I am currently following Harold N. Shapiro's 1950 paper On primes in arithmetic progressions I. Particularly, I was trying to prove Lemma 9 and became stuck at the first step that
$$ \sum_{\substack{dd'\le x\\dd'\equiv B\pmod A}}\mu(d)\log^2d'=\sum_{\substack{d\le x\\\gcd(d,A)=1}}\mu(d)\sum_{\substack{d'\le x/d\\ d'\equiv d^{-1}B\pmod A}}\log^2d' $$
where $d^{-1}$ is the arithmetic inverse to $d$ modulo $A$. I wonder why we can immediately conclude $\gcd(d,A)=1$. Is it that $dd'\equiv B\pmod A$ has no solution when $\gcd(d,A)>1$?
It was presumed that $\gcd(B,A)=1$, so we can invert the congruence to get
$$ B^{-1}dd'\equiv1\pmod A\tag1 $$
Now, all we need is to determine what conditions need to be imposed on $d$ in order for $d'$ to have solution.
It is known that the equation $ax+by=1$ has integer solution $(x,y)$ if and only if $\gcd(a,b)=1$. If $x_0,y_0$ is a pair of solution, then the general solution to this equation can be written as $x=x_0+bt$ and $y_0+at$. This implies that the congruence
$$ ax\equiv1\pmod b $$
has unique solution modulo $b$ if and only if $(a,b)=1$. Consequently, there exists $d'$ satisfying the congruence (1) if and only if $\gcd(B^{-1}d,A)=1$. Because $\gcd(B^{-1},A)=1$ we deduce that $\gcd(d,A)=1$.
Consequently, we see that the identity is automatically true.