Definition.
$$ f(n) := \sum_{d \mid p_n\#} \mu(d)\sum_{r^2 = 1\mod d}\lfloor\frac{p_{n+1}^2 - 2 - r}{d}\rfloor $$
where $p_{n+1}$ is the $(n+1)$th prime number. And where it is understood that each solution $r$ in the inner summation ranges over $\{0, \dots, d-1\}$ (the usual residues modulo $d$).
Conjecture. The function $f(n)$ has no fixed point, i.e. there exists no $n \in \Bbb{N}$ such that $f(n) = n$.
Attempt. The strategy of this attempt is to first express each floor $\lfloor \frac{z}{d}\rfloor$ as $\frac{ z - [z]_d}{d}$ where $[z]_d$ means the usual residue of $z$:
$$ N := p_{n+1}^2 - 2 \\ \ \\ n = f(n) = \sum_{d \mid p_n\#} \mu(d)\sum_{r^2 = 1\mod d}\frac{ N- r -[N-r]_d}{d} \\ \iff \\ n =\left(\sum_{d \mid p_n\#} \mu(d)\sum_{r^2 = 1\mod d}\frac{ N- r}{d}\right) - \left(\sum_{d \mid p_n\#} \mu(d)\sum_{r^2 = 1\mod d}\frac{ [N-r]_d}{d}\right) $$
Now multiply everything by $p_n\#$, giving:
$$ p_n\#\cdot n = N\cdot p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{1}{d} \right) -p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d}\frac{r}{d} \right) - g(n) $$
where $g(n) := p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{[N -r]_d}{d}\right)$. We know that:
$$ \sum_{r^2 = 1\mod d} \frac{1}{d} = \frac{1}{d}\begin{cases}2^{\omega(d)}, \text{ if } 2 \nmid d \\ 2^{\omega(d) - 1}, \text{ if } 2 \mid d \end{cases} $$ That is, by square-freeness of each $d$ being a divisor of $p_n\#$, that is the count of the solutions to $r^2 = 1 \pmod d$. Thus:
$$ N\cdot p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{1}{d} \right) = N \cdot p_n\# \left( \sum_{d \mid \frac{p_n\#}{2}}\frac{(-2)^{\omega(d)}}{d}- \frac{1}{2}\sum_{d \mid \frac{p_n\#}{2}}\frac{(-2)^{\omega(d)}}{2d}\right) $$
Since $\mu(d) = (-1)^{\omega(d)}$ for square-free $d$. And we've we've handled the case statement by breaking up into two sums each over odd divisors of $p_n\#$. This then equals:
$$ \frac{3\cdot N \cdot p_n\#}{4} \prod_{\text{ prime } q \mid \frac{p_n\#}{2}}(1 - \frac{2}{q}) =\frac{3\cdot N \cdot p_n\#}{4} \prod_{\text{ prime } q \mid \frac{p_n\#}{2}}(\frac{q - 2}{q}) \\ \ \\ \ = \frac{3 \cdot N \cdot p_n\#}{4 \frac{ p_n\#}{2}} \prod_{\text{ prime } q \mid \frac{p_n\#}{2}}(q-2) \\ \ \\ = \frac{3 \cdot N}{2}\prod_{\text{ prime } 3 \leq q \leq p_n}(q - 2) $$
The next summation term is:
$$ p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d}\frac{r}{d} \right) = p_n\#\left(\frac{1}{2} + \sum_{2 \lt d \mid p_n\#}\mu(d) \sum_{r^2 = 1 \pmod d, \ \\ r \lt \frac{d}{2}}\frac{r + (d - r)}{d} \right)\\ \ = p_n\#\left(\frac{1}{2} +\sum_{2 \lt d \mid p_n\#}\mu(d)\frac{1}{2}|\{r^2 = 1 \mod d\}|\right) \\ =p_n\#\left(\frac{1}{2} + \sum_{1 \lt d \mid \frac{p_n\#}{2}}(-1)^{\omega(d)}2^{\omega(d)}- \frac{1}{2}\sum_{1 \lt d \mid \frac{p_n\#}{2}}(-1)^{\omega(d)}2^{\omega(d)}\right) \\ = p_n\#\left(\frac{1}{2} + \frac{1}{2}\sum_{1 \lt d \mid \frac{p_n\#}{2}}(-2)^{\omega(d)} \right) \\ \ = \frac{p_n\#}{2}\left( \sum_{d \mid \frac{p_n\#}{2}}(-2)^{\omega(d)}\right) = \frac{p_n\#}{2}\prod_{\text{ prime } 3 \leq q \leq p_n}(1-2) \\ = \frac{p_n\#}{2}(-1)^{n - 1} $$
Back to our original problem, together with the transformed formulae, we want to prove that the following equation has no solution $n \in \Bbb{N}$:
$$ p_n\# \cdot n =\\ \left( \frac{3 \cdot N}{2}\prod_{\text{ prime } 3 \leq q \leq p_n}(q - 2)\right) -\left(\frac{p_n\#}{2}(-1)^{n - 1}\right) - p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{[N -r]_d}{d}\right) $$
Question.
How can the last summation-term:
$$ p_n\#\left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{[N -r]_d}{d}\right) $$ be handled similarly to how I've rearranged the other two summation-terms?
Here's one method I just thought of:
$$ \sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{[N -r]_d}{d} \\ \ = \sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \frac{[N +r]_d}{d} \\ = \sum_{d \mid p_n\#}\mu(d)\left(\sum_{r^2 = 1 \mod d \\ [N]_d + r \lt d} \frac{[N]_d + r}{d} + \sum_{r^2 = 1 \mod d \\ [N]_d + r \geq d} \frac{[N]_d + r - d}{d}\right) \\ = \left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d}\frac{[N]_d + r}{d}\right) - \left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d \\ [N]_d + r \geq d} \frac{d}{d}\right) \\ = \left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d}\frac{[N]_d + r}{d}\right) - \left(\sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d} \lfloor\frac{[N]_d + r}{d} \rfloor\right) $$
where we've used $r$ in place of $-r$ and such a thing is valid for each divisor $d$ because $r^2 = 1 \mod d \iff (-r)^2 = 1 \mod d$ for all $d \geq 1$.
In the first summation-term we've already handled same exact piece in the OP of:
$$ \sum_{d \mid p_n\#}\mu(d)\sum_{r^2 = 1\mod d}\frac{r}{d} = \frac{(-1)^{n - 1}}{2} $$