Reduced positive definite binary quadratic form

A positive definite integral form $[a,b,c]$ is reduced when $|b|\le a\le c$, with $b\ge 0$ if $|b|=a$ or $a=c$. Every proper equivalence class has a unique reduced representative under this boundary convention.

For a positive definite integral form $[a,b,c]$, a unimodular shear can replace $b$ by its representative $b^{\prime }$ with $|b^{\prime }|\le a$. If the resulting $c^{\prime }=(b^{\prime 2}-D)/(4a)$ is smaller than $a$, swapping the variables replaces the leading coefficient by the smaller positive integer $c^{\prime }$. Repeating terminates and produces $|b|\le a\le c$; determinant-one sign changes enforce the boundary convention. Thus every positive definite integral binary quadratic form is properly equivalent to a reduced form.

For a reduced form of negative discriminant $D$, the inequalities imply $3a^2\le |D|$. One can therefore enumerate all classes by checking the finite setand retaining integral $c=(b^2-D)/(4a)$ with $a\le c$ and the boundary sign convention.

The reduced positive definite binary quadratic form bounds $|b|\le a\le c$ and $b^2-4ac=-39$ imply $1\le a\le \sqrt{13}$. Enumerating $a=1,2,3$ and using the nonnegative boundary sign convention gives the four triples $(a,b,c)=(1,1,10),(2,1,5),(2,-1,5),(3,3,4)$. Thus the number of reduced forms is four.

The reduced-form bound $3a^2\le 32$ leaves $a=1,2,3$. Checking $c=(b^2+32)/(4a)$ and the boundary sign convention gives precisely the displayed forms. The middle one is imprimitive and represents the prime two. Odd represented primes are $1$ or $3$ modulo eight by the discriminant criterion for prime representation by a binary quadratic form. The first form represents only $1$ modulo eight among odd integers, and the last only $3$, so a prime is represented by $3x^2+2xy+3y^2$ exactly when it is $3$ modulo eight.

For negative discriminant $-67$, a reduced positive definite binary quadratic form satisfies $1\le a\le \sqrt{67/3}<5$ and $b$ odd with $|b|\le a$. Checking $4a\mid b^2+67$ leaves only $(a,b,c)=(1,1,17)$ under the boundary convention. The reduction algorithm for a positive definite binary quadratic form therefore puts every form in this one proper equivalence of binary quadratic forms class. Its unimodular shears give infinitely many distinct forms $x^2+(2n+1)xy+(n^2+n+17)y^2$.

The reduced positive definite triples are $(1,0,11)$, $(2,2,6)$ and $(3,\pm 2,4)$. The imprimitive middle form represents only even numbers; the two last forms have identical value sets after reversing one variable.

The two reduced positive definite integral forms of discriminant $-35$ areFor a prime $p>7$, the first represents $p$ exactly whenand the second represents $p$ exactly when both Legendre symbols are $-1$.

For odd $n$ prime to 35, proper representation by a positive definite binary quadratic form of discriminant $-35$ is equivalent to $-35$ being a square modulo every prime divisor of $n$. A primitive representing vector extends to a determinant-one integer matrix, producing a form with leading coefficient $n$ and middle coefficient $b$ satisfying $b^2\equiv -35(\mathrm{mod}4n)$. Conversely this congruence constructs such a form. Its local roots lift to prime powers and combine by the Chinese remainder theorem. Every class reduces to one of $x^2+xy+9y^2$ and $3x^2+xy+3y^2$. The condition is local at each prime divisor; a product Jacobi symbol equal to one alone is insufficient.

The class number problem asks for number fields or discriminants having a prescribed class number.

For a negative discriminant $D\equiv 0$ or $1(\mathrm{mod}4)$, the class number $h(D)$ is the number of proper equivalence classes of primitive positive definite integral binary quadratic forms of discriminant $D$. Equivalently, it is the number of reduced primitive forms of discriminant $D$.

For an odd prime number $q$, the primitive positive definite formshave discriminant $-8q$ and lie in distinct proper equivalence classes. The first represents one, whereas the second has minimum positive value two.

If $q\equiv 3(\mathrm{mod}8)$ is prime and $h(-8q)=2$, then for every prime $p>q$,for some integers $x,y$ exactly when $p\equiv \pm 1(\mathrm{mod}8)$ and $p$ is a quadratic residue modulo $q$.

Let $d$ be odd and have $k$ distinct prime divisors. Assigning each complete prime-power factor of $d$ to one of two factors gives $2^k$ ordered coprime factorizations $d=ac$. Interchanging $a$ and $c$ identifies pairs, and each unordered factorization gives a different reduced primitive form $[\min (a,c),0,\max (a,c)]$ of discriminant $-4d$. Hence