Re: Q:Certain group of composites and their prime factors.
- --- In firstname.lastname@example.org, "Dan" <30pack@s...> wrote:
>Are consecutive solutions to a quadratic equation f(1)=11, f(2)=13,...
> Primes of this form --
> 11 +2,+4,+6,+8,+10,+12,+14,..+n
>For any solution f(a)=p*q, there will exist the solution sequences f
> Where if at any point the sum is a composite then these composites
> will not have any of these primes in their prime factors.
(a+k*q)==0(modq) and f(a+k*p)==0(modp) for all integer k. By
quadratic reciprocity, there will also be a second set of solutions f
(a+k*q+b)==0(modq) and f(a+k*p+c)==0(modp). Generally, either no
solution sequence exists, or two distinct sequences exist. The only
way a single sequence occurs is if the factor in question divides the
second derivative of the quadratic.
In your example above, with 2nd derivative=2, all factors have two
solution sequences. The quadratic progression is symmetrical in that
f(0)=f(1), f(-1)=f(2),f(-2)=f(3),... Thus, any prime factor, q,
appearing in the progression at f(a), where a is the lowest positive
index for which f(a)==0(modq), will also appear at f(-a+1). Thus,
the next occurence of f(x)==0(modq) for positive a<x<a+q appears at
x=q-a+1 (for your example sequence). Note that the two smallest
solutions appear 2a-1 places apart in the sequence (one with negative
index, one with positive index), thus the two smallest solutions with
positive index appear q-2a+1 places apart (which added to a, gives q-
a+1 as stated prior).
Of interest, relative to quadratics with 2nd derivative=2, f(a)*f(a+1)
=f(a+f(a)). Generalizations of this principle to quadratics with 2nd
derivative other than 2, can be made to characterize and calculate
other composites which must appear as members of the quadratic
> 2,3,5,7,19,29,37,61,71,73,89,113..?? . There are many more, but IThese are merely the primes which do not appear in the quadratic
> end it @ prime 113.
progression as factors. We are guaranteed to have infinitely many
such "missing" prime factors as long as our quadratic progression
does not contain 0 as an integral solution (generally, we also assume
it is not trivially multiplied through by a common factor)
When zero appears in a quadratic progression, all positive integers
appear as factors. To prove this, simply set f(0)=0 where it appears
and then it is always true that f(a)==0(mod ABS[a]).
>It's all relative and well defined by quadratic reciprocity. In
> I am not sure if this is true of false!
> If it is true, why?
fact, if you would like to create a quadratic progression for which
11 is the only prime factor less than 113, consider f(x)=x^2-x+11
(your example sequence), and then take three equally spaced solutions
and assign them as three consecutive solutions of a new quadratic
progression given by g(k) such that,
where m = 13*17*23*31*41*43*47*53*59*67*79*83*97*101*103*107*109
Note that the quadratic progression given by g(k) is a subset of that
given by f(x).
The sequence of 3 consecutive seed values completely defines g(k) and
can be expressed in form g(k)=A*k^2+B*k+C by solving
From there, you could take any g(a) not divisible by 11 and define a
new partial set of the previous progression, and derive a new
quadratic based on consecutive values g(a),g(a+11), g(a+22). This
newly derived quadratic will yield a progression of integers for
which none have a prime factor less than 113.
It's appealing as a means to guarantee a sequence without small
factors, but the values grow enormous quickly, the equations become
unweildly, and in the scheme of things, you're not really improving
your probability of finding a prime by very much.