3 Jul
2006
3 Jul
'06
2:08 p.m.
No, I don't have a proof, I do have an argument following from a solid conjecture. We know that for n >= 2, there exists a prime p with n < p < 2n. It also appears that for n >= 4, there exists prime p == 3 (mod 4) with n < p < 2n. I cannot prove this myself, but I am sure it is true. This conjecture implies that for n >= 7, there is a prime p == 3 (mod 4) with [n/2] < p <= n. This p occurs with exponent 1 in the prime factorization of n, so that n! cannot be of the form x^2+y^2.