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Let $m$ be a positive integer. We say that $m$ can be represented by a binary quadratic form

$$\displaystyle f(x,y) = ax^2 + bxy + cy^2, a,b,c \in \mathbb{Z}$$

if there exist integers $u,v$ such that $m = f(u,v)$. We say that $m$ is primitively representable if $u,v$ can be chosen so that $\gcd(u,v) = 1$. Note that both notions are preserved under $\text{SL}_2(\mathbb{Z})$-equivalence, so we may say that $m$ is representable by the class $[f]$ of $f$.

Recall that for a fixed integer $D$, the set of $\text{SL}_2(\mathbb{Z})$-equivalence classes of binary quadratic forms of discriminant $D$ forms a group: this was proved by Gauss, and this phenomenon is now known as the Gauss composition laws. Denote by $H(D)$ the form class group of discriminant $D$ with respect to Gauss composition.

Suppose that $f$, or rather the equivalence class $[f]$, is a $4$th power in the form class group $H(D)$ of discriminant $D$: that is, there exists a form $g$ of discriminant $D$ such that $[g]^4 = [f]$, where the operation is Gauss composition. Let $H_4(D)$ denote the subgroup of $H(D)$ consisting of 4-th powers.

Is it known how to determine whether an integer $m$ belongs to the following set?

$$\displaystyle \{n \in \mathbb{Z} : \exists [f] \in H_4(D) \text{ such that } n \text{ is primitively representable by } [f]\}?$$

Using genus theory, the question with 4th powers replaced with squares can be answered as follows. An equivalence class $[f] \in H(D)$ is a square, i.e., there exists $[g] \in H(D)$ such that $[f] = [g]^2$, if and only if $[f]$ is in the principal genus. Using this, one shows that $m$ is representable by a square in $H(D)$ if and only if the sum

$$\displaystyle \sum_{c | D} \left(\frac{m}{c}\right) \ne 0.$$

Edit: By the following paper provided by Will Jagy ( zakuski.math.utsa.edu/~kap/Estes_Pall_1973.pdf), we have $[f] \in H_4(D)$ if and only if $[f]$ is in the principal spinor genus; see https://en.wikipedia.org/wiki/Spinor_genus

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  • $\begingroup$ If I recall correctly, the analogous question for the subgroup of squares is answered by genus theory: $m$ is represented by the square of a form of discriminant $D<0$ if and only if the Legendre symbol $(\frac{m}{p})=1$ for all odd $p|D$, plus an analogous modulo 4 or 8 if $D$ is even. (Caveat, maybe more complicated if $D$ is not fundamental?) $\endgroup$ – Stopple Aug 14 at 17:51
  • $\begingroup$ @Stopple that's correct. Do you think it's helpful to include this fact in the body of the question? $\endgroup$ – Stanley Yao Xiao Aug 14 at 18:21
  • $\begingroup$ For me, more historical context is helpful $\endgroup$ – Stopple Aug 14 at 18:29

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