# [Proof] Complete square

1. Dec 16, 2011

### Karamata

1. The problem statement, all variables and given/known data
Proof that for $$n\geq 5$$ expression $$\sum_{i=1}^{n}i!$$ can't be a complete square

3. The attempt at a solution
Mathematical induction maybe?
$$n=5$$ $$\sum_{i=1}^{5}i!=1+2!+3!+4!+5!=152$$ OK
$$n\rightarrow n+1$$ $$\sum_{i=1}^{n+1}i!=\sum_{i=1}^{n}i!+(n+1)!$$
...
One guy said that
But, I don't understand him.

Sorry for bad English.

2. Dec 16, 2011

### Oster

1+2+6+24+120 = 153 right?
subsequent factorials are all divisible by 10, so dividing by 5 will always give the remainder 3.