NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1
i.e.
n! + prime(n) != m^k


I assume "integral" really meant integer since no differential is given.

n=3

then

prime(3) = 2

substituting:

n! + prime(n)! = 3! + 2! = 8

2^3 = 8

Therefo

n! +prime(n)! = m^k

where n = 3, m =2 and k = 3

All integers and k1.

If I read you correctly, you just proved his conjecture is false. So all
those pointy headed mathematicians are running about trying to prove it to
be true. Should we tell them or let them run about? BTW, a simple, yet
elegant method on your part.

Mike

"Charles Momsen" wrote in message
...

NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1
i.e.
n! + prime(n) != m^k


I assume "integral" really meant integer since no differential is given.

n=3

then

prime(3) = 2

substituting:

n! + prime(n)! = 3! + 2! = 8

2^3 = 8

Therefo

n! +prime(n)! = m^k

where n = 3, m =2 and k = 3

All integers and k1.