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A natural number $p$ is prime if and only if it has exactly two solution sets $\{x,y\}$ to the equation $$\frac1p = \frac1x + \frac1y, \text{ or equivalently, }p=\frac{xy}{x+y}.$$

Those solution sets are $\{p,p(p+1)\}$ and $\{2p\}$ (where $x=y=2p$).

Is this a known fact about prime numbers? If so, could somebody provide a reference?

A proof is given below.

Proof: $(\implies)$ Suppose $p$ is prime. Then if $$p=\frac{xy}{x+y},$$ we can express $x=p^ak$ and $y=p^bl$ for $k$ and $l$ coprime to $p$. Without loss of generality, suppose $0\leq a\leq b$ and $1\leq b$. Then we have $$p=p^b\frac{kl}{k+lp^{b-a}}.$$

Suppose that $b>a$. We then have that $p\not\mid k+lp^{b-a}$, and therefore $b=1$ and $a=0$. Therefore $kl=k+lp$, which implies $k(l-1)=lp$. Thus, $p\mid l-1$.

Consider any natural number $m\mid k$. Then $m\mid k(l-1)=lp$. Since $p\not\mid k$, we have $p\not\mid m$. Thus, $m\mid l$.

Similarly, for any natural $n\mid l$, we have $n\mid l(k-p)=k$.

So $l$ and $k$ have the exact same divisors, and thus $l=k$. Thus,

$$k^2=k+kp=k(p+1),$$ meaning $k=p+1$. So $x=p+1$, which implies $y=p(p+1)$. This is the first solution set.

Next, consider the case in which $b=a$. Then $$p=p^b\frac{kl}{k+l},$$ so $$p^{1-b}=\frac{kl}{k+l}.$$

Since $k$ and $l$ are coprime to $p$, it follows $b>0$. We also have $$\frac{1}{p^{b-1}}=\frac1k+\frac1l,$$ implying that $$\frac{1}{p^{a+b-1}}=\frac{1}{p^{2b-1}}=\frac1x+\frac1y,$$ so $2b-1=1$ and thus $a=b=1$. Finally, that implies $$p=p\frac{kl}{k+l},$$ so $kl=k+l$ and therefore $k=l=2$. This gives the solution $x=y=2p$. So if $p$ is prime then it has two solution sets.

$(\Longleftarrow)$ If $p$ is not prime, then $p=qr$ where $q$ is prime and $r\neq 1$. Then $$\frac1p=\frac1{p+r}+\frac1{q(p+r)}$$ gives us another solution set.

MathTrain
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    It is probably known. It follows from the fact that the number of solutions to $\frac{1}{n}=\frac{1}{x}+\frac{1}{y}$ is exactly the number of divisors of $n^2$. – Wojowu Jan 31 '21 at 23:07
  • Agreeing with Wojowu here. I would certainly have interest in any proof that leaned on this aspect of primality more than other definitions, but the characterization itself is known. – leslie townes Jan 31 '21 at 23:09
  • I agree, it's almost certainly known. Would you know where to look for a reference? – MathTrain Jan 31 '21 at 23:11
  • Use the method in the linked dupe to show that the solutions correspond to factorizations of $,p^2.,$ Counting these factorizations shows that there are precisely two $\iff p,$ is prime (by unique factorization). There are many prior instances of this idea, e.g. see here. Please search before posting to avoid such duplication. – Bill Dubuque Jan 31 '21 at 23:43

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This is a known fact, and here follows a simple proof.

Let $p$ be a prime. Suppose $x \ge y \ge 1$ are integers such that $$\frac{1}{p} = \frac{1}{x}+ \frac{1}{y}$$ Clearly this equation implies that $x,y >p$. Multiplying everything by $xyp$ we get $$xy-px-py=0$$ $$xy-px-py+p^2=p^2$$ $$(x-p)(y-p)=p^2$$ $(x-p)$ and $(y-p)$ are two positive integers whose product is $p^2$: there are only two possibilities:

  1. $x-p=p^2$ and $y-p=1$: this gives us the solution $(x;y)=(p^2+p;p+1)$

  2. $x-p=p$ and $y-p=p$: this gives us the solution $(x;y)=(2p;2p)$

Crostul
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    Please strive not to post more dupe answers to dupes of FAQs. See my comment on the question for how to locate such dupes. – Bill Dubuque Jan 31 '21 at 23:41