On the equation \(1!^k+2!^k+\cdots+n!^k=x^2\)
Consider the equation given in the title of this paper, where parameters \(n\), \(k\), and \(x\) are positive integers, and ignore the trivial case \(n=1\). One readily verifies that triples \((n,k,x)\) given by \((2,3,3)\), \((3,1,3)\), and \((3,3,15)\) are solutions. The author conjectures that these are the only nontrivial solutions. Indeed, the author shows that no further solutions with \(n=2\) or \(n\geq 4\) are to be found, leaving only the equation \(1+2^k+6^k=x^2\). Moreover, it is shown that the number of solutions to this equation is finite, with \((k,x)=(1,3)\) and \((3,15)\) being two of them, and that if any further solutions exist, they must have \(k\equiv 3\pmod 4\).
scientific article; zbMATH DE number 1960293