[proofplan]
The two assertions are independent. The congruence $b \equiv d \pmod{2}$ is immediate from $d = b^2 - 4ac$: reducing modulo $4$ gives $d \equiv b^2 \pmod{4}$, which forces $d$ and $b$ to have the same parity. For the size bound, we use the reduced inequality $|b| \leq a \leq c$ to estimate the expression $|d| = 4ac - b^2$ from below. Replacing $c$ by $a$ (valid because $a \leq c$) and $b^2$ by $a^2$ (valid because $|b| \leq a$) gives $|d| \geq 3a^2$. Solving for $a$ produces $a \leq \sqrt{|d|/3}$. Positive definiteness enters through $d < 0$, so $|d| = -d$ and every quantity above is legitimately positive.
[/proofplan]
[step:Establish the parity congruence $b \equiv d \pmod 2$]
By definition of the discriminant,
\begin{align*}
d = b^2 - 4ac.
\end{align*}
Reducing modulo $4$, the term $4ac$ vanishes and we obtain $d \equiv b^2 \pmod{4}$. In particular, $d$ and $b^2$ have the same parity. Since $b^2$ and $b$ have the same parity (because the parity of an integer is preserved under squaring: $b$ even $\iff b^2$ even, $b$ odd $\iff b^2$ odd), it follows that
\begin{align*}
b \equiv b^2 \equiv d \pmod{2}.
\end{align*}
This proves $b \equiv d \pmod{2}$.
[/step]
[step:Establish the size bound $|b| \leq a \leq \sqrt{|d|/3}$ using the reduced inequalities]
Since $f$ is reduced, the defining inequalities give $|b| \leq a \leq c$. Since $f$ is positive definite, $a > 0$ and $d < 0$, so $|d| = -d = 4ac - b^2$. We estimate this from below:
\begin{align*}
|d| = 4ac - b^2 &\geq 4a \cdot a - b^2 \qquad \text{(since } c \geq a \text{ and } 4a > 0\text{)} \\
&= 4a^2 - b^2 \\
&\geq 4a^2 - a^2 \qquad \text{(since } b^2 \leq a^2 \text{ from } |b| \leq a\text{)} \\
&= 3a^2.
\end{align*}
Hence $3a^2 \leq |d|$, i.e. $a^2 \leq |d|/3$. Since $a > 0$, taking square roots (a monotone operation on non-negative reals) yields
\begin{align*}
a \leq \sqrt{\tfrac{|d|}{3}}.
\end{align*}
Combined with the hypothesis $|b| \leq a$ we obtain the chain
\begin{align*}
|b| \leq a \leq \sqrt{\tfrac{|d|}{3}}.
\end{align*}
[guided]
We are given that $f = (a, b, c)$ is a reduced positive definite BQF with discriminant $d = b^2 - 4ac < 0$. By the [Definition of Reduced Form](/theorems/???) we have
\begin{align*}
|b| \leq a \leq c,
\end{align*}
and positive definiteness (together with [Discriminant Determines Definiteness](/theorems/1728)) gives $a > 0$, $c > 0$, $d < 0$. In particular $|d| = -d$, and we can compute
\begin{align*}
|d| = -(b^2 - 4ac) = 4ac - b^2.
\end{align*}
This is a **positive** quantity, which is the first sanity check: the right-hand side $4ac - b^2$ is indeed positive when $d < 0$.
Now we estimate $|d|$ from below by replacing each factor by a smaller quantity. The key replacements are:
- $c \geq a$ (from reducedness), so $4ac \geq 4a \cdot a = 4a^2$. The coefficient $4a > 0$ is crucial for preserving the inequality direction.
- $b^2 \leq a^2$ (from $|b| \leq a$, squaring the inequality of non-negative numbers).
Combining:
\begin{align*}
|d| = 4ac - b^2 \geq 4a^2 - b^2 \geq 4a^2 - a^2 = 3a^2.
\end{align*}
So $|d| \geq 3a^2$. Dividing by $3$ (positive) gives $a^2 \leq |d|/3$, and taking square roots gives $a \leq \sqrt{|d|/3}$. Reducedness already supplies $|b| \leq a$, so we chain the inequalities:
\begin{align*}
|b| \leq a \leq \sqrt{\tfrac{|d|}{3}}.
\end{align*}
[/guided]
[/step]
[step:Combine the two bounds into the stated conclusion]
The congruence $b \equiv d \pmod{2}$ follows from the first step, and the size bound $|b| \leq a \leq \sqrt{|d|/3}$ from the second. Together they give exactly the stated theorem:
\begin{align*}
|b| \leq a \leq \sqrt{\tfrac{|d|}{3}}, \qquad \text{and} \qquad b \equiv d \pmod{2}.
\end{align*}
[/step]
