Pete Inglesby from Wolverhampton; Patrick Snow, age 16, Dame Alice Owens School, Potters Bar; Yatir Halevi, age 18, Maccabim-Reut High-School, Israel; Andrei Lazanu, age 12, School 205, Bucharest, Romania all solved this problem. Here is Pete's solution:

Consider numbers of the form

$$ u(n) = 1! + 2! + 3! +...+ n!. $$

How many such numbers are perfect squares?

In fact only when $n=1$ or $n=3$ does $u(n)$ give a perfect square (1 and 9 respectively).

The solution lies in the fact that all factorials above $5!$ are multiples of 10. This is because they all have 2 and 5 as prime factors. For $n> 5$

n!
=1 ×2×3×4×5×¼×n
= 2×5(1×3×4×¼×n)
= 10(1×3×4×¼n).

When adding a multiple of 10, to another integer, k, the last digit of k will remain the same. Suppose k = 10j + l and m = 10n, with l and n integers, then k + m = 10(j + n) + l.

So when adding all factorials equal to or higher than 5! then you are adding a multiple of ten, so the last digit of the sum will be the same. This is important because the last digit of u(4) is a 3. We have u(4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 26 = 33. Therefore for any u(n) with n > 5 the last digit will be 3. For any integer p = (10a + b) with a and b integers and b < 10, p2 = (100a2 + 20ab + b2). Therefore the last digit is of p2 is only dependent on b2.
b b 2 Last digit of b 2
0 0 0
1 1 1
2 4 4
3 9 9
4 16 6
5 25 5
6 36 6
7 49 9
8 64 4
9 81 1

This table shows that the squares of all integers between 0 and 9, and therefore all integers, end in either 1, 4, 5, 6 or 9.

No perfect squares end in 3. Therefore, u(n) with n > 4 can never be a perfect square as it has been shown to always end in 3.

All that remains is to find u(n) for all integers of 4 or less:

u(1) = 1! = 1;

u(2) = 1! + 2! = 3;

u(3) = 1! + 2! + 3! = 9;

u(4) = 1! + 2! + 3! + 4! = 33.

Only u(1) and u(3) are perfect squares, and so they are the only sums of factorials to be perfect squares.