https://aops.com/community/p11396651 Supercali wrote: Proof outline: The only solutions are $(x,y)=(1,1)$. The given conditions imply that $x|y^2-y+1$ and $y|x^2-x+1$, as well as $gcd(x,y)=1$. So we get $$xy| x^2+y^2-x-y+1$$After that it is an easy infinite descent.

jasperE3 wrote: Find all pairs of positive integers $(x,y)$ so that $$\frac{(x^2-x+1)(y^2-y+1)}{xy}\in\mathbb N.$$ Without the loss of generality,assume that $x \ge y$ and proceed. By the fact that $x|y^2-y+1$ and $y|x^2-x+1$, as well as $gcd(x,y)=1$. So we get $$xy| x^2+y^2-x-y+1$$And then factorise to get some congruencies to get $(1,1)$ to be the only solution

Vieta jumping method

The answer is $(1,1)$ only which clearly works. The conditions imply that $$\frac{x^2+y^2-x-y+1}{xy}:=k \in \mathbb{Z}^+.$$By AM-GM we have $$k-1 \geq\frac{xy-x-y+1}{xy}=\frac{(x-1)(y-1)}{xy}.$$Thus if $k=1$ then we need $x=1$ or $y=1$, else the last fraction is positive, which then clearly forces $(x,y)=(1,1)$. Now suppose $k>1$ has a solution, and consider the minimal solution to the equation with $x\geq y$. The given equation rewrites as $$x^2-(ky+1)x+(y^2-y+1)=0,$$so aside from $x$ there is some other solution $x'$ to this quadratic with $x' \geq x\geq y$, hence $y^2-y+1=xx'\geq y^2$, or $y \leq 1 \implies y=1$. This forces $(x,y)=(1,1)$ as before, which means $k=1$: contradiction. $\blacksquare$