# Multiplicative property of integers which can be represented by the principal binary quadratic form

Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. We say $D = b^2 - 4ac$ is the discriminant of $F$. Let $m$ be an integer. If $m = ax^2 + bxy + cy^2$ has a solution in $\mathbb{Z}^2$, we say $m$ is represented by $F$.

My question Is the following proposition true? If yes, how do we prove it?

Proposition Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Let $F$ be the principal form of discriminant $D$(for the definition, see this question). Suppose $m$ and $n$ are represented by $F$, then $mn$ is also represented by $F$.

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Any thoughts about the three answers that have been up for two days now? –  Gerry Myerson Sep 8 '12 at 6:25
@GerryMyerson I think Mr Lubin's answer is the clearest. However, I think his answer can be perfected by considering the order $R$ of a suitable quadratic number field whose($R$'s) discriminant is $D$. –  Makoto Kato Sep 9 '12 at 0:35

If $D\equiv0 {\pmod 4}$, it is true. Let $D=-4n$. Then, the principal form is $x^2+ny^2$.

Now,, if $m=x^2+ny^2$ and $n=z^2+nw^2$, then $mn=(x^2+ny^2)(z^2+nw^2)=(xz \pm nyv)^2+n(xw\mp yz)^2$, which is of the required form.

I am trying to think of a similar identity for the odd case, but I am not sure if it exists.

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The correct viewpoint makes the proposition obvious, because of the relation of these forms to the Norm forms of quadratic integer rings. If $n$ is squarefree and congruent to $2$ or $3$ modulo $4$, then the ring of integers in $K=\mathbb{Q}(\sqrt n)$ is $\mathbb{Z}[\sqrt n]$, and the Norm of $A+B\sqrt n$ is $A^2-nB^2$. If $n$ is congruent to $1$ modulo $4$, then the ring of integers is $\mathbb{Z}[\rho]$, where $\rho=(1+\sqrt n)/2$, and with integral basis $\{1, \rho\}$, and the Norm of $A+B\rho$ is $A^2+AB +B^2(1-n)/4$.

But the Norm from $K$ down to $\mathbb{Q}$ is multiplicative; thus if $r$ and $s$ are norms from upstairs, so is $rs$.

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OP has commented on your answer (as a comment on the original question). –  Gerry Myerson Sep 9 '12 at 1:36

Editing in response to comment, and taking the opportunity to reformat.

Fortuon did $D\equiv0\pmod4$, I'll do $D\equiv1\pmod4$.

Let $\omega=(1+\sqrt D)/2$ and $\omega'=(1-\sqrt D)/2$. Note $$F(x,y)=x^2+xy+((1-D)/4)y^2=(x+\omega y)(x+\omega'y)$$

If $F(a,b)=m,F(c,d)=n$, $$(a+\omega b)(c+\omega d)=g+\omega h,\quad(a+\omega' b)(c+\omega' d)=g+\omega' h$$ where $g=ac+bd(D-1)/4$ and $h=ad+bc+bd$. So $$F(g,h)=(g+\omega h)(g+\omega'h)=(a+\omega b)(c+\omega d)(a+\omega' b)(c+\omega' d)=F(a,b)F(c,d)=mn$$

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Could you explain "So $f(g,h)=mn$"? –  Makoto Kato Sep 9 '12 at 0:29