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Statement

Equation (1) of the paper (p. 893) is $n!+1=p_k^{a}p_{k+1}^{b}$ with $a\ge0$, $b\ge0$ and $p_{k-1}\le n<p_k$, where $p_k$ is the $k$th prime; see the Theorem.

Lemma (p. 893, unnumbered, quoted). "In equation (1), one has $ab\neq0$."

The Lemma is stated under the paper's standing assumption, made just before Section 2 (p. 893), that the solution of (1) has $n\ge12$; the proof uses $n\ge12$ to find two primes in $[n+1,2n]$ and ends in a contradiction with it. Read with that hypothesis: if $n\ge12$ and $p_{k-1}\le n<p_k$, then $n!+1$ is not of the form $p^{a}$ with $p\in\{p_k,p_{k+1}\}$. Without it the statement fails: $4!+1=5^2$ with $p_2=3\le4<5=p_3$.

Source. F. Luca, On a conjecture of Erdős and Stewart, Math. Comp. 70 (2001), no. 234, 893--896, DOI 10.1090/S0025-5718-00-01178-9; Section 2, the Lemma on printed p. 893, its proof on pp. 893--894, read in the journal's printing recorded on the source card.

Read depth. Claims checked: the statement and the standing assumption were read clause by clause on the page image. The proof was read for its structure and not checked.

Proof pointer

Pp. 893--894. Suppose $n!+1=p^{a}$ with $p\in\{p_k,p_{k+1}\}$ and write $a=2^{i}a_1$ with $a_1$ odd. The $2$-adic valuation of $p^{a}-1$ is at most $\log_2(p_{k+1}+1)+\log_2 a$ (display (3)); $p^{a}<n^{n}$ and $p>n$ give $a<n$, and two primes in $[n+1,2n]$ for $n\ge12$ give $p_{k+1}+1\le2n$, so $\operatorname{ord}_2(n!)<2\log_2 n+1$ (display (5)). Against $\operatorname{ord}_2(n!)\ge n-\log_2(n+1)$ (display (6), which the paper cites as Lemma 1 of its reference [1]) this forces $n\le11$.

Dependencies

The lower bound (6) for $\operatorname{ord}_2(n!)$, cited on p. 894 as "Lemma 1 in [1]", the paper's reference to Y. Bugeaud and M. Laurent, J. Number Theory 61 (1996), 311--342; and the fact, used without reference, that $[n+1,2n]$ contains at least two primes for $n\ge12$.

Bears on

• Problem 1058: the Lemma excludes, for $n\ge12$, the solutions in which $n!+1$ is a power of a single one of $p_k$, $p_{k+1}$. It enters the problem only through Case 2 of the proof of the Theorem (p. 895), which answers it; on its own it does not decide the problem.