Primitive Inverse Pythagorean Triplets

Note that $\begin{aligned}\frac{1}{x^2}+\frac{1}{y^2}&=\frac{1}{z^2} \\z&=\frac{xy}{\sqrt{x^2+y^2}}\end{aligned}$

Since $z$ is an integer, so must $\sqrt{x^2+y^2}$ be an integer. Therefore, option C is definitely true.

Now, since $z$ is an integer, it also follows that $\sqrt{x^2+y^2}$ 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 $\begin{aligned}z&=\frac{dm\cdot dn}{\sqrt{d^2m^2+d^2n^2}}\\ &=\frac{dmn}{\sqrt{m^2+n^2}}\end{aligned}$

Due to how we defined $d$ , $\sqrt{m^2+n^2}$ cannot divide $mn$ . Now, since the left-hand side is an integer, we have that $d=k\sqrt{m^2+n^2}$ for an integer $k$ . Thus, $\begin{aligned}x&=dm=km\sqrt{m^2+n^2}\\ y&=dn=kn\sqrt{m^2+n^2}\\ z&=\frac{xy}{x+y}=kmn\end{aligned}$

Now, as $\gcd{(x,y,z)}=1$ , we must choose $k=1$ . Then, $\begin{aligned}x&=m\sqrt{m^2+n^2}\\ y&=n\sqrt{m^2+n^2}\\ z&=mn\end{aligned}$

Obviously, $\sqrt{m^2+n^2}$ is an integer. Using what we know from studying Pythagorean triplets, we can say that, for some $s$ , $t$ , and $r$ , $\begin{aligned}m&=r(s^2-t^2)\\ n&=2rst\end{aligned}$

Using this to rewrite $x$ , $y$ , and $z$ : $\begin{aligned}x&=r^2(s^2+t^2)(s^2-t^2)\\ y&=2r^2st(s^2+t^2)\\ z&=2r^2st(s^2-t^2)\end{aligned}$

Again, set $r=1$ so that $\gcd{(x,y,z)}=1$ : $\begin{aligned}x&=(s^2+t^2)(s^2-t^2)\\ y&=2st(s^2+t^2)\\ z&=2st(s^2-t^2)\end{aligned}$

Now, by inspection, we see that we can produce primitive inverse Pythagorean triplets by choosing $s$ and $t$ carefully. For example, setting $t=s-1$ always result in $\gcd{(x,y,z)}=1$ . This demonstrates that there are infinite primitive inverse Pythagorean triplets. Thus, option A is true.

With the above formulas, we can show that $\frac{xz}{y}$ , $\frac{yz}{x}$ , and $\frac{xy}{z}$ must be perfect squares. Thus, option B is also true.