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Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$. Let $f(x, y) = ax^2 + bxy + cy^2 \in \mathfrak{F}$. Let $\alpha = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$. We write $f^\alpha(x, y) = f(px + qy, rx + sy)$. Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $SL_2(\mathbb{Z})$ acts on $\mathfrak{F}$ from right.

Let $f, g \in \mathfrak{F}$. If $f$ and $g$ belong to the same $SL_2(\mathbb{Z})$-orbit, we say $f$ and $g$ are equivalent.

Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$. 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 $ax^2 + bxy + cy^2$. 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 $ax^2 + bxy + cy^2$.

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

Proposition Let $ax^2 + bxy + cy^2 \in \mathfrak{F}$. Suppose its discriminant is not a square. Let $m$ be an integer. Then $m$ is properly represented by $ax^2 + bxy + cy^2$ if and only if there exist integers $l, k$ such that $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent.

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    @Makoto: You have the advantages of already knowing the context and having recently written the post, so that *you* can easily ignore everything that isn't the actual question. t.b.'s comment lets me put words to a problem I've had with your questions previously: a large fraction of your questions I never finish reading, because they drag on and on and I lose interest before I actually get to the content of the question.2012-09-08

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I cannot understand the comment of William Jagy, but I think this question is in fact quite easy, and I shall employ the method using topographs as in Cnonway's Sensual Quadratic Form.
But, once one knows what a topograph is, and some basic properties if that concept, this becomes an easy exercise.
A topograph of a quadratic form is like haha.
Here the lettres mean the values represented by a form $f$. And we represent $m$ there. The condition that $m$ is properly represented implies that the representation must occur in the graph(see Conway's little book). Then, taking $h=m+x-y$, we shall find that the form $f$ is equivalent with $\langle m,h,x\rangle$, as required(see the book again). (For an example of the graph, see the answer of Will Jagy.)
Conversely, if there is such a form $\langle m,h,x\rangle$ equivalent with $f$, then $f$ should also properly represent $m$, as the other form does.
Feel free to tell me where the errors are, if any; thanks in advance.
P.S. The diagram above is hand-made, and is in fact modelled on the form $X^2+4XY+Y^2$.

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Lemma 1 Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$. Let $\alpha = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$. Then $f^\alpha(x, y) = f(px + qy, rx + sy) = kx^2 + lxy + my^2$, where

$k = ap^2 + bpr + cr^2$

$l = 2apq + b(ps + qr) + 2crs$

$m = aq^2 + bqs + cs^2$.

Proof: Clear.

Proof of the proposition Let $f(x, y) = ax^2 + bxy + cy^2$.

Suppose $m$ is properly represented by $f(x, y)$. There exist integers $p, r$ such that gcd$(p, r) = 1$ and $m = f(p, r)$. Since gcd$(p, r) = 1$, there exist integers $s, r$ such that $ps - rq = 1$. By Lemma 1, $f(px + qy, rx + sy) = mx^2 + lxy + ky^2$, where

$m = ap^2 + bpr + cr^2$

$l = 2apq + b(ps + qr) + 2crs$

$k = aq^2 + bqs + cs^2$.

Hence, $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent.

Conversely suppose $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent. There exists integer $p, q, r, s$ such that $ps - rq = 1$ and $f(px + qy, rx + sy) = mx^2 + lxy + ky^2$. Letting $x = 1, y = 0$, we get $f(p, r) = m$. Since $ps - rq = 1$, gcd$(p, r) = 1$. Hence $m$ is properly represented by $ax^2 + bxy + cy^2$.