if $37x^2-113y^2=p$ is solvable.with p a odd prime. What's the set of all $p$? Does it have a formula?
2026-04-24 02:35:24.1776998124
What's the set p,if $37x^2-113y^2=p$ is solvable,with p a prime
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A book I recommend for this material is Duncan A. Buell, Binary Quadratic Forms and USED COPIES.
This discriminant has class number two. This form is alone in its genus. Also, the principal form represents $-1,$ so there is no distinction between representing $p$ and $-p.$ This form represents $37$ and $113$ and all primes that are nonresidues $\pmod {37}$ and $\pmod {113},$ such as $5,17,19,23,29,\ldots.$ Note that $(2| 113) = 1.$
There are just two classes in this discriminant, and they are in different genera, the form $37 x^2 - 113 y^2$ and the principal form $x^2 - 4181 y^2. $ As $$ 489181478614821790^2 - 4181 \cdot 7565365622404889^2 = -1, $$ any number $n$ is represented by one of these forms if and only if $-n$ is represented by the same form.
Here are the Lagrange cycles. The first coefficient in each triple is a number that is represented primitively. A primitive representation is one with $\gcd(x,y)=1.$ For example, we can solve $x^2 - 4181 y^2 = 4$ with both $x,y$ odd, so that $\gcd(x,y)=1$ in this case, as $$ 985522780227^2 - 4181 \cdot 15241460455^2 = 4. $$ Alright, for each triple $\langle a,b,c \rangle$ in the cycles below, the indicated forms are "reduced." The discriminant is $\Delta = 4 \cdot 37 \cdot 113 = 16724$ so that $\sqrt \Delta \approx 129.32.$ It follows from the definition of reduced that $|a|, b, |c| < 130.$ It is a theorem of Lagrange, Theorem 85 on page 111 of Introduction to the Theory of Numbers by Leonard Eugene Dickson, that any number $n$ that is primitively represented by one of these forms, with $|n| < (1/2) \sqrt \Delta \approx 64.66,$ occurs as the first coefficient of a form in that cycle.