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Determine the smallest natural number $n$ such that $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$ has as solutions exactly $15$ ordered pairs of natural $(x,y)$. I found out that this problem is not an obscure one and, using an algorithm, I found out that the $n=12$ is the smallest number to respect the property. I did some computations but I don't think this helps so I'll just ask for any help in writing a mathematical proof for this. Natural number solutions to $\frac{xy}{x+y}=n$ (equivalent to $\frac 1x+\frac 1y=\frac 1n$)

The last link didn't help too much.

Bill Dubuque
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  • For any particular $n$, it's not hard to count. Wlog, letting $x≤y$ then we see that $x≤2n$. So just search. – lulu Apr 12 '21 at 16:00
  • Seems to me all integers will have an infinite number of pairs. Let $x$ be any real number not equal to $0$ or $n$ and let $y = \frac 1{\frac 1n - \frac 1x}$. – fleablood Apr 12 '21 at 16:00
  • Of course $\frac 1{\frac 1n - \frac 1x}$ need not be an integer. – GEdgar Apr 12 '21 at 16:09
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    Cross multiplying by $xy$, $x+y=\frac{xy}{n}\to xy-nx-ny=0$. Then use https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick – TheBestMagician Apr 12 '21 at 16:16

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From the linked answer, we know that solutions of $\frac1x+\frac1y=\frac1n$ correspond bijectively with factors of $n^2$; we want $15$ solutions, so $\tau(n^2)=15$.

For $N=\prod_ip_i^{a_i}$, $\tau(N)=\prod_i(1+a_i)$. Since we have a square, all the $a_i$ are even; since $15=3×5$, $n^2$ is of the form $p^2q^4$ for $p,q$ distinct primes and $n=pq^2$. We now check $p=2,q=3$ and $p=3,q=2$ and arrive at the answer $n=12$.

Parcly Taxel
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