OT - do you have the solution?
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.
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