Positive Definite Binary Quadratic Forms and Reduction — Examples

Let $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T_k=\left(\begin{array}{cc}1& k\\ 0& 1\end{array}\right)$. Direct substitution gives $f\mid S=(25,-214,458)$, $(f\mid S)\mid T_4=(25,-14,2)$, $((f\mid S)\mid T_4)\mid S=(2,14,25)$, then after $T_{-3}$ one gets $(2,2,1)$, after another $S$ one gets $(1,-2,2)$, and after $T_1$ one gets $(1,0,1)$.

The final form $(1,0,1)$ is reduced because $\mathrm{\mid }0\mathrm{\mid }\le 1\le 1$ and the boundary sign condition is automatic.

By repeated use of the right-action law [L1], the composite matrix is $S{T}_{4}S{T}_{-3}S{T}_1=\left(\begin{array}{cc}3& 4\\ -13& -17\end{array}\right)$, which has determinant $1$, so the single displayed substitution is exactly the product of the six moves in step 1.1.

Thus the explicit reduction algorithm indeed carries $(458,214,25)$ to the reduced form $(1,0,1)$.

Here $a\le \sqrt{4\mathrm{/}3}<2$ by [L1], so the positive integer $a$ must be $1$.

The discriminant equation gives $-4=b^2-4c$, so $4c=b^2+4$. Since the form is reduced, $\mathrm{\mid }b\mathrm{\mid }\le 1$; the choices $b=\pm 1$ make $c=5\mathrm{/}4$, not an integer, while $b=0$ gives $c=1$. Thus the only reduced possibility is $(1,0,1)$.

The form $(1,0,1)$ is primitive, so by [L2] there is exactly one proper-equivalence class of primitive positive-definite forms of discriminant $-4$. Hence [L3] gives $h(-4)=1$.

Here $a\le \sqrt{8\mathrm{/}3}<2$, so the positive integer $a$ must be $1$.

The discriminant equation gives $-8=b^2-4c$, so $4c=b^2+8$. Reducedness gives $\mathrm{\mid }b\mathrm{\mid }\le 1$; the choices $b=\pm 1$ make $c=9\mathrm{/}4$, not an integer, while $b=0$ gives $c=2$. Thus the only reduced possibility is $(1,0,2)$.

The form $(1,0,2)$ is primitive, so by [L2] there is exactly one proper-equivalence class of primitive positive-definite forms of discriminant $-8$. Hence [L3] gives $h(-8)=1$, and every such form is properly equivalent to $(1,0,2)$.

If $p=x^2+2y^2$, then reducing modulo $8$ shows $x^2\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}8)$ and $2y^2\equiv 0$ or $2(\mathrm{m}\mathrm{o}\mathrm{d}8)$, because squares modulo $8$ are $0,1,4$ and $p$ is odd. Hence $p\equiv 1$ or $3(\mathrm{m}\mathrm{o}\mathrm{d}8)$.

Conversely, suppose $p\equiv 1$ or $3(\mathrm{m}\mathrm{o}\mathrm{d}8)$. Then [L4] and [L5] give $\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=1$, so [L6] yields $\left(\frac{-2}{p}\right)=1$. By [L7], there is an integer $t$ with $t^2\equiv -2(\mathrm{m}\mathrm{o}\mathrm{d}p)$.

Set $c=(t^2+2)\mathrm{/}p$, which is an integer by step 1.2, and define $f=(p,2t,c)\mathrm{.}$ Then $\mathrm{\Delta }(f)=(2t{)}^2-4pc=4t^2-4(t^2+2)=-8,$ and $f(1,0)=p$, so $f$ represents $p$. If a divisor $d$ of $p$, $2t$, and $c$ were greater than $1$, then $d=p$ because $p$ is prime; but $p\mid 2t$ would force $p\mid t$, hence $p\mid t^2+2$, so $p\mid 2$, impossible for an odd prime. Thus $f$ is primitive. Since its leading coefficient is $p>0$ and its discriminant is $-8<0$, [L3] makes $f$ positive definite. Therefore [L1] shows that $f$ is properly equivalent to $x^2+2y^2$, and then [L2] shows that $x^2+2y^2$ represents $p$.

Steps 1.1 and 2.1 prove the claimed criterion.

Here $a\le \sqrt{20\mathrm{/}3}<3$, so $a$ is $1$ or $2$.

If $a=1$, then $-20=b^2-4c$, so $4c=b^2+20$. Reducedness gives $\mathrm{\mid }b\mathrm{\mid }\le 1$. The values $b=\pm 1$ make $c=21\mathrm{/}4$, not an integer, while $b=0$ gives $c=5$, yielding $(1,0,5)$.

If $a=2$, then $-20=b^2-8c$, so $8c=b^2+20$. Reducedness gives $\mathrm{\mid }b\mathrm{\mid }\le 2$. The values $b=0,\pm 1$ give no integer $c$, while $b=\pm 2$ gives $c=3$; the boundary rule forces $b=2$. Thus the only reduced possibility with $a=2$ is $(2,2,3)$.

The two forms of steps 2.1 and 2.2 are primitive and distinct, and [L2] shows that no other reduced primitive form of discriminant $-20$ exists. Hence [L3] gives $h(-20)=2$.

Here $a\le \sqrt{23\mathrm{/}3}<3$, so $a$ is $1$ or $2$.

If $a=1$, then $-23=b^2-4c$, so $4c=b^2+23$. Reducedness gives $\mathrm{\mid }b\mathrm{\mid }\le 1$. The value $b=0$ gives no integer $c$, while $b=\pm 1$ gives $c=6$; the boundary rule forces $b=1$. Thus $(1,1,6)$ is the only reduced possibility with $a=1$.

If $a=2$, then $-23=b^2-8c$, so $8c=b^2+23$. Reducedness gives $\mathrm{\mid }b\mathrm{\mid }\le 2$. The values $b=0,\pm 2$ give no integer $c$, while $b=\pm 1$ gives $c=3$, and both signs are allowed because $\mathrm{\mid }b\mathrm{\mid }<a$. Thus the reduced possibilities are $(2,1,3)$ and $(2,-1,3)$.

The three forms from steps 2.1 and 2.2 are primitive and distinct, and [L2] shows that no other reduced primitive form of discriminant $-23$ exists. Hence [L3] gives $h(-23)=3$.

The discriminants are $0^2-4\cdot 1\cdot 6=-24$ and $0^2-4\cdot 2\cdot 3=-24$.

The form $f$ represents $1$, since $f(1,0)=1$.

The form $g$ does not represent $1$: if $g(x,y)=1$, then $2x^2+3y^2=1$, but $y\mathrm{\ne }0$ would force the left-hand side to be at least $3$, while $y=0$ would give $2x^2=1$, impossible in integers.

Since $f$ and $g$ do not represent the same integers, [L1] shows that they are not properly equivalent.

Both forms are reduced: for each one, $\mathrm{\mid }\pm 1\mathrm{\mid }\le 2\le 3$, and no boundary clause is violated because $\mathrm{\mid }b\mathrm{\mid }<a<c$.

For every integers $x,y$, one has $g(x,y)=2x^2-xy+3y^2=f(x,-y)$. Thus $g$ represents exactly the integers that $f$ represents, and conversely.

The reduced triples are distinct because $(2,1,3)\mathrm{\ne }(2,-1,3)$. Therefore [L2] forbids proper equivalence between $f$ and $g$.

The matrix $N=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ has determinant $-1$, and direct substitution gives $f\mid N=(2,-1,3)=g$.

Thus allowing determinant $-1$ merges the two distinct proper classes from [L1]: the forms are equivalent under a unimodular substitution, but not under a determinant-one unimodular substitution.

Both forms have discriminant $60$, since $6^2-4(-6)(1)=60$ and $6^2-4(1)(-6)=60$.

The matrix $M=\left(\begin{array}{cc}-8& 7\\ -7& 6\end{array}\right)$ has determinant $1$, and direct substitution gives $f\mid M=g$. Thus $f$ and $g$ are properly equivalent.

Under Granville's positive-discriminant convention, a form of discriminant $d>0$ is reduced when $0<\sqrt{d}-b<2\mathrm{\mid }a\mathrm{\mid }<\sqrt{d}+b$. Since $7<\sqrt{60}<8$, one has $0<\sqrt{60}-6<2$ and $13<\sqrt{60}+6<14$. Hence $0<\sqrt{60}-6<12<\sqrt{60}+6$ for $f$ and $0<\sqrt{60}-6<2<\sqrt{60}+6$ for $g$, so both are reduced in that convention.

The two reduced forms $f$ and $g$ are distinct, yet step 1.2 puts them in one proper-equivalence class. So the uniqueness statement [L1], which is true for positive-definite forms, does not extend to positive discriminant.