For which positive integer values of n is 1! + 2! + ... + n! a square?
The answer is when n = 1 or 3.
The first four cases are by direct calculation.
Let S(n) = 1! + 2! + ... + n! for n = 1, 2, ...
S(1) = 1, a square
S(2) = 1 + 2 = 3, not a square
S(3) = 3 + 6 = 9, a square
S(4) = 9 + 24 = 33, not a square
When n >= 5, n! is a multiple of 5.
That means that when S(n) is divided by 5, it will leave the
same remainder as when S(4) is divided by 5, namely 3.
However, no square can leave a remainder of 3 when divided by 5 (just
try sqaring the five possible cases 5k, 5k+1, 5k+2, 5k+3 and 5k+4).
Hence, S(n) is never a square for n >= 5, and so the only cases for
which it is a square are n = 1 and n = 3.