randy@umcp-cs.UUCP (03/22/84)
Since the time this problem was posted, I have seen only one response.
In case you've forgotten, the problem was to prove that there are
no n>1 and k>1 such that
2
n! = k
Certainly, I believe it given the following conjecture:
For any integer m > 1, there exists a prime p such that m/2 < p <= m.
Can anyone prove this conjecture or offer pointers to a proof, or
solve the problem some other way?
--
Randy Trigg
...!seismo!umcp-cs!randy (Usenet)
randy%umcp-cs@CSNet-Relay (Arpanet)gmf@uvacs.UUCP (03/24/84)
From
Elementary Theory of Numbers
by W. Sierpinski (Warsaw, 1964):
p. 138:
"Corollary 4. For natural numbers > 1 number n! is not a k-th power
with k > 1 being a natural number."
Sierpinski gets this from a theorem of Tchebycheff, p. 137: "If n is
a natural number > 3, then between n and 2n-2 there is at least one
prime number." This in turn is obtained from: "If n is a natural
number > 5, then between n and 2n there are at least two different
prime numbers" (p. 137).
G. Fisher