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.

Another result from the "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." part is that the resultant m^k must have certain limitations, up to n.

It can be proved by induction that 1!+2!+...n!<(n+1)! for all positive integers n. So if 1!+2!+...n!=m^k, then m^k=1!+...a! mod (a+1)!

For example,
m^k=1 mod 2,
m^k=(1+2) mod 6
m^k=(1+2+6) mod 24

Without finding an explicit formula for 1!+2!+...n!, it becomes hard to utilize this for larger n though.

Posted by Gamer on 2007-02-01 23:46:35