- Can I interpret Dickson's conjecture as saying, that for any number of

distinct arithmetical progressions of positive integer terms, providing

some terms in each AP are prime, that there is a common integer

multiplier for the constant difference in each AP, such that the

resulting terms are all prime?

Thanks folks.

Bill Sindelar - On 2/6/07, w_sindelar@... <w_sindelar@...> wrote:
> Can I interpret Dickson's conjecture as saying, that for any number of

Well, I'm no expert, but I think that sounds almost but not quite right.

> distinct arithmetical progressions of positive integer terms, providing

> some terms in each AP are prime, that there is a common integer

> multiplier for the constant difference in each AP, such that the

> resulting terms are all prime?

> Thanks folks.

> Bill Sindelar

Let me try to say it in math.

> distinct arithmetical progressions of positive integer terms,

OK, so far so good.

a_1 n + d_1

a_2 n + d_2

...

a_k n + d_k

> providing some terms in each AP are prime,

That's not enough.

For instance, suppose that you take

2n+3

2n+5

2n+7

Well, clearly they'll all be prime pretty often, since they all cover

each odd number > 7.

But there's equally clearly no value of n that makes them simultaneously prime.

The "obvious" criterion is that if you take the product of the first

term of each arithmetic progression, call it p_1, and so on with p_2,

p_3, and so on,

then there must be no prime p that is a divisor of p_n for ALL n.

In my 2n+3, 2n+5, 2n+7 example, the product

(2n+3)(2n+5)(2n+7) = 8n^3+60n^2+142n+105 is "obviously" always

divsible by 3, so it's no good.

This is the part where you need to be very careful in your statement

-- what is it that rules out certain sets of values?

I think a good statement is that the sequence of products of terms of

your arithmetic progressions must have GCD = 1.

> that there is a common integer

In other words, a particular value of n,

> multiplier for the constant difference in each AP,

> such that the

so that a_1 n + b_1, ..., a_k n + b_k,

> resulting terms are all prime?

are all primes for that one value of n.

Indeed.

And I think that if there's always one such value, there must always

be infinitely many such values, because once you've found one such n,

you can re-set the b's (by adding a_i * n to each b_i) and then

generate another value of n, and another ...

But this is quite a powerful conjecture! Pretty much every unsolved

problem about primes (twin primes, arithmetic progressions, etc, etc)

will be solved if this conjecture is proven.

--Joshua Zucker - Bill Sindelar wrote:
> Can I interpret Dickson's conjecture as saying, that for any number

No, as Joshua Zucker explained.

> of distinct arithmetical progressions of positive integer terms,

> providing some terms in each AP are prime, that there is a common

> integer multiplier for the constant difference in each AP, such

> that the resulting terms are all prime?

A simple counter example is 3n+1 and 3n+2 (one of them is always

even).

See http://primes.utm.edu/glossary/page.php?sort=DicksonsConjecture

for an exact statement of Dickson's conjecture:

Given a family of linear functions with integer coefficients a_i > 1

and b_i:

a_1*n + b_1, a_2*n + b_2, a_k*n + b_k,

then there are infinitely many integers n > 0 for which these are

simultaneously prime unless they "obviously cannot be" (that is,

unless there is a prime p which divides the product of these for all

n).

(The exact statement is not "obviously cannot be", but the

explanation of that: "unless there is a prime p which divides the

product of these for all n".)

A brute force way to test whether there is such a prime p without

using modular arithmetic:

For each prime p <= k, test whether p divides the product of the

linear functions for all n from 0 to p-1.

The condition of the conjecture is satisfied iff there is no such p.

In the above example, 2 divides (3n+1)*(3n+2) for both n=0 and n=1,

so the condition is not satisfied.

Zucker gave the example 2n+3, 2n+5, 2n+7.

3 divides (2n+3)*(2n+5)*(2n+7) for n = 0, 1, 2.

Note however that a case of simultaneuos primes may be possible if

one of the primes is p itself. For example, 2n+1, 2n+3, 2n+5 are all

prime for n=1, although 3 always divides their product. (If n=0 is

allowed then Zucker's example gives all primes).

A more powerful conjecture about polynomials which includes Dickson's

as a special case is Schinzel's Hypothesis H:

http://primes.utm.edu/glossary/page.php?sort=HypothesisH

--

Jens Kruse Andersen