Re: Small prime divisors of very large numbers
- --- In firstname.lastname@example.org,
"Mike Oakes" <mikeoakes2@...> wrote:
> Then b(q,n+1) = b(q,n) mod p,Err, Mike, how does the second line follow from the first?
> so q^b(q,n+1) = q^b(q,n) mod p,
If x = y mod (p-1)
then q^x = q^y mod p,
in my book.
Has one of us made a big boob?
- Dear all,
I come back to this topic, looking for a working "pow_mod".
but 2^2^2 % 5 = 16 % 5 = 1.
(and idem for 2^2^2^2 - of course.).
David gave some other code, in
Below is a Pari-GP procedure "pmod(a,m)" to compute
a^(a^(a^ ... ^(a[k-1]^a[k]) ... ) modulo m
where the modulus "m" need not be prime.
Although this may work for prime m
(at least, pmod([2,2,2],5) = Mod(1,5) as should),
= Mod(1, 4)
which is most certainly wrong.
So, to put it short, has anyone a working pmod() in his "library" ?
Thanks in advance!
--- In email@example.com, "David Broadhurst"
I consider that his code should not be trusted.