0
$\begingroup$

I would like to prove the following statement:

If $s>1$ is a positive integer and

  • $s\equiv0$ modulo 3 and
  • $s\equiv0$ modulo 4 and
  • $2s+1$ is prime

then $2s+1 = x^{2}+24y^{2}$ for some nonzero positive integers $x$ and $y$.

For example consider the number $36$. Then $36\equiv0$ modulo 3 and $36\equiv0$ modulo 4 and $2*36+1=73$ is prime. Then we can write $73=x^{2}+24y^{2}$. Here we take $x=7$ and $y=1$. We can check that $7^{2}+24*1^{2} = 49+24=73$.

24 is an Idoneal number. According to this paper the infinitude of primes of the form $x^{2}+24y^{2}$ has been established. Now I am certain that $12|s$ and so we can write $s=12t$ for some integer $t$. Then $2s+1 =2(12t)+1=24t+1$ which is congruent to $1$ modulo $24$. If we can establish the statement above then $24t+1=x^{2}+24y^{2}$. Compare this with A107008 and read Sloane's comment and question.

The sequence of $s$ that solve the statement are $36,48,96,120,156,168,204,...$. This sequence is not in Sloane's database. The sequence $t$ that solve the statement above are $3,4,8,10,13,14,17,...$ which are numbers $t$ such that $24t+1$ is prime, A107008

$\endgroup$
3
  • 1
    $\begingroup$ Knowing Sloane, it was not a question, it was more a comment about how the sequences in the database should be labelled $\endgroup$
    – Will Jagy
    Mar 2 '16 at 20:04
  • $\begingroup$ Well no update to his comment in 8 years. But now we know. $\endgroup$ Mar 2 '16 at 20:59
  • 1
    $\begingroup$ I looked at your recent question on the modular group. For your interests, I think you would get some real benefit from Buell, Binary Quadratic Forms, Classical Theory and Modern Computations. I have worked up some wonderful software (C++) using just the simple suggestions in his book. springer.com/us/book/9780387970370 $\endgroup$
    – Will Jagy
    Mar 2 '16 at 21:29
1
$\begingroup$

It's true, of course. The class group of discriminant $-96$ has four classes (binary forms up to equivalence) $$ x^2 + 24 y^2, $$ $$ 4 x^2 + 4 xy+7y^2, $$ $$ 5 x^2 + 2 xy + 5 y^2, $$ $$ 3 x^2 + 8 y^2. $$ Each form is alone in its genus, which can be seen by congruences $\pmod 3$ and $\pmod 8.$

Any prime $p$ such that $(-96|p) = 1$ can be written as one of these forms. We also need to check the primes that divide $96,$ as $2$ is not represented by a primitive form, but $3$ is. When $p \equiv 1 \pmod {24},$ we can express $p = x^2 + 24 y^2.$ When $p \equiv 7 \pmod {24},$ we can express $p = 4x^2 +4xy + 7 y^2.$ When $p \equiv 5 \pmod {24},$ we can express $p = 5x^2 +2xy + 5 y^2.$ When $p=3$ or $p \equiv 11 \pmod {24},$ we can express $p = 3x^2 + 8 y^2.$

Most of the students here seem to ask about using the geometry of numbers, Minkowski's bound and so on, to find such expressions. I like this better: given $p \neq 2,3$ and Legendre symbol $(-96|p) = 1,$ we can solve $$ b^2 \equiv -96 \pmod p. $$ By insisting that this $b$ be even, possibly by switching to $p - b$ if we began with $b$ odd, we can solve $$ b^2 \equiv -96 \pmod {4p}. $$ This means $$ b^2 = -96 + 4 p t, $$ or $$ b^2 - 4 p t = -96.$$ That is, the binary quadratic form $\langle p,b,t \rangle$ or $$ f(x,y) = p x^2 + b x y + t y^2 $$ has discriminant $-96.$ Gauss reduction takes this to one of the four reduced forms above, and since $p \equiv 1 \pmod {24},$ it is actually equivalent to $x^2 + 24 y^2.$ The two by two reduction matrix, in $SL_2 \mathbb Z,$ inverted, tells us how to express $p = u^2 + 24 v^2.$

$\endgroup$
3
  • $\begingroup$ I think I follow the argument. I guess I should know that $(-96|2s+1)=1$? if I am off and not following let me know. But with that said your argument looks good. Also is it worth submitting the sequence $s= 36,48,96,120,156,168,204,....$ this is more of an opinion question but just curious. Thanks. $\endgroup$ Mar 2 '16 at 20:39
  • 1
    $\begingroup$ @AnthonyHernandez as $p \equiv 1 \pmod 3$ and $p \equiv 1 \pmod 8,$ we have $(-1|p) = 1,$ $(2|p ) = 1,$ and $(3 | p) = 1.$ So $(32|p) = 1$ and $(-96|p) = 1.$ $\endgroup$
    – Will Jagy
    Mar 2 '16 at 20:45
  • $\begingroup$ Cool! The origin of the question comes from a finite geometry of arrangements. $\endgroup$ Mar 2 '16 at 20:51

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.