representation of integers by equivalent integral binary quadratic forms

Theorem 1.

If $F\mathrm{,}G$ are equivalent integral binary quadratic forms, then $F$ and $G$ represent the same set of integers.

Proof.

Write $G(x,y)=F(\alpha x+\beta y,\gamma x+\delta y)$ where

Then $m=G(r,s)\Rightarrow m=F(\alpha r+\beta s,\gamma r+\delta s)$, so if $G$ represents $m$, so does $F$. Since the matrix has determinant 1, it is invertible and its inverse is another integer matrix, so the reverse statement follows as well. ∎

Lemma 2.

$F$ properly represents an integer $m$ if and only if $F$ is properly equivalent to a form $m\mathit{}{x}^{\mathrm{2}}\mathrm{+}B\mathit{}x\mathit{}y\mathrm{+}C\mathit{}{y}^{\mathrm{2}}$.

Proof.

$\Leftarrow$: It is obvious by the above that $F$ represents $m$; the problem is to show that it represents $m$ properly. Write $G(x,y)=m{x}^2+Bxy+C{y}^2$; then $G(x,y)=F(\alpha x+\beta y,\gamma x+\delta y)$, where $\alpha \delta -\beta \gamma =1$. Then $m=G(1,0)=F(\alpha ,\gamma )$. But clearly $(\alpha ,\gamma )=1$ since otherwise we cannot have $\alpha \delta -\beta \gamma =1$. So $F$ represents $m$ properly.
$\Rightarrow$: Write $F(p,q)=m$, where $(p,q)=1$. Since $(p,q)=1$, we can find integers $r,s$ such that $ps-qr=1$, and then

$$\begin{array}{c}F(px+ry,qx+sy)=a{(px+ry)}^2+b(px+ry)(qx+sy)+c{(qx+sy)}^2\hfill \\ \hfill =(a{p}^2+bpq+c{q}^2)x^2+(2apr+bps+bqr+2cqs)xy+(a{r}^2+brs+c{s}^2)y^2\\ \hfill =F(p,q)x^2+(2apr+bps+bqr+2cqs)xy+F(r,s)y^2=m{x}^2+Bxy+C{y}^2\end{array}$$

∎

Definition 1.

If $F$ is a binary quadratic form, its discriminant, $\mathrm{\Delta }(F)$ is $b^2-4ac$.

Note that $\mathrm{\Delta }(F)$ is always either congruent to $0$ or $1$ mod 4, and that $b$ is even (odd) exactly when $\mathrm{\Delta }(F)\equiv 0(1)(mod4)$.

Theorem 3.

If $F\mathrm{,}G$ are equivalent integral quadratic forms, then $\mathrm{\Delta }\mathit{}\mathrm{(}F\mathrm{)}\mathrm{=}\mathrm{\Delta }\mathit{}\mathrm{(}G\mathrm{)}$.

Proof.

For any form $F$, define

Then

$$2F(x,y)=(xy)M_F\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$$

Note further that $\mathrm{\Delta }(F)=-det(M_F)$.

Now in our particular case, if $G(x,y)=F(\alpha x+\beta y,\gamma x+\delta y)$, then

Hence

But $\mathrm{\Delta }(F)=-det(M_F)$, so since ,

∎

Note that this proof shows that applying a set of transformations amount to multiplying by the transform matrix on the left and its transpose on the right.

Example: In the previous example, note that $\mathrm{\Delta }(F)=1-4\cdot 1\cdot 6=-23$, and $\mathrm{\Delta }(G)={51}^2-4\cdot 82\cdot 8=2601-2624=-23$.

The converse of this theorem is not true - that is, there are forms of the same discriminant that represent different numbers. For example, $x^2+5y^2$ and $2x^2+2xy+3y^2$ both have discriminant $-20$, yet the second form represents $2$ while the first clearly does not. However, equivalence classes of forms under arbitrary (proper or improper) equivalence represent disjoint sets of primes:

Theorem 4.

Let $p$ be an odd prime. Suppose $F\mathrm{,}G$ both represent $p$ and $\mathrm{\Delta }\mathit{}\mathrm{(}F\mathrm{)}\mathrm{=}\mathrm{\Delta }\mathit{}\mathrm{(}G\mathrm{)}$. Then $F$ and $G$ are equivalent (but perhaps not properly equivalent).

Proof.

Since $p$ is prime, $F$ obviously represents $p$ properly. So $F\sim p{x}^2+bxy+c{y}^2$. Note that the transformation $(x,y)\mapsto (x+dy,y)$ results in a form whose middle term is $2pd+b$, so by an appropriate choice of $d$ we can arrange that . Similarly, $G\sim p{x}^2+b^{\prime }xy+c^{\prime }{y}^2$ with . Note also that since $b^2-4pc=b^{\prime 2}-4p{c}^{\prime }$, it follows that $b\equiv b^{\prime }(2)$ (i.e. $b,b^{\prime }$ have the same parity).

Since $\mathrm{\Delta }(F)=\mathrm{\Delta }(G)$, we see that $b^2-4pc=b^{\prime 2}-4p{c}^{\prime }\iff b^2\equiv b^{\prime 2}(p)\iff b\equiv \pm b^{\prime }(p)$, so $b=\pm b^{\prime }+kp$ for some $k$. Since $b,b^{\prime }$ have the same parity and $p$ is odd, $k$ is even; since , $k=0$ (since otherwise $b,b^{\prime }$ would be separated by at least $2p$, which is impossible).

We are left with two cases. If $b=b^{\prime }$, then $\mathrm{\Delta }(F)=\mathrm{\Delta }(G)$ implies that $c=c^{\prime }$ and hence $F\sim G$. If $b=-b^{\prime }$, then again $\mathrm{\Delta }(F)=\mathrm{\Delta }(G)$ implies that $c=c^{\prime }$. Then $F$ and $G$ are equivalent via the transformation $(x,y)\mapsto (x,-y)$. ∎

Note that $F(x,y)=a{x}^2+bxy+c{y}^2$ and $G(x,y)=a{x}^2-bxy+c{y}^2$ are always improperly equivalent via the transformation $(x,y)\mapsto (x,-y)$. They are sometimes properly equivalent, and sometimes not. For example, $2x^2+2xy+3y^2$ and $2x^2-2xy+3y^2$ are properly equivalent while $3x^2+2xy+5y^2$ and $3x^2-2xy+5y^2$ are not. (See the article on reduced integral binary quadratic forms for details).

In summary, we have proved the following:

$$F,G\text{ equivalent }\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}F,G\text{ represent the same set of integers }$$

$$F,G\text{ equivalent }\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}\mathrm{\Delta }(F)=\mathrm{\Delta }(G)$$

$$\mathrm{\Delta }(F)=\mathrm{\Delta }(G)\text{ and }F,G\text{ both represent some odd prime }p\mathit{\hspace{1em}}\Rightarrow \mathit{\hspace{1em}}F\text{ and }G\text{ are equivalent}$$

We conclude with the following lemma and corollary, which provide concrete criteria for when an integer is representable by a class of forms.

Lemma 5.

If $D\mathrm{\equiv }\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{(}\mathrm{4}\mathrm{)}$ is an integer, and $m$ is an odd integer relatively prime to $D$, then $m$ is properly represented by a primitive form of discriminant $D$ if and only if $D$ is a quadratic residue $\mathrm{mod}m$.

Proof.

If $F(x,y)$ properly represents $m$, then by the preceding lemma, we may assume that $F(x,y)=m{x}^2+bxy+c{y}^2$. Then $D=b^2-4mc$, being the discriminant of $F$, so that $D\equiv b^2(D)$. Conversely, if $D\equiv b^2(D)$, we may assume $D\equiv b(2)$ (if they have different parities, replace $b$ by $b+m$; since $m$ is odd, the condition now holds and $D\equiv {(b+m)}^2(D)$ as well). Since $D\equiv 0,1(4)$, it follows that $D\equiv b^2(4)$ and thus $D\equiv b^2(4m)$. Hence $D=b^2-4mc$ for some integer $c$. But then $m{x}^2+bxy+c{y}^2$ represents $m$ and has discriminant $D$; it is primitive since $\gcd (m,b)=\gcd (m,D)=1$. ∎

Corollary 6.

Let $n$ be an integer, and $p$ an odd prime not dividing $n$. Then $\mathrm{\left(}\frac{\mathrm{-}n}{p}\mathrm{\right)}\mathrm{=}\mathrm{1}$ if and only if $p$ is represented by a primitive form of discriminant $\mathrm{-}\mathrm{4}\mathit{}n$.

Proof.

By the preceding lemma, $p$ is represented by a primitive form of discriminant $-4n$ if and only if

$$1=\left(\frac{-4n}{p}\right)=\left(\frac{-n}{p}\right)$$

∎

Title	representation of integers by equivalent integral binary quadratic forms
Canonical name	RepresentationOfIntegersByEquivalentIntegralBinaryQuadraticForms
Date of creation	2013-03-22 19:18:48
Last modified on	2013-03-22 19:18:48
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Topic
Classification	msc 11E12
Classification	msc 11E16
Related topic	integralbinaryquadraticforms