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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 $F$. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $F$ is primitive.

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 $F$. If $m = ax^2 + bxy + cy^2$ has a solution $(s, t)$ such that gcd$(s, t) = 1$, we say $m$ is properly 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 $F = ax^2 + bxy + cy^2$ be a primitive binary quadratic form of discriminant $D$. Let $m \neq 0$ be an integer. There exists an integer $n$ which is properly represented by $F$ and gcd($n, m) = 1$.

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    I must say -- this question is very important and useful. One example of where it comes in handy is in deriving the Dirichlet composition of two quadratic forms.2018-05-29

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The answer to your question is YES.

Put $Q(x,y)=ax^2+bxy+cy^2$. Let $p_1,p_2, \ldots ,p_r$ be the prime divisors of $m$. Note that

$ Q(1,0)=a,\ Q(0,1)=c, \ Q(1,1)=a+b+c $

If $Q$ is primitive, those three numbers are not all divisible by $p_i$. So for each $i$, there are two integers $x_i$ and $y_i$ such that $Q(x_i,y_i) \not\equiv 0 \ ({\sf mod} \ p_i)$ (and in fact, we can take $(x_i,y_i)$ to be one of the three $(0,1),(1,0),(1,1)$).

By the Chinese remainder theorem, there is an integer $x$ such that $x\equiv x_i \ ({\sf mod} \ p_i)$ for all $i$, and an integer $y$ such that $y\equiv y_i \ ({\sf mod} \ p_i)$ for all $i$. Then

$ \forall i,\ Q(x,y) \equiv Q(x_i,y_i) \not\equiv 0 \ ({\sf mod} \ p_i) $

So $n=Q(x,y)$ is divisible by none of the $p_i$ and is therefore coprime to $m$.

Finally, to ensure that the representation is proper, replace $(x,y)$ with $(x’,y’)=(\frac{x}{g},\frac{y}{g})$ where $g=gcd(x,y)$. Then $Q(x',y')=\frac{Q(x,y)}{g^2}$ is still coprime to $m$.

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    @OldJohn : you’re welcome, I didn’t even realize what typos you corrected.2012-10-05