# representation of integers by equivalent integral binary quadratic forms

###### Proof.

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

 $\det\left(\begin{array}[]{cc}\alpha&\gamma\\ \beta&\delta\end{array}\right)=\pm 1$

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 $mx^{2}+Bxy+Cy^{2}$.

###### Proof.

$\Leftarrow$: It is obvious by the above that $F$ represents $m$; the problem is to show that it represents $m$ properly. Write $G(x,y)=mx^{2}+Bxy+Cy^{2}$; 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

 $\displaystyle F(px+ry,qx+sy)=a(px+ry)^{2}+b(px+ry)(qx+sy)+c(qx+sy)^{2}\\ \displaystyle=(ap^{2}+bpq+cq^{2})x^{2}+(2apr+bps+bqr+2cqs)xy+(ar^{2}+brs+cs^{2% })y^{2}\\ \displaystyle=F(p,q)x^{2}+(2apr+bps+bqr+2cqs)xy+F(r,s)y^{2}=mx^{2}+Bxy+Cy^{2}$

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

###### Theorem 3.

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

###### Proof.

For any form $F$, define

 $M_{F}=\left(\begin{array}[]{cc}2a&b\\ b&2c\end{array}\right)$

Then

 $2F(x,y)=(x\text{ }y)M_{F}\left(\begin{array}[]{c}x\\ y\end{array}\right)$

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

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

 $2G(x,y)=(\alpha x+\beta y\text{ }\gamma x+\delta y)M_{F}\left(\begin{array}[]{% c}\alpha x+\beta y\\ \gamma x+\delta y\end{array}\right)=(x\text{ }y)\left(\begin{array}[]{cc}% \alpha&\gamma\\ \beta&\delta\end{array}\right)M_{F}\left(\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta\end{array}\right)\left(\begin{array}[]{c}x\\ y\end{array}\right)$

Hence

 $M_{G}=\left(\begin{array}[]{cc}\alpha&\gamma\\ \beta&\delta\end{array}\right)M_{F}\left(\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta\end{array}\right)$

But $\Delta(F)=-\det(M_{F})$, so since $\det\left(\begin{array}[]{cc}\alpha&\gamma\\ \beta&\delta\end{array}\right)=\det\left(\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta\end{array}\right)=\pm 1$,

 $\Delta(G)=-\det(M_{G})=-\det\left(\begin{array}[]{cc}\alpha&\gamma\\ \beta&\delta\end{array}\right)\det(M_{F})\det\left(\begin{array}[]{cc}\alpha&% \beta\\ \gamma&\delta\end{array}\right)=-\det(M_{F})=\Delta(F)$

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

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 $F,G$ both represent $p$ and $\Delta(F)=\Delta(G)$. Then $F$ and $G$ are equivalent (but perhaps not properly equivalent).

###### Proof.

Since $p$ is prime, $F$ obviously represents $p$ properly. So $F\sim px^{2}+bxy+cy^{2}$. 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 $-p. Similarly, $G\sim px^{2}+b^{\prime}xy+c^{\prime}y^{2}$ with $-p. Note also that since $b^{2}-4pc=b^{\prime 2}-4pc^{\prime}$, it follows that $b\equiv b^{\prime}\pod{2}$ (i.e. $b,b^{\prime}$ have the same parity).

Since $\Delta(F)=\Delta(G)$, we see that $b^{2}-4pc=b^{\prime 2}-4pc^{\prime}\Leftrightarrow b^{2}\equiv b^{\prime 2}% \pod{p}\Leftrightarrow b\equiv\pm b^{\prime}\pod{p}$, 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 $-p, $k=0$ (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 $\Delta(F)=\Delta(G)$ implies that $c=c^{\prime}$ and hence $F\sim G$. If $b=-b^{\prime}$, then again $\Delta(F)=\Delta(G)$ implies that $c=c^{\prime}$. Then $F$ and $G$ are equivalent via the transformation $(x,y)\mapsto(x,-y)$. ∎

Note that $F(x,y)=ax^{2}+bxy+cy^{2}$ and $G(x,y)=ax^{2}-bxy+cy^{2}$ 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:

 $\displaystyle F,G\text{ equivalent }\quad\Rightarrow\quad F,G\text{ represent % the same set of integers }$ $\displaystyle F,G\text{ equivalent }\quad\Rightarrow\quad\Delta(F)=\Delta(G)$ $\displaystyle\Delta(F)=\Delta(G)\text{ and }F,G\text{ both represent some odd % prime }p\quad\Rightarrow\quad F\text{ and }G\text{ are equivalent}$

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 $D\equiv 0,1\pod{4}$ 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  $\mod m$.

###### Proof.

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

###### 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 RepresentationOfIntegersByEquivalentIntegralBinaryQuadraticForms 2013-03-22 19:18:48 2013-03-22 19:18:48 rm50 (10146) rm50 (10146) 7 rm50 (10146) Topic msc 11E12 msc 11E16 integralbinaryquadraticforms