Selection Examination A

If $x-2=0 \Leftrightarrow x=2$, then $y=1$ and we have the solution $(x,y)=(2,1)$.

If $x-2 \neq 0$, then, since $p$ is prime, from the equation $p(x-2)=x(y-1)$ it follows that $y \neq 1$ and $p|x$ or $p|y-1$. Therefore we have the cases:

* If $p|x$, then $x = px'$, where $x'$ positive integer, and then:
$$\begin{align*} p(px'-2) &= px'(y-1) \Leftrightarrow px' - 2 = x'(y-1) \Leftrightarrow x'(p-y+1) = 2 \\ &\Leftrightarrow x' = 1, p-y+1 = 2 \quad \eta \quad x' = 2, p-y+1 = 1 \\ &\Leftrightarrow x' = 1, y = p-1 \quad \eta \quad x' = 2, y = p \Leftrightarrow x = p, y = p-1 \quad \eta \quad x = 2p, y = p \\ &\Leftrightarrow (x,y) = (p,p-1) \quad \eta \quad (x,y) = (2p,p) \end{align*}$$

* If $p|y-1$, then $y-1 = py'$, where $y'$ is a positive integer and then
$$\begin{align*} p(x-2) &= xpy' \Leftrightarrow x-2 = xy' \Leftrightarrow x(1-y') = 2 \\ &\Leftrightarrow x = 1, 1-y' = 2 \quad \eta \quad x = 2, 1-y' = 1 \\ &\Leftrightarrow x = 1, y' = -1 \text{ (impossible) or } x = 1, y' = 0 \text{ (impossible).} \end{align*}$$

Moreover, if we are given $x+y=21$, then we have:

* Considering the solutions $(x,y) = (p,p-1)$, we find $2p-1=21 \Leftrightarrow p=11$, and so $(x,y,p) = (11,10,11)$.

* Considering the solutions $(x,y) = (2p,p)$, then $3p=21 \Leftrightarrow p=7$, and hence we find $(x,y,p) = (14,7,7)$.