density of squarefree numbers in $\mathbb{Z}$ that are 1 mod 4 It's common exercise to show the "density" of square-free numbers in $\mathbb{Z}$ is $\frac{6}{\pi^2}$ which we could say is $6 \times \frac{1}{3^2} = \frac{2}{3}$ or possibly $ 6 \times (\frac{7}{22})^2 = \frac{197}{242}$.  In fact $\frac{3}{5}, \frac{5}{8}$ are also good estimates...
What about fundamental discriminants in number theory:


*

*$D$ is square-free and $D \equiv 1 \text{ mod } 4$

*$D = 4\times m$ with $m$ square-free and $m \equiv 2,3\text{ mod }4 $


Is there some easy Euler-product calculation we can do such as:
$$ \frac{1}{\zeta(2)} = \prod_p \left( 1 - \frac{1}{p^2} \right) = \frac{6}{\pi^2}$$
I am not even sure which primes to use.  It sems to make a huge difference if $d$ is + or -...
We can compute the densities separately in steps #1 or #2 or can they be unified into a single step?



*

*Two problems about Squarefree numbers
 A: You have essentially answered the question.  The first thing to note is that the density of odd squarefree numbers is 4/$\pi^2$ which is easily seen from the Dirichlet generating series for odd squarefree numbers which is $ \prod_{p\neq 2} (1 + 1/p^s)$.  It is easy to see that this is $\zeta(s)/\zeta(2s)$ with the 2-Euler factor removed, from which the residue at s=1 is determined as 4/$\pi^2$.  
With this in hand, let us count the positive fundamental discriminants less than $x$.  Let $\chi$ denote the non-trivial Dirichlet character (mod 4).  The number of squarefree $D\equiv 1$ (mod 4) is then $\sum_{D \leq x \, odd} \frac{1+\chi(D)}{2} \mu^2(D)$.  The key fact needed now is that the Dirichlet series $\sum_{D}\frac{\chi(D)\mu^2(D)}{D^s}$ equals $(1-2^{-s})^{-1}L(s,\chi)/\zeta(s)$ which is analytic for $\Re(s)\geq 1$ so that using the standard Tauberian theorem or otherwise, we get $\sum_{D\leq x} \mu^2(D) \chi(D) = o(x)$.  So the density of positive discriminants $\equiv 1$ (mod 4) is $2/\pi^2$.  
The other cases are similarly treated.  The $D$ of the form $4m$ with $m\equiv 2$ (mod 4) and $m$ squarefree and $D\leq x$ is the same as counting the number of $m_1$ $\leq x/8$ with $m_1$ squarefree and odd is asymptotic to $\frac{4}{\pi^2} (x/8)$ which gives a density of $1/2\pi^2$.
For the case $m\equiv 3$ (mod 4), we get by a similar analysis density of $1/2\pi^2$.  Thus the total density of positive fundamental discriminants is $3/\pi^2$.  
A similar analysis for negative fundamental discriminants gives a density of $3/\pi^2$.  
