1. ## Quickie #12

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

2. Originally Posted by Soroban
For what $\displaystyle n$ is $\displaystyle \sum^n_{k=1} k!$ a square?

Certainly for 1, 3, although for all n in general I'd start by using the fact

n^2 = summation(2k-1, k = 1...n)

and go from there.

3. I have a feeling 20 is one...

4. Originally Posted by anthmoo
I have a feeling 20 is one...
No, 20 is not one. I checked with Maple.

You end up with 33sqrt(2.3519995354562*10^15)

5. 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.