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Claim. A. M. Naciri, On the Brocard–Ramanujan equation with 7-free integers and prime powers, Integers 25 (2025), #A71, Theorem 1: (i) for every $k\geq2$ the equation $n!+1=x^2$ has only finitely many solutions $(n,x)$ with $x\pm1$ a $k$-free number, and the only possible solutions with $x\pm1$ 7-free are $(4,5)$, $(5,11)$ and $(7,71)$; (ii) for every $l\geq2$ it has only finitely many solutions with $x\pm1$ having fewer than $l$ prime divisors, and the only possible solution with $x\pm1$ a prime power is $(4,5)$. The proof treats $x-1$ and $x+1$ alike, so in each case one of the two cofactors carries the condition. It bounds a divisor of $n!$ through Chebyshev's bounds for $\pi(n)$ and Legendre's formula, compares with $n!\geq(n/e)^n$, and finishes with Berndt and Galway's search, which found no further solution with $n\leq10^9$.

Covers. The problem's assertion for two families of solutions of $n!+1=x^2$: the solutions with $x+1$ or $x-1$ 7-free are exactly $(4,5)$, $(5,11)$ and $(7,71)$, and the only solution with $x+1$ or $x-1$ a prime power is $(4,5)$. Not covered: every other solution. The finiteness statements for $k$-free cofactors and for cofactors with fewer than $l$ prime divisors settle no instance. Both cases rest on Berndt and Galway's search to $n\leq10^9$. For prime powers the paper's bound $n\leq4e^2\cdot9^8$ is about $1.27\times10^9$, not below $10^9$ as printed; the same argument with $\omega(x\pm1)=1$ in place of $l=2$ gives $n\leq4e^2\cdot5^4<2\cdot10^4$, so the statement stands.

Acceptance. Refereed: Integers 25 (2025), #A71 (received 15 January, accepted 15 July, published 15 August 2025); library card naciri_2025. The site's commentary on a problem it has not settled is not acceptance, so reviewed is not listed.

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