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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    $\begingroup$ I think you're asking questions just in the same way you used to a very few weeks ago: you give no reasons, insights, background, etc. and it just looks as if you were bored and tried to do something at the computer. This kind of things may cause downvoting. For example, have you already tried some simple examples, say with $\,D=1\,,\,4\,$? Perhaps something with $\,b=0\,$...? $\endgroup$ – DonAntonio Sep 5 '12 at 9:32
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    $\begingroup$ @DonAntonio "This kind of things may cause downvoting." Writing motivation for a question is not required in this site. Please read the FAQ. $\endgroup$ – Makoto Kato Oct 2 '12 at 21:00
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    $\begingroup$ @DonAntonio The time delay has nothing to do with the validity of our comments. $\endgroup$ – Makoto Kato Oct 3 '12 at 2:57
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    $\begingroup$ Well, that's your opinion, not mine. My opinion was clearly stated in my first comment. You now do whatever you want. $\endgroup$ – DonAntonio Oct 3 '12 at 3:02
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    $\begingroup$ @DonAntonio It's not my opinion. The FAQ does not say writing motivation is necessary. $\endgroup$ – Makoto Kato Oct 3 '12 at 3:05

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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  • $\begingroup$ Could you explain why gcd$(x, y) = 1$? $\endgroup$ – Makoto Kato Oct 4 '12 at 8:03
  • $\begingroup$ @MakotoKato : see my update at the end. $\endgroup$ – Ewan Delanoy Oct 4 '12 at 8:25
  • $\begingroup$ Please correct $Q(x, y) = ax^2 + bxy + y^2$ and $Q(0, 1) = b$. Then I will accept your answer. Thanks. $\endgroup$ – Makoto Kato Oct 4 '12 at 17:32
  • $\begingroup$ I think there is also a missing $c$ in the definition of $Q$? $\endgroup$ – Old John Oct 4 '12 at 18:06
  • $\begingroup$ @OldJohn That's one of the corrections I'm asking: "Please correct $Q(x, y) = ax^2 + bxy + y^2$". $\endgroup$ – Makoto Kato Oct 4 '12 at 19:56

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