For what $\displaystyle n$ is $\displaystyle \sum^n_{k=1} k!$ a square?

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- Jan 15th 2007, 12:17 PMSorobanQuickie #12
For what $\displaystyle n$ is $\displaystyle \sum^n_{k=1} k!$ a square?

- Jan 15th 2007, 12:41 PMAfterShock
- Jan 15th 2007, 12:47 PManthmoo
I have a feeling 20 is one...

- Jan 15th 2007, 12:50 PMAfterShock
- Jan 16th 2007, 11:25 AMSoroban
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.