): infinitely many primes $p$ have the same properties for $p^r$.
• Theorem 2 (p. 311): the same for the exponent $2r$, $r$ odd with $3\nmid r$; the page records why the printed sketch of part (ii) does not prove it as written.
• Corollary 2 (p. 312; proof p. 318): infinitely many primes $p$ with the properties of Theorem 2 for $p^{2r}$.

Read status. Claims checked for the four results above, read clause by clause on the print; the proofs of Theorem 1(i) and Theorem 2(i) were read for their structure, the plus-sign proofs are the paper's sketches, and the finite verification of Lemma 3 was not rerun.

Bears on

• Problem 1113: Theorem 1(ii) and Corollary 1(ii) give, for each odd $r$, infinitely many Sierpiński numbers $k^r$, including $p^r$ with $p$ prime, and the construction of p. 315 yields the seven-prime covering set deduced above; Theorem 2(ii) and Corollary 2(ii) assert the same for $k^{2r}$ with $3\nmid r$, on a printed argument that does not, as written, supply a cover. No result here gives a Sierpiński number without a finite covering set, and none decides the problem.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.