6
$\begingroup$

A SPECIFIC CASE:

Any prime number can be classified as either $p \equiv 1 \pmod 3$ or $p \equiv 2 \pmod 3$.

If $p = 3$ or $p = 1 \pmod 3$, then the prime $p$ can be represented by the quadratic form $ x^2 + 3y^2, x,y \in \mathbb Z.$

But what if $p \equiv 2 \pmod 3$?

Is there a quadratic form $ax^2+bxy+cy^2$ such that $p= ax^2+bxy+cy^2, $ when $p \equiv 2 \pmod 3$ where $x,y, a, b,c \in \mathbb Z$?

GENERAL CASE:

The general question is, is there a set of quadratic forms which represent all prime numbers?

We will classify the prime numbers, say, by $m$. Any prime is defined by $p \equiv i \pmod m$ where $1 \leq i\leq m-1$.

In above example, $i \in \{1, 2\}, m=3$. Let, the set of quadratic forms is $A$, then the number of elements in $A$ is at-least $(m-1)$.

QUESTION:

For a given $m$ can we find a set $A$ such that any prime $p$ can be represented by one of the quadratic form of $A$ ?

If it is possible then how? If there is a condition on $m$, what is it?

Does the question has any relation to the following theorem ?

enter image description here

One can answer only the specific case, if they wish to do so.

EDIT:

Is there a finite set of (preferably irreducible) binary quadratic forms such that every prime is represented by at least one of the forms in the set?

$\endgroup$
  • 2
    $\begingroup$ I'm fairly sure that for any primitive quadratic form and any $m>0$, the form represents infinitely many primes which are $1\pmod m$. This should follow from Chebotarev's density theorem. Therefore there cannot be such a form which only represents non-1 residues modulo a number. $\endgroup$ – Wojowu Oct 11 at 23:08
  • 3
    $\begingroup$ Actually on second reading, you didn't ask for the form to only represent the forms of specific residue. In that case there are some trivial examples, like the form $xy$, or $x^2-y^2$. I'm not sure if you can do with only irreducible forms. $\endgroup$ – Wojowu Oct 11 at 23:11
  • 4
    $\begingroup$ If a quadratic form represents $p$, then the discriminant $b^2 -4ac$ of the form is a perfect square modulo $p$. (Proof: The discriminant is invariant under change of variables. Because the form represents $p$, we can change variables so that $a=p$.) So indeed no non-split form represents all primes and sets like the set of primes congruent to $2$ mod $3$ can never be represented. $\endgroup$ – Will Sawin Oct 11 at 23:25
  • 4
    $\begingroup$ You can certainly get all primes if you are willing to accept overlap between the different quadratic forms. $\endgroup$ – Will Sawin Oct 11 at 23:37
  • 3
    $\begingroup$ I read the question as permitting overlap. I think the question is, is there a finite set of (preferably irreducible) binary quadratic forms such that every prime is represented by at least one of the forms in the set? $\endgroup$ – Gerry Myerson Oct 12 at 4:02
14
$\begingroup$

Every prime $p$ is represented by at least one of the following quadratic forms: $x^2+y^2$, $x^2+3y^2$, $3x^2-y^2$:

  • if $p=2$ or $p\equiv 1\pmod{4}$, then $p$ is represented by $x^2+y^2$;
  • if $p=3$ or $p\equiv 1\pmod{3}$, then $p$ is represented by $x^2+3y^2$;
  • if $p\equiv 11\pmod{12}$, then $p$ is represented by $3x^2-y^2$.

This follows from Lemma 2.5, Corollary 2.6, (page 26) in Cox: Primes of the form $x^2+ny^2$ coupled with the fact that $x^2+y^2$, $x^2+3y^2$, $3x^2-y^2$, $x^2-3y^2$ represent all integral binary quadratic forms of discriminant lying in $\{-4,\pm 12\}$.

Added. More generally, if an odd number of discriminants multiply to a square, then the quadratic forms of those discriminants together represent all primes coprime to those discriminants. In the example above, the discriminants were the elements of $\{-4,\pm 12\}$, and we could do without the form $x^2-3y^2$. See also this related post.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ I gave him $x^2 + y^2, x^2+ 2 y^2, x^2 - 2 y^2$ at math.stackexchange.com/questions/3820129/… and suggested he read Dickson's little 1929 book Intro to.. It would appear he refuses to do that. The claim I made there, no finite set of positive binaries could work, is something for which I lack a proof. $\endgroup$ – Will Jagy Oct 12 at 16:37
  • 2
    $\begingroup$ posted a new question here on MO about the finiteness claim. $\endgroup$ – Will Jagy Oct 12 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.