- Aug 6, 2012
>> If x and y are relatively prime, the conjecture would be modified to

More succint? Yes. As strong? No. Mark's conjecture characterizes

>> this: If A is the set of all numbers of the form 5x^2 + 5xy + y^2, with

>> x and y relatively prime, then A contains *only* and *all* of those

>> numbers which are +/- 1 mod 10 primes raised to an odd power, the prime

>> 5 to the first power, and the composites that can be made from their

>> products.

>

> Perhaps. I thought my statement more succinct though.

(a slightly different) set A fully (i.e. he describes what *is* and

what *is not* in it), while yours provides a condition which sufficient,

but not necessary for an integer to be member of A.

>>> I bet it could be proven.

Depending on how understands Mike's statement, it might be

provable, or it might be false :-)

Mike says that:

(1) A contains +/- 1 (mod 10) primes raised to odd powers.

(2) Prime 5 raised to first power.

(3) Composites made from products of (1) and (2).

(4) Nothing other than required by (1), (2) and (3).

The (potential) problem lies in interpretation of (3).

In order to simplify the notation, let F(x,y) = 5x^2 + 5xy + y^2.

Since F(3,4) = 11^2, 11^2 belongs to Mike's set. Since it's not

covered by conditions (1) and (2), it must have been introduced by

condition (3) (applied to numbers 11 and 11, which belong to the

set by (1)). The problem is that condition (3) cannot be applied

the same way to 5 and 5 (as introduced by condition (2)); the number

25 can not be expressed as F(x,y) with relatively prime x and y.

My take on the more rigorous phrasing goes like this:

The set A = { F(x,y) | gcd(x,y) = 1 } contains integer T

if and only if:

- All (prime) factors of T are of the form +/- 1 (mod 10), OR

- T is divisible by 5 and all (prime) factors of (T/5) are of

the form +/-1 (mod 10).

If we omit the gcd(x,y) = 1 condition, the set B = { F(x,y) }

will be the same as set A, plus multiplication by any square.

> Perhaps again, but I think its status will remain

There is no "presupposing" about qfbsolve(). The good doctor

> 'empirical observation' for quite some time. Since

> the qfbsolve procedure presupposes that it is true,

> and the good doctor declined to comment upon it.....

David did provide a hint that qfbsolve() works for ALL primes

P of the form described above and also provided a pointer to

the relevant book. Considering that Pari/GP is open-source,

you could also look at the guts of qfbsolve(), see how (easy)

and why (difficult) it works with your own eyes :-)

If I recall correctly, it uses a modified Cornacchia

algorithm (http://en.wikipedia.org/wiki/Cornacchia%27s_algorithm)

to either find a representation of any prime P or to declare

that prime P cannot be represented by that quadratic form.

The good doctor also provided a handy trick to deal with products:

If S = F(a,b) and T = F(c,d), then ST = F(ad - bc, 5ac + 5bc + bd).

This, together with the trivial observations F(1,0) = 5 and

F(0,y) = y^2 shows that the set B contains all the numbers

described above (i.e. the "if" part of the equivalence).

It takes a bit more effort to do the same for set A; mainly

because the "product" rule doesn't preserve the "relatively

prime" property, but unless I overlooked something, this part

should be doable too, without too much effort.

Finally, let's dispel some of the misconceptions related to qfbsolve:

> qfbsolve(Q,p)

Nope. The qfbsolve algorithm works and there is a proof

>

> [snip]

>

> The algorithm used runs in probabilistic polynomial

> time in p (through the computation of a square root of D

> modulo p); it is polynomial time in D if Q is imaginary,

> but exponential time if Q is real (through the computation

> of a full cycle of reduced forms). In the latter case,

> note that bnfisprincipal provides a solution in heuristic

> subexponential time in D assuming the GRH.

> --

>

> As to the status of proofs of conjectures relating

> to qfbsolve, my guess is that many of them are still

> empirical.

that it works for all primes and all quadratic forms (giving

you either the representation or a telling you with

certainty that one doesn't exist). What we do not know is how

*long* it takes to find the solution. In case of "nice" forms

(negative discriminant) we know it runs in polynomial time.

In case of "ugly" forms (positive discriminant, such as

the currently discussed form), it can take longer. The runtime

is still finite, though -- in fact, it's bounded by

an exponential function.

So no, there is no "still empirical" here -- the existence

has been proved constructively (via a resonably fast algorithm).

Peter - << Previous post in topic Next post in topic >>