Prove of $\prod_{d|n} (\mu(d)(\mu(d) + 3) + 4) = 4^{d(n)}$

Question

Found an interesting relation: $$\prod_{d|n} (\mu(d)(\mu(d) + 3) + 4) = 4^{d(n)}$$ where $\mu(n)$ is a Möbius function and $d(n)$ is a divisors count.

I think this should be something known. The prove I know is a bit tricky and can be found in http://oeis.org/A262804

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Based on nice answers, just the other formula:

$$\prod_{d|n} \left(\mu(d) \left(\mu(d) \left(\mu(d) + \frac{a(a-1)^2}{2}\right) + \frac{ a^3 - a-2}{2}\right)+ a^2\right) = a^{2d(n)}$$

Answer 1

Hint: $$\begin{align}\mu(d)(\mu(d)+3)+4&=\begin{cases}4&\mu(d)=0\\2&\mu(d)=-1\\8&\mu(d)=1\end{cases}\\&=2^{2+\mu(d)}\end{align}$$

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Given $m,p$ we have that if $a=\frac{m(p-1)^2}2,b=\frac{m(p^2-1)}2,c=mp ,$ then $$\prod_{d\mid n}\left(a\mu^2(d)+b\mu(d)+c\right)=(mp)^{d(n)}$$

The case $m=p=2$ is your case, and is the only case where $a=1,$ at where $m,p$ are integers.

When $m=1,p=3,$ you get $a=2,b=4,c=3.$

Answer 2

Hint: For $n > 1 , \sum_{d\mid n } \mu (d) = 0 $.

Hence, $ \sum_{d\mid n } ( \mu (d) + 2) = 2 d(n)$.

The result follows.

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Note that for $n= 1$ the relation isn't true.
In particular, for $ n = 1$, $ \sum_{d\mid n } \mu (d) = 1 $, and in a similar manner we conclude that the RHS $= 2^{ 2 + 1 }$.