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:$$\frac{2}{p}=\frac{1}{m}+\frac{1}{n} \text{ where } m>n \geq 1 \text{ and } m,n \in \mathbb{Z}$$Furthermore, this representation is unique. The proof is simple enough and yet quite elegant. Here it is:$$\begin{align} \frac{2}{p}&=\frac{1}{m}+\frac{1}{n}\\2mn&=np+mp\\2mn-np-mp&=0\\4mn-2np-2mp&=0\\4mn-2np-2mp+p^2&=0+p^2\\(2m-p)(2n-p)&=p^2 \end{align}$$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:$$2m-p=p^2 \text{ and } 2n-p=1\\ m=\frac{p(p+1)}{2} \text{ and } n=\frac{p+1}{2}$$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 \( \dfrac{2}{97}=\dfrac{1}{4753}+\dfrac{1}{49} \). 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:
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