[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Classic Math gone wrong...Re: (n!+1)^(1/2)
> >For any number n, if the square root of (n!)+1 is an integer, it is also
> >prime. (This is interesting, but rather useless in practice)
>
>For any number a, 1<a<=n, n! mod a == 0; therefore, n!+1 mod a == 1. n!+1
>is prime. Prime numbers don't have integral square roots.
You're getting things missed up with the classic proof that there is
no largest prime number. This doesn't hold in general. Try a=5.
5!=5*4*3*2*1=120. 120+1=121. 121=11*11.
The classic proof goes:
Is there a largest prime number?
If there is then collect all primes, p1...pn and multiply them
together p=p1*p2*...*pn. p+1 is not divisible by p1...pn. Therefore
p+1 is a prime. Therefore there is no largest prime number.
>
>
>Scott Collins | "That's not fair!" -- Sarah
> | "You say that so often. I wonder what your basis
> 408.862.0540 | for comparison is." -- Goblin King
>................|....................................................
>BUSINESS. fax:974.6094 R254(IL5-2N) [email protected]
>Apple Computer, Inc. 5 Infinite Loop, MS 305-2D Cupertino, CA 95014
>.....................................................................
>PERSONAL. 408.257.1746 1024:669687 [email protected]