Re: Another prime number theorem?



luiroto@xxxxxxxxx wrote:
> Thomas Baruchel wrote:
> > Le 13-06-2005, Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> a écrit :
> > > m is prime if and only if Product_(k=2 to k=m-1)
> > > sin(Pi*m/k) <> 0 .
> >
> > well ; you mean that m/k is not an integer for any k
> > smaller than m...
> > This is a very well known assertion ; in fact it is
> > the definition of a prime number ;-)
>
> Naturally.
> Besides, as the Sin(x) is a complicated function, it
> is simpler to express the theorem in the following form:
> "m is prime if and only if, the sum (k = 2 to sqr(m)) of
> INT(m/k - INT(m/k)) = 0
>
> The function INT(x) being the integer part of x.

Not quite; you have two errors in your statement:

(1) INT(x - INT(x)) is 0 for _any_ real number x:
INT (x) <= x < INT (x) + 1
0 <= x - INT (x) < 1
so INT (x - INT (x)) = 0

(2) If you remove the outer INT, it's still not
right; but if you replace sum with prod, it is,
because you want (m/k - INT(m/k)) to be zero for
at least one k, not all of them.

Once again, the correctedversion is:

THEOREM. m is prime iff
prod(k=2..sqrt(m), m/k - INT(m/k)) > 0.

--- Christopher Heckman

.

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