Difference between two products

Question

Let $q$ be a square free natural number. Can the difference $$ \prod_{p \mid q} (p^2+1) - \prod_{p \mid q} (p-1)^2 $$ be estimated in terms of $q$? What would be the correct order of the difference in terms of $q$? Is the difference $\asymp q^2$? Can something be said about $$ \prod_{p \mid q} \frac{p^2+1}{p+1} - \prod_{p \mid q} \frac{(p-1)^2}{p+1}, $$ in the same spirit?

Answer 1

Such difference equals $$ q^2\left[\prod_{p\mid q}\left(1+\frac{1}{p^2}\right)-\prod_{p\mid q}\left(1-\frac{1}{p}\right)^2\right]\tag{1} $$ and by Euler's product $$ \prod_{p\in\mathcal{P}}\left(1+\frac{1}{p^2}\right)=\frac{\zeta(2)}{\zeta(4)}=\frac{15}{\pi^2}\tag{2}$$ while $$ \prod_{p\leq x}\left(1-\frac{1}{p}\right)^2\approx\frac{C}{\log(x)^2}\tag{3}$$ hence your difference is by $\frac{15}{\pi^2}q^2$, but can be as small as $2q$ if $q$ is a prime.