Determining all odd primes that can be expressed in the form $x^2+xy+5y^2$

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Determine all odd primes that can be expressed in the form $x^2+xy+5y^2$.

Its discriminant $d=-19$. And it is in its reduced form. But how to approach to find all such odd primes. any suggestions?

Please provide a hint based on quadratic forms only, as I am doing an elementary number theory course.

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This is the norm form of the number field $ K = \mathbf Q((1 + \sqrt{-19})/2) $, which has class number $ 1 $. Therefore, the primes represented by this form are precisely the primes that are either split or ramified in $ \mathcal O_K $. $ 2 $ is inert, and an odd prime is split if and only if $ -19 $ is a quadratic residue modulo that prime, which, by quadratic reciprocity, comes down to determining the residue class of that prime modulo $ 4 \times 19 = 76 $.

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$\mathbb{Q}(\sqrt{-19})$ is one of the few imaginary quadratic fields with class number one: in our case, $x^2+xy+5y^2$ is the only reduced binary quadratic form of discriminant $-19$. It follows that the numbers represented by such a quadratic form give a semigroup, and the primes represented by such a quadratic form are the primes for which $-19$ is a quadratic residue, i.e., by quadratic reciprocity, the primes belonging to some arithmetic progressions $\!\!\pmod{76}$.

For a wonderful reference, see D.A. Cox - Primes of the form $x^2+ny^2$, or these notes by M. Bates.

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If $(-19|p) = 1$ for an odd prime $p,$ this means that there is a solution to $\beta^2 \equiv -19 \pmod p.$ If $\beta$ is even, replace it by $\beta \mapsto p - \beta,$ which is now odd. We now have $$\beta^2 \equiv -19 \pmod {4p}.$$ This is a good thing. $$ \beta^2 = -19 + 4pt $$ for some (nonzero) integer $t.$ Or, $$ \beta^2 - 4pt = -19. $$ This means that the positive quadratic form $$ \langle p, \beta, t \rangle $$ has discriminant $-19.$ A finite sequence of reduction steps takes this to a reduced form. As the only reduced form is $\langle 1,1, 5 \rangle,$ we have produced a 2 by 2 matrix of integers $R$ with determinant $1.$ With $H$ the Hessian matrix of $$ p x^2 + \beta x y + t y^2 $$ and $G$ the Hessian matrix of $x^2 + xy + 5 y^2,$ we have $$ R^t H R = G. $$ Name $$ Q = R^{-1}, $$ this is also determinant $1,$ with $$ Q^t G Q = H. $$ The left column of $Q$ gives a representation of $p$ by $x^2 + xy + 5 y^2.$