Let $f(x,y)$ be a binary quadratic form with co-prime integer coefficients. We say that $f$ is a proper subform of $g(x,y)$ if there exists an integer matrix $A = \left(\begin{smallmatrix} a_1 & a_2 \\ a_3 & a_4 \end{smallmatrix}\right)$ with $|\det A| > 1$ such that

$$\displaystyle f(x,y) = g(a_1 x + a_2 y, a_3 x + a_4 y).$$

For example, the form $f(x,y) = 4x^2 + 4xy + 5y^2$ is a proper subform of $g(x,y) = x^2 + y^2$, since

$$\displaystyle 4x^2 + 4xy + 5y^2 = (2x + y)^2 + (2y)^2.$$

If $f$ is a proper subform, then the discriminant of $f$ is divisible by $\det(A)^2$, so it is not square-free. My question is the converse: suppose that $\Delta(f)$ is divisible by an odd square $m^2$. Is $f$ a proper subform of another form $g$?


Yes. It is about equivalence over $SL_2 \mathbb Z.$ Your form $f$ (primitively) represents some value not divisible by the fixed prime $p.$ Indeed, from original coefficients $\langle a,b,c \rangle$ we know that at least one of $a,c, a+b+c$ is not divisible by $p.$ We may therefore demand $a \neq 0 \pmod p$ in $\langle a,b,c \rangle.$

For an integer $t$ we have the equivalent form $$ \langle a, \; \;b + 2 a t, \; \;c + b t + a t^2 \rangle. $$ As $p$ is odd and $a \neq 0 \pmod p,$ there exists some $\delta$ such that $b + 2 a \delta \equiv 0 \pmod p.$

At this point, we now have $\langle a,b,c \rangle$ with $a \neq 0 \pmod p$ and $b \equiv 0 \pmod p.$ You have stated that $$ b^2 - 4 a c \equiv 0 \pmod {p^2}. $$ It follows that $c \equiv 0 \pmod {p^2}.$

We may now rewrite $$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle.$$ The descent is $$\langle \alpha, \; \;\beta p, \; \;\gamma p^2 \rangle \mapsto \langle \alpha, \; \;\beta, \; \;\gamma \rangle.$$

As far as a factor of $4,$ taking $\Delta = b^2 - 4 a c,$ we can descend when $\Delta \equiv 4 \pmod {16},$ as $x^2 +3y^2 \mapsto x^2 + xy+ y^2,$ and $\Delta \equiv 0 \pmod {16},$ as $x^2 +20y^2 \mapsto x^2 + 5y^2.$ No descent for the others, for example $x^2 + y^2$ or $x^2 + 2 y^2.$

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