For what $\displaystyle n$ is $\displaystyle \sum^n_{k=1} k!$ a square?
Hello, AfterShock!
You found the only two solutions . . .
Note that squares must end in: $\displaystyle 0,\,1,\,4,\,5,\,6,\,9$
List the first few sums:
$\displaystyle \begin{array}{cccc} 1! & = & \boxed{1}\\ 1!+2! & = & 3\\ 1!+ 2!+3! & = & \boxed{9}\\ 1!+2!+3!+4! & = & 33\end{array}$
Since $\displaystyle 5!,\,6!,\,7!,\,\hdots$ all end in $\displaystyle 0$,
. . all subsequent sums will end in $\displaystyle 3$
. . and hence cannot be squares.
Therefore: $\displaystyle n = 1,\,3$ are the only solutions.