## 990Factor Record

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• Apr 25, 2001
Hi to all,
I conclude the discovery of irreducible a priori `factorised'
closed forms with a bit more theory and an interesting question.

We already have that q ( = 2*p + 1) / (2p)! + 1.
Now if we look for primes of the form q = 3*p + 1 then we can use
Wilson's theorem to obtain (q-1)! == -1 mod q or
(3p)! + 1 == 0 mod q or q / (3p)! + 1
Similarly we can look for primes of the form q = X*p +1 and
the new `factored' form is q / (X*p)! + 1 . We can even look
for primes of the form
q = X*p + Y X, and Y integers. Then the factored form is
q / (X*p + (Y-1))! + 1

So we now have a `large' number of new primes to look for, of the
form q = X*p + Y
and each new record prime of this form (Generalised Sophie Gemain)
will enable a still larger number to be factored `a priori.'

My question is this , can anyone prove there are in fact an
infinite number of primes of the form q = 2*p +1 ? (Sophie
Germain) or indeed of the form q = X*p + Y?

From David Burton's book if p and q are twin primes there is a
formula,
p*q / 4((p-1)! + 1) + p which is a `doubly factored' form
p*q / f(p)
And p and q are twin primes. So Phil Carmody and David Underbakke
can claim a new world record factorisation by plugging their record
twin into this form!

Finally, to end on a lighter note and also with another question and
possibly a new standard. H.E. Rose writes in his famous book
p12 `For a typical 500 digit integer it would take more than a
lifetime to find its factors.' My question is, how long would it
take to factor ANY random 500 digit integer today? Perhaps this can
be called the `Rose500' benchmark in honour of H.E. Rose.

Regards,
Paul Mills,
Kenilworth,
England.
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