The primes $2s+1$ with the constraint that $s$ satisfy certain congruence relations and Euler's idoneal numbers.

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

• Knowing Sloane, it was not a question, it was more a comment about how the sequences in the database should be labelled Mar 2 '16 at 20:04
• Well no update to his comment in 8 years. But now we know. Mar 2 '16 at 20:59
• 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 Mar 2 '16 at 21:29

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 $p = u^2 + 24 v^2.$
• 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. Mar 2 '16 at 20:39
• @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 $(-96|p) = 1.$ Mar 2 '16 at 20:45