58th Ukrainian National Mathematical Olympiad

Factor the left-hand side of the equation: $$(y{x}^2-x{y}^2)+(yp-xp)=p\Rightarrow yx(x-y)-p(x-y)=p\Rightarrow (yx-p)(x-y)=p.$$ The last equation is possible in several cases.

Case 1. $yx-p=1$, $x-y=p$. Then, we get a quadratic equation: $x=y+p$ $$y(y+p)-p=1\Rightarrow y^2+yp-p-1=0.$$ Since $y=1$ is the solution to this equation, then another solution is $y=-p-1$, but it is not a positive integer. Hence, we obtain that $x=p+1$. Putting these values to the initial equation, we see that a tuple $(p+1;1;p)$ satisfies the statement for arbitrary prime number $p$.

Case 2. $yx-p=-1$, $x-y=-p$. Then, we get a quadratic equation: $y=x+p$ $$x(x+p)-p=-1\Rightarrow x^2+xp-p+1=0.$$ Since $x\in \mathbb{N}$, the discriminant of the last equation must be the square of an integer number: $$D=p^2-4(-p+1)=p^2+4p-4.$$ Let us see which $p$ satisfies the conditions: $$(p+1{)}^2<p^2+4p-4<(p+2{)}^2.$$ Right inequality holds for any $p$, while the left one can be re-written: $$p^2+2p+1<p^2+4p-4\iff 2p>5\iff p\ge 3.$$ Therefore, we must check separately the case $p=2$, whereas for any other prime $p$ the discriminant is not the square of an integer number. For $p=2$, we get $D=8$, which is also not the square of an integer, which means such positive integer $x$ does not exist.

Case 3. $yx-p=p$, $x-y=1$. Then, we obtain the equations: $x=y+1$ and $yx=2p$. From the 1st equation, $x>y$, from the 2nd, $y=1$ and $x=2p$ or $y=2$ and $x=p$. If we combine these conditions, then we obtain $y=1,x=2,p=1$ – contradiction, or $y=2,x=p=3$ – solution.

Case 4. $yx-p=-p$, $x-y=-1$. Obviously, the first equation yields the contradiction.