A Super Nice Equation

Observe that since $\begin{aligned} \left(\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{z-x}\right)^2 &=\frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(y-z)^2}+2\left(\frac{1}{(x-y)(y-z)}+\frac{1}{(y-z)(z-x)}+\frac{1}{(z-x)(x-y)}\right) \\ &=\frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(y-z)^2}+2\left(\frac{(z-x)+(x-y)}{(x-y)(y-z)(z-x)}+\frac{1}{(z-x)(x-y)}\right) \\ &= \frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(y-z)^2}+2\left(\frac{z-y}{(x-y)(y-z)(z-x)}+\frac{1}{(z-x)(x-y)}\right) \\ &=\frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(y-z)^2}+2\left(-\frac{1}{(x-y)(z-x)}+\frac{1}{(z-x)(x-y)}\right) \\ &= \frac{1}{(x-y)^2}+\frac{1}{(y-z)^2}+\frac{1}{(y-z)^2} \end{aligned}$ thus, since 2021 is clearly not a square of a rational number, we're done. $\blacksquare$

Remark: This problem is adapted from 2014 Baltic Way.

Suppose $x,y,z \in \mathbb{Q}$ , let $a=\frac{1}{x-y}, b=\frac{1}{y-z}, c=\frac{1}{z-x},$ then $a,b,c \in \mathbb{Q}$ . Because $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ , we can write $c=-\frac{ab}{a+b}$ So $a^2+b^2+\frac{a^2b^2}{(a+b)^2}=2021$

Setting $s=a+b$ and $p = ab$ this writes as $s^2-2p+ \frac{p^2}{s^2} = 2021$ or (multiplied by $s^2$ and rearranged):

$p^2-2ps^2 +s^4 -2021s^2=0$ Solving for $p$ : $p = s^2 \pm s\sqrt{2021 }$ Note that s cannot be 0, because p, a, and b would be 0 too. $\frac{p-s^2}{s}=\pm\sqrt{2021}$ Since the left hand side is rational and the right hand side is irrational, this is impossible. Conclusion: x, y, and z cannot all be rational.