1!+2!+3!=9=32.
Prove that k=3 is the only case of the sum 1!+2!+3!+...k!
resulting in an integer power of an integer number.
(In reply to
The easy half by Jer)
My previous comment is basically dealing with mod 10 to rule out squares.
I haven't found a method that works for all prime powers, but with different moduli one can rule out different powers.
For example, using mod 7, it's easy to rule out cubes.
|
Posted by Jer
on 2015-02-25 10:43:51 |