[proofplan] The proof is a two-way implication. The forward direction inspects the discriminant $d = b^2 - 4ac$ modulo $4$: the term $4ac$ vanishes, so $d \equiv b^2 \pmod{4}$, and the squares modulo $4$ are exactly $\{0, 1\}$. The converse is a construction: for each residue class we exhibit an explicit form. When $d \equiv 0 \pmod{4}$ we use $(1, 0, -d/4)$; when $d \equiv 1 \pmod{4}$ we use $(1, 1, (1-d)/4)$. In each case the coefficients are integers precisely because of the congruence on $d$, and a direct computation confirms the discriminant. [/proofplan] [step:Reduce the discriminant modulo $4$ and enumerate quadratic residues] Let $f = (a, b, c)$ be any BQF, so $a, b, c \in \mathbb{Z}$. By definition, \begin{align*} \operatorname{disc}(f) = b^2 - 4ac. \end{align*} Since $4ac \equiv 0 \pmod{4}$, \begin{align*} \operatorname{disc}(f) \equiv b^2 \pmod{4}. \end{align*} The squares modulo $4$ are computed from the four residue classes: $0^2 \equiv 0$, $1^2 \equiv 1$, $2^2 \equiv 0$, $3^2 \equiv 1 \pmod{4}$. Hence $b^2 \in \{0, 1\} \pmod{4}$, and therefore $\operatorname{disc}(f) \equiv 0$ or $1 \pmod{4}$. This proves the forward direction: every discriminant of a BQF is $\equiv 0$ or $1 \pmod{4}$. [/step] [step:For $d \equiv 0 \pmod 4$, exhibit the form $(1, 0, -d/4)$ with discriminant $d$] Suppose $d \equiv 0 \pmod{4}$, so we may write $d = 4k$ for a unique $k \in \mathbb{Z}$, namely $k = d/4$. Define the BQF \begin{align*} f_0: \mathbb{Z}^2 &\to \mathbb{Z} \\ (x, y) &\mapsto x^2 - \tfrac{d}{4}\, y^2, \end{align*} so $f_0 = (1, 0, -d/4)$. The coefficients lie in $\mathbb{Z}$ because $d/4 \in \mathbb{Z}$ by hypothesis. Computing: \begin{align*} \operatorname{disc}(f_0) = 0^2 - 4 \cdot 1 \cdot (-d/4) = d. \end{align*} Hence a BQF of discriminant $d$ exists in this case. [/step] [step:For $d \equiv 1 \pmod 4$, exhibit the form $(1, 1, (1-d)/4)$ with discriminant $d$] Suppose $d \equiv 1 \pmod{4}$. Then $1 - d \equiv 0 \pmod{4}$, so $(1 - d)/4 \in \mathbb{Z}$. Define \begin{align*} f_1: \mathbb{Z}^2 &\to \mathbb{Z} \\ (x, y) &\mapsto x^2 + xy + \tfrac{1-d}{4}\, y^2, \end{align*} so $f_1 = (1, 1, (1-d)/4)$, and all coefficients are integers. Computing: \begin{align*} \operatorname{disc}(f_1) = 1^2 - 4 \cdot 1 \cdot \tfrac{1-d}{4} = 1 - (1 - d) = d. \end{align*} Hence a BQF of discriminant $d$ exists in this case as well. [/step] [step:Combine the two directions to conclude] The forward direction, established in the first step, shows that every BQF has discriminant $\equiv 0$ or $1 \pmod{4}$. The two construction steps show, conversely, that every $d \in \mathbb{Z}$ with $d \equiv 0$ or $1 \pmod{4}$ is realised as the discriminant of some BQF (namely $f_0$ or $f_1$, according to the residue class of $d$). The two directions together give the claimed biconditional: \begin{align*} \exists\, f \text{ a BQF with } \operatorname{disc}(f) = d \iff d \equiv 0 \text{ or } 1 \pmod{4}. \end{align*} [/step]