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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?

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    $p$ are primes or just divisors? – 2017-02-21
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    @Exodd $p$ means a prime divisor of $q$. – 2017-02-21
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    Are multiple prime factors included up to their count of multiplicity, or are they only counted once? For example, if $q=8$, does $p=2$ once or 3 times? – 2017-02-21
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    @Paul: q is squarefree, so q=8 isn't considered here. – 2017-02-21
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    Note that $\prod_{p \mid q} (p^2+1)=\sum_{d \mid q} d^2$. ($p$ is prime, $m$ is not) The same can be done for another product and it will be not hard to see that this is proportional to $q^2$. – 2017-02-21
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    @didgogns I suppose you meant $\prod_{p \mid q} (p^2+1) = \sum_{d|q} d^2$. Can you elaborate your arguments? For instance how do you treat the other product? – 2017-02-21
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    @user41481 Thanks for pointing it out. – 2017-02-21

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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.

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    It has much lower lower bounds, consider the case that $q$ is a prime. – 2017-02-21
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    @didgogns: correct. Fixed. – 2017-02-21
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    @JackD'Aurizio Can you say something 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? – 2017-02-22
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    @user41481: in the same spirit, compute $\prod_{p\mid q}(1+p)$ by yourself. What you did, i.e. accepting my answer, then modifying your question adding a second question and un-accepting my answer, is not very fair by my standards. – 2017-02-22
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    @JackD'Aurizio.. I am sorry, that was by mistake. – 2017-02-27