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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 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 $\,b=0\,$...?2012-09-05
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    @DonAntonio "This kind of things may cause downvoting." Writing motivation for a question is not required in this site. Please read the FAQ.2012-10-02
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    You're answering a comment to one of YOUR questions almost one month after I posted it? I don't even remember this one...2012-10-03
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    @DonAntonio The time delay has nothing to do with the validity of our comments.2012-10-03
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    Well, that's your opinion, not mine. My opinion was clearly stated in my first comment. You now do whatever you want.2012-10-03
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    @DonAntonio It's not my opinion. The FAQ does not say writing motivation is necessary.2012-10-03
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    I don't care and, apparently, neither do al the numerous people that downvoted so many of your questions. This seems to be common sense but, again, you do what your want.2012-10-03
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    @DonAntonio You are saying that you don't care the rules of this site. Remember that you were heavily downvoted because of voting to close my thread without proper reason. http://meta.math.stackexchange.com/questions/4936/voting-to-close-a-question-because-one-thinks-it-is-extremely-specialised2012-10-03
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    You keep on writing nonsense, @Makoto Kato, and that's a pity. I never said I don't care about this site' rules but about the site's FAQ not stating something. It bewilders me you can't say the difference between these two, and I'm closing this ridiculous interchange of comment right now. Have a nice day.2012-10-03
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    @DonAntonio The FAQ does not say writing motivation is necessary. So it is not consistent with the rules of the site to downvote a question because the motivation is not written.2012-10-04
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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

1 Answers 1

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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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    Could you explain why gcd$(x, y) = 1$?2012-10-04
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    @MakotoKato : see my update at the end.2012-10-04
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    Please correct $Q(x, y) = ax^2 + bxy + y^2$ and $Q(0, 1) = b$. Then I will accept your answer. Thanks.2012-10-04
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    I think there is also a missing $c$ in the definition of $Q$?2012-10-04
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    @OldJohn That's one of the corrections I'm asking: "Please correct $Q(x, y) = ax^2 + bxy + y^2$".2012-10-04
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    @ewan I hope you don't mind, but I corrected two typos in your answer, so that your answer can be accepted - please let me know if my edits are not what you intended.2012-10-04
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    @OldJohn : you’re welcome, I didn’t even realize what typos you corrected.2012-10-05