We know that the number of squarefree integers $\le x$ that are coprime to $A$ is
$$
Q\_A(x) = x \prod\_{p|A} \left(1-\frac{1}{p}\right) \prod\_{p \nmid A} \left(1-\frac{1}{p^2}\right) + O(\sqrt{x}).
$$
Do we have explicit upper and lower bounds on $Q\_A(x)$?



Explicit bounds on number of squarefree numbers coprime to a certain number
I am assuming that explicit refers to the error term? In this case you can write $$Q\_A(x)=\sum\_{d\mid A}\mu(d)\sum\_{k\leq \sqrt x \atop \gcd(A,k)=1}\mu(k)\left[\frac{x}{dk^2}\right],$$ where $[t]$ denotes the largest integer $\leq t $. Note that $[t]=0$ when $0\leq t<1$ hence only the terms with $dk^2 \leq x $ make a contribution. Using $[t]=t+\theta(t)$ for some real number $ \theta(t) $ which satisfies $|\theta(t)|\leq 1$, we get $$\left|Q\_A(x)-\sum\_{d\mid A}\mu(d)\sum\_{k\leq \sqrt x \atop \gcd(A,k)=1}\mu(k) \frac{x}{dk^2}\right| \leq \sum\_{d\mid A}|\mu(d)| \sum\_{1\leq k\leq \sqrt {x/d} } 1,$$ where we ignored $\gcd(A,k)=1$ in the error term. Now we have $$ \sum\_{d\mid A}|\mu(d)| \sum\_{1\leq k\leq \sqrt {x/d} } 1\leq \sqrt x \epsilon(A) ,$$ where $$\epsilon(A):=\sum\_{d\mid A}\frac{|\mu(d)|}{\sqrt d }=\prod\_{p\mid A}\left(1+\frac{1}{\sqrt p }\right).$$ Hence, $$ 
xM-\epsilon(A) \sqrt x \leq Q\_A(x)\leq xM+\epsilon(A) \sqrt x,$$ where $$M=\sum\_{d\mid A}\frac{\mu(d)}{d}\sum\_{k\leq \sqrt{ x/d} \atop \gcd(A,k)=1} \frac{\mu(k)}{k^2}.$$ We can simplify $M$ to look into an Euler product: first adding the missing terms $k>\sqrt {x/d} $ we introduce an error term that is in absolute value $$\leq \sum\_{d\mid A} |\mu(d)|\sum\_{k>\sqrt {x/d} } \frac{1}{k^2}\leq \sum\_{d\mid A} |\mu(d)|\int\_{\sqrt{x/d}/2}^\infty \frac{\mathrm d u }{u^2} =\frac{2\epsilon(A)}{\sqrt{x}}.$$ The full series then becomes $$ \sum\_{d\mid A}\frac{\mu(d)}{d}\sum\_{k=1 \atop \gcd(k,A)=1}^\infty \frac{\mu(k)}{k^2}=\prod\_{p|A} \left(1-\frac{1}{p}\right) \prod\_{p \nmid A} \left(1-\frac{1}{p^2}\right),$$ hence, $$\prod\_{p|A} \left(1-\frac{1}{p}\right) \prod\_{p \nmid A} \left(1-\frac{1}{p^2}\right) -\frac{2\epsilon(A)}{\sqrt{x}}\leq M\leq \prod\_{p|A} \left(1-\frac{1}{p}\right) \prod\_{p \nmid A} \left(1-\frac{1}{p^2}\right)+\frac{2\epsilon(A)}{\sqrt{x}}.$$ Combining with the previous work we get $$ \left|Q\_A(x)-x\prod\_{p|A} \left(1-\frac{1}{p}\right) \prod\_{p \nmid A} \left(1-\frac{1}{p^2}\right)\right|\leq 3\epsilon(A) \sqrt x . $$ Assuming the Riemann hypothesis one can prove an exponent better than $1/2$ in $x^{1/2}$ and without Riemann hypothesis this can be improved slightly to something like $x^{1/2}/\exp(c\sqrt{\log x } )$ but it will be technical to find the implied constant as it depends on the zero free region of the Riemann zeta.


**Remark**: A bound for $\epsilon(A)$ may be found as follows: for $y=\tau(A)^{2/3} $ we have $$\epsilon(A) \leq \sum\_{d\leq y}1+\sum\_{d\mid A}1/\sqrt y \leq y+\tau(A)/\sqrt y=2 \tau(A)^{2/3}.$$


