Primitive Harmonic Triplets

Note that $\begin{aligned}\frac{1}{x}+\frac{1}{y}&=\frac{1}{z} \\z&=\frac{xy}{x+y}\end{aligned}$

Now, since $z$ is an integer, $x+y$ must divide $xy$ . In other words, there must exist a certain integer $d\ge 2$ which divides both $x$ and $y$ . Choose $d$ such that it is the greatest common factor of $x$ and $y$ , or $d=\gcd{(x,y)}$ .

Then, rewrite the above expression with $\begin{aligned}x&=dm\\ y&=dn\end{aligned}$

This gives $\begin{aligned}z&=\frac{dm\cdot dn}{dm+dn}\\ &=\frac{dmn}{m+n}\end{aligned}$

Due to how we defined $d$ , $m+n$ cannot divide $mn$ . Now, since the left-hand side is an integer, we have that $d=k(m+n)$ for an integer $k$ . Thus, $\begin{aligned}x&=dm=km(m+n)\\ y&=dn=kn(m+n)\\ z&=\frac{xy}{x+y}=kmn\end{aligned}$

Now, as $\gcd{(x,y,z)}=1$ , we must choose $k=1$ . Then, $\begin{aligned}x&=m(m+n)\\ y&=n(m+n)\\ z&=mn\end{aligned}$

Using the above, we can choose any coprime integers $m,n$ to get a primitive harmonic triplet. Thus, there are an infinite amount of primitive harmonic triplets.

Now, considering the fact that $z=mn$ , where $m,n$ must be coprime, realize that any two coprime factors of $z$ correspond to a primitive harmonic triplet. The number $2^{5}3^{4}5^{3}7^{2}11$ can be factored into $2^{4}=16$ coprime pairs (equal to $2^{\text{number of prime factors}-1}$ ). As the order of the pairs matter (and the fact that coprime integers can never be equal to each other), multiply this number by $2$ to get $\boxed{32}$ .