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

  • $\begingroup$ Your Equation (3) follows from Hermite's Identity (see the link in my answer). $\endgroup$ Aug 22, 2018 at 2:44
  • $\begingroup$ You claims (1) and (2) also follow from Hermite's Identity (ignoring the case $n=1$ so $p_1=2$). $\endgroup$ Aug 22, 2018 at 2:50
  • 1
    $\begingroup$ can you explain the notation: What is $p_n$ (is it the $n$-th prime?)? what is $\delta(n,1)$? $\endgroup$
    – Lior B-S
    Aug 22, 2018 at 6:14
  • $\begingroup$ Sorry $$\delta \left( x,y \right) =\cases{1&$x=y$\cr 0&$x\neq y$\cr}$$ and yes as always $p_n$ is the $n^{th}$ prime number $\endgroup$ Aug 22, 2018 at 14:48

1 Answer 1


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.

  • $\begingroup$ Well I'm not sure that they follow from the identity you provided, in that I feel as if they are both equivalent statements, I suppose yes, this identity is indeed more general and so one could say that one follows from the other $\endgroup$ Aug 22, 2018 at 2:57
  • $\begingroup$ But still this undoubtable a valid answer and a new identity I was unaware of, so thankyou for your time this will be a lot of help for me. $\endgroup$ Aug 22, 2018 at 2:58
  • $\begingroup$ Sorry is the proof supplied on Wikipedia the only one you are aware of for Hermite's Identity or is there any others you would be able to refer me to @Batominovski? $\endgroup$ Aug 22, 2018 at 15:28
  • $\begingroup$ That is the only proof I know. And I don't think any other proofs will be very different from the one given in Wikipedia. $\endgroup$ Aug 22, 2018 at 15:30
  • $\begingroup$ Sure it's not so much as I was expecting a drastically different collection of statements, it's more that I grasp a better understanding at a much faster rate if I look at how the problem was approached from a number of different authors, but thankyou for the response none the less. $\endgroup$ Aug 22, 2018 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.