<s>[INST] Prime Number Congruence Modulo 8 Proof I want to rigorously prove that: 
$$p_n=2\Biggl(\Bigl\lfloor \frac{p_n+1}{8}\Bigr\rfloor+\Bigl\lfloor \frac{p_n+3}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+5}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+7}{8} \Bigr\rfloor\Biggr)-1+\delta(n,1) \quad\quad\quad\quad\quad\quad(0)$$
So far what has convinced me is the observations as follows:
$$\frac{p_n-5}{2}-2\Biggl(\Bigl\lfloor \frac{p_n-1}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n-5}{8} \Bigr\rfloor\Biggr)+\frac{1}{2}\delta(n,1) \in {\{0,1}\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$$
$$\frac{p_n+1}{2}-2\Biggl(\Bigl\lfloor \frac{p_n+1}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+5}{8} \Bigr\rfloor\Biggr)-\frac{3}{2}\delta(n,1) \in {\{0,1}\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$
$$\Bigl\lfloor \frac{n+1}{8} \Bigr\rfloor+\Bigl\lfloor \frac{n+3}{8} \Bigr\rfloor+\Bigl\lfloor \frac{n+5}{8} \Bigr\rfloor+\Bigl\lfloor \frac{n+7}{8} \Bigr\rfloor=\Bigl\lfloor \frac{n+1}{2} \Bigr\rfloor \,\,\forall n \in \mathbb N\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,(3)$$
$(3)$ explains why  the RHS of $(0)$ must be odd, and $(0)$ being the sum of the expressions in lemmas $(1)$ & $(2)$ * show why   the RHS of $(0)$ is equal to $p_n$, but this is as far as I can get without a text reference to something specifically relevant.
*I have manipulated these based on considerations of what a congruence relation implies, having the property of translation
 [/INST] Here is an alternative way.  It is well known that
$$\lfloor nx\rfloor =\sum_{i=1}^n\,\left\lfloor x+\frac{i-1}{n}\right\rfloor\text{ for all }x\in\mathbb{R}\text{ and }n\in\mathbb{Z}_{>0}\,.$$
Take $n:=8$ and $x:=\dfrac{p}{8}$ if $p$ is an odd prime.  Thus, we get
$$p=\Biggl\lfloor 8\left(\frac{p}{8}\right)\Biggr \rfloor =\sum_{i=1}^8\,\left\lfloor\frac{p+i-1}{8}\right\rfloor=\sum_{i=1}^8\,\left\lfloor\frac{p+i}{8}\right\rfloor-1\,.$$
Now, since $p\equiv 1\pmod{2}$, we obtain
$$\left\lfloor\frac{p+2i-1}{8}\right\rfloor=\left\lfloor\frac{p+2i}{8}\right\rfloor\text{ for }i=1,2,3,4\,.$$
This shows that
$$p=2\,\sum_{i=1}^4\,\left\lfloor\frac{p+2i-1}{8}\right\rfloor-1\text{ for an odd prime }p\,.$$
You just have to check the case $p=2$ separately.
</s>