Determine all possible triplets of integers (n,m,k) satisfying 1!+2!+3!+...+n!=m^k, where n, m and k are greater than 1.

In order to be a perfect power, the expression must be 0 mod m. Factorials from m on will be 0 mod m, so 1!...(m-1)! must be 0 mod m.

Also, the factorial expression can be written as 1*(2*(3*...*(n+1)...+1)+1) which equals 1!+2!+3!+...n!

The inside equals a perfect square, thus = 1!+2!...+(n-3)!+nē(n-2)!

Edited on February 1, 2007, 11:27 pm

Posted by Gamer on 2007-02-01 23:25:38