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Zbl 1171.11020
Kihel, Omar; Luca, Florian
Variants of the Brocard-Ramanujan equation.
(English)
[J] J. ThÃ©or. Nombres Bordx. 20, No. 2, 353-363 (2008). ISSN 1246-7405

It is expected that the only solutions to the Diophantine equation $$n!+1=y^2\tag 1$$ are $(y,n)=(5,4),(11,5),(71,7)$ (the Brocard-Ramanujan problem). {\it M. Overholt} [Bull. Lond. Math. Soc. 25, No. 2, 104 (1993; Zbl 0805.11030)] showed that the weak form of Szpiro's conjecture (a special case of the ABC conjecture) implies that (1) has only finitely many solutions. Generalizing Overholt's result (and {\it A. Dabrowski}'s result [Nieuw Arch. Wiskd., IV. Ser. 14, No. 3, 321--324 (1996; Zbl 0876.11015)]), the second author showed that the full ABC conjecture implies that the Diophantine equation $n!=P(x)$ has only finitely many solutions [Glas. Mat. (3) 37, No. 2, 269--273 (2002; Zbl 1085.11023)]. The authors discuss some variants of the above Diophantine equations. Here are the main results of this paper (we write $k\times n$ to mean that $k$ does not divide $n)$. \par Theorem 1. The Diophantine equation $x^p\pm y^p=\prod^n_{k\times n,k=1}k$ admits only finitely many integer solutions $(x,y,p,n)$ with $p\ge 3$ a prime number and $\gcd(x,y)=1$. \par Theorem 2. Let $P\in{\Bbb Q}[x]$ be a polynomial of degree $\ge 2$. The ABC conjecture implies that the equation $P(x)=\prod^n_{k \times n,k=1}k$ has only finitely many solutions $(x,n)$, where $x$ is a rational number and $n$ is a positive integer. \par They also prove the following generalization of a result of {\it A. Dabrowski} [loc. cit.]. \par Theorem 3. Let $A$ be an integer which is not a perfect square. Then the Diophantine equation $x^2-A= \prod^n_{k \times n,k=1}k$ has only finitely many solutions.
[Andrzej Dabrowski (Szczecin)]
MSC 2000:
*11D85 Representation problems of integers
11D61 Exponential diophantine equations

Keywords: Brocard-Ramanujan problem; Diophantine equation involving factorials; ABC conjecture

Citations: Zbl 0876.11015; Zbl 1085.11023; Zbl 0805.11030

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