# Equivalent integral quadratic forms properly represent the same integers

Definitions:

An integral quadratic form (IQF) is some instance of $$f(x,y)=ax^2+bxy+cy^2$$, where $$a,b,c \in \mathbb{Z}$$.

Let $$f(x,y),g(x,y)$$ denote IQFs. We say $$f(x,y)$$ and $$g(x,y)$$ are properly equivalent, denoted by $$f(x,y)∼g(x,y)$$, if there exist integers $$p,q,r,s$$ such that $$f(x,y)=g(px+qy,rx+sy)$$ and $$ps−qr=1$$

We say $$f(x,y)$$ represents $$m \in \mathbb{Z}$$ if there exist $$x_0,y_0 \in \mathbb{Z}$$ such that $$f(x_0,y_0)=m$$. We say $$f(x,y)$$ properly represents $$m$$ if it represents $$m$$ and $$gcd(x_0,y_0)$$.

Claim to be proven

If $$f(x,y) \sim g(x,y)$$ and $$f(x,y)$$ properly represents $$m$$, then $$g(x,y)$$ properly represents $$m$$.

Attempt

I have been able to show that $$\sim$$ is an equivalence relation. Then, I assume $$f(x,y) \sim g(x,y)$$ and $$f(x_0,y_0)=m$$ where $$gcd(x_0,y_0)=1$$. Since $$f(x,y) \sim g(x,y)$$, we have $$f(x_0,y_0)=g(px_0+qy_0,rx_0+sy_0)=m$$, so $$g(x,y)$$ represents $$m$$. All that is left is to show that $$gcd(px_0+qy_0,rx_0+sy_0)=1$$. But I am not sure how to proceed. I have tried thinking of $$p,q,r,s$$ as a $$2 \times2$$ matrix with determinant $$1$$ acting on the $$2 \times 1$$ vector $$[x_0,y_0]$$. I think that something to do with rotations should be involved, but I am not sure how to properly express it such that the desired result, $$gcd(px_0+qy_0,rx_0+sy_0)=1$$, is obtained. Any help is appreciated.

## 2 Answers

with integer matrices with determinant $$\pm 1,$$ the gcd is preserved because the inverse of the matrix also has integer elements. From $$f(x,y)=g(px+qy,rx+sy)$$ we define $$u = px+qy \; , \; \; \; v = rx+sy$$

Taking $$g = \gcd (x,y)$$ we see that $$g|u$$ and $$g| v.$$ Therefore $$g | \gcd ( u,v)$$ Suppose we name $$h = \gcd ( u,v).$$ So far we have $$g|h.$$

We take the inverse matrix to arrive at $$x = su-qv \; , \; \; \; y = -ru +p v$$

Well $$h$$ divides both $$x,y$$ so $$h | g.$$ But $$g|h.$$ So $$h = g$$

Summary: $$(x,y)=(px+qy,rx+sy)$$ as ideals, so $$px+qy$$ and $$rx+sy$$ are coprime since $$x$$ and $$y$$ are coprime.

All the detail:

If $$ps-qr=1$$, then $$(px+qy)s-(rx+sy)q=x(ps-qr)=x$$.

If $$ps-qr=-1$$, then $$-(px+qy)s+(rx+sy)q=x(-ps+qr)=x$$.

Hence $$x$$ is a linear combination of $$px+qy,rx+sy$$. So $$x=c_1(px+qy)+c_2(rx+sy)$$ for some $$c_1,c_2 \in \mathbb{Z}$$

Similarly, we can show $$y$$ is a linear combination of $$px+qy,rx+sy$$. So $$y=c_3(px+qy)+c_4(rx+sy)$$ for some $$c_3,c_4 \in \mathbb{Z}$$

By assumption $$gcd(x,y)=1$$. By Bezout's Identity there exist $$a,b$$ such that $$ax+by=1$$.

Using the above:

$$a(c_1(px+qy)+c_2(rx+sy))+b(c_3(px+qy)+c_4(rx+sy))=1$$

$$ac_1(px+qy)+ac_2(rx+sy)+bc_3(px+qy)+bc_4(rx+sy)=1$$

$$ac_1(px+qy)+bc_3(px+qy)+ac_2(rx+sy)+bc_4(rx+sy)=1$$

$$(ac_1+bc_3)(px+qy)+(ac_2+bc_4)(rx+sy)=1$$

Let $$c_5 :=ac_1+bc_3$$, $$c_6:=ac_2+bc_4$$. Then the above becomes

$$c_5(px+qy)+c_6(rx+sy)=1$$

Let $$d=gcd(px+qy,rx+sy)$$.

Then $$(px+qy)=du$$ and $$rx+sy=dw$$ for some $$u,w \in \mathbb{Z}$$.

Substituting into the above equation yields:

$$c_5(du)+c_6(dw)=1$$

$$d(c_5u+c_6w)=1$$

So $$d$$ divides $$1$$, and since $$gcd$$ is positive by definition, $$d=1$$. Hence $$gcd(px+qy,rx+sy)=1$$, as desired.