Representations of some primes as $3x^2-4y^2$?

Question

I am looking for (elementary) proofs (idea of the proofs is also OK) or references to proofs of the followings:

$$ p\equiv11\pmod{12}\longrightarrow p=3x^2-4y^2 $$

Any help appreciated.

Answer 1

It occurs to me that this distinction should be made: we take primes positive. If $p \equiv 11 \pmod {12},$ we have an expression $3 x^2 - 4 y^2 = p.$ However, if prime $q \equiv 1 \pmod {12},$ we have an expression $3 x^2 - 4 y^2 = -q.$

I will give the outline of the method I like, which is in Buell, Binary Quadratic Forms: Classical Theory and Modern Computations.

Your form, $\langle 3,0,-4 \rangle$ is integrally equivalent to the "reduced" form $\langle 3,6,-1 \rangle$ as well as the form $\langle -1,0,12 \rangle.$ The only other class in this discriminant is its negative, $\langle -3,0,4 \rangle$ or reduced $\langle -3,6,1 \rangle$ as well as the form $\langle 1,0,-12 \rangle.$

$\langle \alpha,\beta,\gamma \rangle$ refers to $$ f(x,y) = \alpha x^2 + \beta x y + \gamma y^2. $$

Oh, it turns out that an indefinite form $\langle \alpha,\beta,\gamma \rangle$ is reduced if and only if both $$ \alpha \gamma < 0 \; \; \; \mbox{AND} \; \; \; \beta > |\alpha+ \gamma|. $$ I have proved this on this site; it is also in a book, still in preparation, by Franz Lemmermeyer.

Now, with your (positive) prime $p \equiv 11 \pmod {12},$ we can calculate Legendre symbol $(48|p)= 1.$ This means there is an integer solution to $\beta^2 \equiv 48 \pmod p.$ If $\beta $ is even, we take $B = \beta.$ If the $\beta$ we found, we take $B = p - \beta.$ Either way, we now have $B$ even, $B^2$ divisible by $4,$ and $$ B^2 \equiv 48 \pmod {4p}. $$ That means that $$ B^2 = 48 + 4pT $$ and $$ B^2 - 4pT = 48. $$ That is, there is a form $$ \langle p, B, T \rangle $$ of discriminant $48.$ We follow the reduction algorithm in Buell, originally due to Gauss and Lagrange, and eventually arrive at $\langle 3,6,-1 \rangle.$ It cannot be the other one, because $x^2 - 12 y^2$ does not represent anything $3 \pmod 4$ or $2 \pmod 3.$ Let's see, a single further step takes $\langle 3,6,-1 \rangle$ to $\langle 3,0,-4 \rangle.$ That is, we have created a two by two matrix of integers, determinant $1,$ that takes $ \langle p, B, T \rangle $ to $\langle 3,0,-4 \rangle.$ The inverse of that matrix takes $\langle 3,0,-4 \rangle$ to $ \langle p, B, T \rangle ,$ and the left hand column shows how to write $3 x^2 - 4 y^2 = p.$

I should probably emphasize how the matrices and the forms (as triples of coefficients) compare: the transformation $$ \langle a,b,c \rangle \mapsto \langle c,-b + 2 c \delta,a- b \delta + c \delta^2 \rangle $$ is the same as the matrix identity $$ \left( \begin{array}{rr} 0 & 1 \\ -1 & \delta \end{array} \right) \left( \begin{array}{rr} 2a & b \\ b & 2c \end{array} \right) \left( \begin{array}{rr} 0 & -1 \\ 1 & \delta \end{array} \right) = \left( \begin{array}{rr} 2c & -b+2c\delta \\ -b+ 2 c \delta & 2a - 2 b \delta + 2 c \delta^2\delta \end{array} \right) $$

Answer 2

Using quadratic number fields the basic idea is this: primes $p \equiv \pm 1 \bmod 12$ split in $K = {\mathbb Q}(\sqrt{3})$. Since $K$ has class number $1$, the prime ideals above $p$ are principal, hence $x^2 - 3y^2 = \pm p$. Now clearly $p \equiv 1 \bmod 12$ implies that the plus sign holds, whereas for $p \equiv 11 \bmod 12$ the minus sign must hold: $x^2 - 3y^2 = -p$. Looking at this equation modulo $4$ will tell you that $x$ is even.