Where $[...]$ are the Iverson brackets. This expression arises from my research on counting the number of reducible quadratics with coefficients constrained by naive height $N \ge 1$ and $r$, $s$, $t$ are integer coefficients of $r x^2 + s x + t$. The term $s^2 \pm 4 r t$ is the quadratic discriminant, thus this problem is counting the number of perfect square discriminants. The $+4 r t$ arises from splitting these sums over $- N \le r, s, t \le + N$ into the domains $1 \le r, s, t \le N$. I have not made any progress in reducing the title sum or rewriting it in a different form.
The second part of this problem is the reduction of the unconstrained sum $$\sum_{r, s, t=1}^{N} \left[\sqrt{s^2 \pm 4 r t} \in \mathbb{Z}\right]$$ and its corresponding asymptotic expansion as $N \rightarrow \infty$. The problem that I posted at Looking for asymp exp as $N \rightarrow \infty$ for $\sum_{r=1}^{N}\sum_{t=1}^{N}\left[\sqrt{rt}\in Z\right]$ w/wo the constraint $GCD(r,t)=1$ does answer some parts of this problem for the case when $s=0$.
If we sum over the three variables $r$, $s$, and $t$ with $\gcd \left(r,s,t \right)=1$ then we have the probability of three integers chosen at random being coprime is
$$\frac{1}{N^3} \sum_{r,s,t = 1}^{N} \left[\gcd \left(r,s,t \right) = 1 \right] = \frac{1}{N^3} \sum_{d = 1}^{N} \mu \left(d \right) {\left\lfloor{\frac{N}{d}}\right\rfloor}^{3} \sim \frac{1}{\zeta \left(3 \right)} + \cdots$$
where $\mu \left(n \right)$ is the Mobius function.
We can write the condition that $\left[{\gcd \left({r, s, t}\right) = 1}\right]$ as
$$\sum_{r, s, t = 1}^{N} \left[{\sqrt{{s}^{2} \pm 4\, r\, t} \in \mathbb{Z}}\right] \left[{\gcd \left({r, s, t}\right) = 1}\right] = \sum_{r, s, t = 1}^{N} \sum_{d \mid \gcd \left({r,s,t}\right)} \left[{\sqrt{{s}^{2} \pm 4\, r\, t} \in \mathbb{Z}}\right] \mu \left({d}\right)$$
