violet matrix
My collegues gave me the start of the proof but I can't finish it:
"case 1) n is even
Then by Bertrand postulate ==> There exists a prime p, n/2 < p < n
So p < n, but since p < n it means that p | n! , since p appears in n!
But we can multiply by 2 and get that n < 2p < 2n ==> n < 2p ==> n < p^2 ( the case where n = 2 can be ignored since 2! = 2 which is not a square number) so by this it means that p^2 DOES NOT divide n.
Now we just have to prove that p^2 DOES NOT DIVIDE n! and this case is done(since it should divide n! if it were a square number)
IDK HERE(they used contradiction I think)
Case 2) n is odd is done the same except that we do for ( n-1 )/2 < p < n -1"
torpid vine