Selection Examination A

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

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

If $p|x$, then $x=p{x}^{\prime }$, where $x^{\prime }$ positive integer, and then: $$\begin{array}{rl}p(p{x}^{\prime }-2)& =p{x}^{\prime }(y-1)\iff p{x}^{\prime }-2=x^{\prime }(y-1)\iff x^{\prime }(p-y+1)=2\\ & \iff x^{\prime }=1,p-y+1=2\eta x^{\prime }=2,p-y+1=1\\ & \iff x^{\prime }=1,y=p-1\eta x^{\prime }=2,y=p\iff x=p,y=p-1\eta x=2p,y=p\\ & \iff (x,y)=(p,p-1)\eta (x,y)=(2p,p)\end{array}$$

If $p|y-1$, then $y-1=p{y}^{\prime }$, where $y^{\prime }$ is a positive integer and then $$\begin{array}{rl}p(x-2)& =xp{y}^{\prime }\iff x-2=x{y}^{\prime }\iff x(1-y^{\prime })=2\\ & \iff x=1,1-y^{\prime }=2\eta x=2,1-y^{\prime }=1\\ & \iff x=1,y^{\prime }=-1\text{ (impossible) or }x=1,y^{\prime }=0\text{ (impossible).}\end{array}$$

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

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

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