Is the set of prime pairs such that $gcd(p−1,q−1)=2$ of positive density? Is the set of prime pairs such that $gcd(p−1,q−1)=2$ of positive density? For example, for $p,q≤10^4$ the answer is approximately $1/2$.
I was wondering if it were possible to use sieve methods and results such as  the Siegel-Walfisz Theorem to give a good approximation of prime pairs of this form.
The motivation for the question is for understanding the order of elements in the group $(\mathbb{Z}/pq\mathbb{Z})^∗≃(\mathbb{Z}/p\mathbb{Z})^∗×(\mathbb{Z}/q\mathbb{Z})^*$.
 A: The answer is 
$$ 
\frac{3}{4} \prod_{p>2} \Big(1 -\frac{1}{(p-1)^2} \Big) = 0.4951\ldots .
$$ 
The product above is also known as the twin prime constant. 
This follows easily from the prime number theorem in arithmetic progressions.  Restricting to odd primes $p$ and $q$ below $N$ we want to count $(p-1,q-1)=2$ which by Mobius inversion can be expressed as 
$$ 
\sum_{\substack{d| (p-1)/2 \\ d| (q-1)/2 } } \mu(d). 
$$ 
Thus we want
$$ 
\sum_{3 \le p, q\le N } \sum_{\substack{d| (p-1)/2 \\ d| (q-1)/2 } } \mu(d) 
= \sum_{d \le N} \mu(d) \sum_{\substack{p, q\le N \\ p\equiv q\equiv 1 \mod{2d}}} 1 .
$$ 
For $d\le (\log N)^3$, use the prime number theorem in APs (Siegel-Walfisz) to see that these terms are 
$$ 
\sim \sum_{d\le (\log N)^3} \frac{\mu(d)}{\phi(2d)^2} \frac{N^2}{(\log N)^2}. 
$$ 
For $d> (\log N)^3$ estimate the sums over $p$ and $q$ trivially by $(N/d)^2$.  So these terms contribute 
$$ 
\ll \sum_{d> (\log N)^3} \frac{N^2}{d^2} \ll \frac{N^2}{(\log N)^3}, 
$$ 
which is negligible.  
So the required density is 
$$ 
\sum_{d\le (\log N)^3} \frac{\mu(d)}{\phi(2d)^2},
$$ 
which tends to the constant given above as $N\to \infty$.  
