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Statement

Notation (p. 893): $p_k$ is the $k$th prime, for $k\ge1$. The paper's equation (1) is

$$n!+1=p_k^{a}p_{k+1}^{b}\quad\text{for some } a\ge0,\ b\ge0 \text{ and } p_{k-1}\le n<p_k. \tag{1}$$

The paper reports from Guy's Unsolved problems in number theory (1994), Problem A2 (its reference [3]), that Erdős and Stewart conjectured that every solution of (1) has $n\le5$.

Theorem (p. 893, unnumbered, quoted). "Equation (1) has no solutions for $n\geq6$."

So a natural number $n\ge6$ in $[p_{k-1},p_k)$ never has $n!+1$ composed only of the primes $p_k$ and $p_{k+1}$. The paper does not list the solutions with $n\le5$; they are $n=2,3,4,5$ ($3=p_2$, $7=p_4$, $25=p_3^2$, $121=p_5^2$), and $n=1$ too ($2=p_1$) once $p_0=1$ is read into the range $p_0\le n<p_1$ (a check made here, with the convention the problem's claim page adopts). Each of them has $ab=0$.

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; the Theorem on printed p. 893, its proof on pp. 893--895, read in the journal's printing recorded on the source card.

Read depth. Claims checked: the statement and equation (1) were read clause by clause on the page image. The proof was read for its structure and not checked; the two computations it reports are not reproduced in the paper and were not rerun here.

Proof pointer

The paper notes (p. 893) that a direct check rules out $5<n\le11$ and then assumes $n\ge12$.

1. The Lemma (Section 2, pp. 893--894): every solution with $n\ge12$ has $ab\ne0$.
2. Section 3 (pp. 894--895): writing $n!=p_{k+1}^{b}\bigl(p_k^{a}-(1/p_{k+1})^{b}\bigr)$, a $2$-adic lower bound for linear forms in two logarithms (Théorème 4 of Bugeaud and Laurent, with $p=2$) bounds $\operatorname{ord}_2(n!)$ from above by an explicit quantity of order $(\log n)^4$ (display (13)); against $\operatorname{ord}_2(n!)\ge n-\log_2(n+1)$ and with $p_k<p_{k+1}<2n$ this gives $n<7\,242\,116$, so $n<p_k<p_{k+1}<7.5\cdot10^6$.
3. Section 4, Case 1 (p. 895), $n>193$: a solution with $3\nmid a$ or $3\nmid b$ gives $n!+1=Ax^3$ with $A=p_k^{\delta_1}p_{k+1}^{\delta_2}$, $\delta_1,\delta_2\in\{0,1,2\}$ not both $0$, and $A$ must be a cubic residue modulo every prime $q\le193$ with $q\equiv1\pmod3$. Since $y$ is a cubic residue modulo $q$ exactly when $y^2$ is, the paper reduces to $A$ of the forms $p_k$, $p_kp_{k+1}$ and $p_k^2p_{k+1}$, and a computer search by A. Flammenkamp found no such $A$ with $193<p_k<p_{k+1}<7.5\cdot10^6$. Hence $3\mid a$ and $3\mid b$, and $n!=x^3-1^3$ with $x=p_k^{a/3}p_{k+1}^{b/3}$ is excluded by the Erdős–Obláth theorem (the paper's Theorem EO).
4. Section 4, Case 2 (p. 895), $n\le193$: by the Lemma $ab>0$, and a second computation found $n!+1\not\equiv0\pmod{p_kp_{k+1}}$ throughout this range.

Not reconstructed here.

Dependencies

• Y. Bugeaud and M. Laurent, Minoration effective de la distance $p$-adique entre puissances de nombres algébriques, J. Number Theory 61 (1996), 311--342, Théorème 4, applied on p. 894 with $\mu=15$, $\nu=10$, $c(\mu,\nu)=18$.
• Theorem EO (p. 895), attributed to P. Erdős and R. Obláth, Acta Szeged 8 (1937), 241--255: the equation $x^p\pm y^p=n!$ has no solutions with $p>2$ prime and $\gcd(x,y)=1$. The print states no exception for trivial solutions such as $2!=1^p+1^p$; none arises in Case 1, where $n>193$.
• The paper's Lemma (p. 893).
• Two computations by A. Flammenkamp, reported in Section 4 (p. 895).

Bears on

• Problem 1058: the problem asks whether only finitely many $n\in[p_{k-1},p_k)$ have $n!+1$ divisible by no primes other than $p_k$ and $p_{k+1}$. Such $n$ are exactly the solutions of (1), so the Theorem puts all of them at $n\le5$ and the answer is yes. The problem's claim page for this paper records how the corpus uses it.