Reciprocals of Primes

I came across an interesting video on YouTube by Michael Penn in which he investigates an interesting property of prime numbers, namely that if $p$ is an odd prime, then:$$\frac{2}{p}=\frac{1}{m}+\frac{1}{n}\text{ where }m>n\ge 1\text{ and }m,n\in \mathbb{Z}$$Furthermore, this representation is unique. The proof is simple enough and yet quite elegant. Here it is:$$\begin{array}{rl}\frac{2}{p}& =\frac{1}{m}+\frac{1}{n}\\ 2mn& =np+mp\\ 2mn-np-mp& =0\\ 4mn-2np-2mp& =0\\ 4mn-2np-2mp+p^2& =0+p^2\\ (2m-p)(2n-p)& =p^2\end{array}$$This is the crucial point. $p^2$ factorises to either $p\times p$ or $p^2\times 1$. However, because $m>n$, then $2m-p>2n-p$ and so we need to look at the factors of $p^2$ and $1$. So, because $p^2>1$, we have:$$\begin{gathered} 2m-p=p^2\text{ and }2n-p=1 \\ m=\frac{p(p+1)}{2}\text{ and }n=\frac{p+1}{2} \end{gathered}$$The $p\times p$ factorisation leads to the trivial case, where $m=n=p$. Figure 1 shows the results for the odd primes up to 97.

Figure 1

Thus we can write ${\displaystyle \frac{2}{97}}={\displaystyle \frac{1}{4753}}+{\displaystyle \frac{1}{49}}$. If a composite number is entered into the above calculations, there will be more than one representation. For example, in the case of 25, we can write:$$\frac{2}{25}=\frac{1}{75}+\frac{1}{15}=\frac{1}{325}+\frac{1}{13}$$The reason is that the possible factorisations are no longer just $p\times p$ and $p^2\times 1$. In the case of 25, we now have 625 x 1 as well as 125 x 5 and of course the trivial 25 x 25. Hence the two solutions shown above. Here is the embedded video:

It was only when reading some of the comments to this video that I came across a connection with the harmonic mean. For two numbers, $m$ and $n$, the harmonic mean is defined to be $\frac{2mn}{m+n}$. If we go back to our original expression, we see that:$$\begin{array}{rl}\frac{2}{p}& =\frac{1}{m}+\frac{1}{n}\\ \frac{2}{p}& =\frac{m+n}{mn}\\ \frac{p}{2}& =\frac{mn}{m+n}\\ p& =\frac{2mn}{m+n}\end{array}$$So $p$ represents the harmonic mean of the numbers $m$ and $n$. For example, the harmonic mean of 49 and 4753 is 97 (referring back to Figure 1). The harmonic mean is one of the Pythagorean means, the other two being the arithmetic mean and the geometric mean. I did mention these means briefly in a post from 13th September 2020 titled Root-Mean-Square and Other Means but I should investigate this topic in more detail.