Problem 1170

Factor the left-hand side of the equation:
$$(yx^2 - xy^2) + (yp - xp) = p \Rightarrow yx(x - y) - p(x - y) = p \Rightarrow (yx - p)(x - y) = p.$$
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:
$$p^2 + 2p + 1 < p^2 + 4p - 4 \Leftrightarrow 2p > 5 \Leftrightarrow p \ge 3.$$
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.