SELECTION EXAMINATION

1. The equation is written: $$\begin{gathered} xy(x+y-10)-3x^2-2y^2+21x+16y=60 \\ \iff (y-3)x^2+(y^2-10y+21)x=2y^2-16y+60 \\ \iff (y-3)x^2+(y-7)(y-3)x=2(y-3)(y-5)+30 \\ \iff (y-3)x^2+(y-7)(y-3)x-2(y-3)(y-5)=30 \\ \iff (y-3)[x^2+(y-7)x-2(y-5)]=30 \\ \iff (y-3)(x^2-7x+xy-2y+10)=30 \\ \iff (y-3)(x^2-4-7x+xy-2y+14)=30 \\ \iff (y-3)((x-2)(x+2)-7(x-2)+(x-2)y)=30 \\ \iff (y-3)(x-2)(x+y-5)=30=1\cdot 2\cdot 3\cdot 5. \end{gathered}$$ Since $x-2$, $y-3$ and $x+y-5$ are integers such that $x-2+y-3=x+y-5$ and moreover $x-2\ge -1$, $y-3\ge -2$ and $x+y-5\ge -3$, we have to consider the following cases: $$\begin{gathered} \{\begin{array}{l}x-2=2\\ y-3=3\\ x+y-5=5\end{array}\text{ or }\{\begin{array}{l}x-2=3\\ y-3=2\\ x+y-5=5\end{array}\text{ or }\{\begin{array}{l}x-2=1\\ y-3=5\\ x+y-5=6\end{array}\text{ or }\{\begin{array}{l}x-2=5\\ y-3=1\\ x+y-5=6\end{array} \\ (x,y)=(4,6)\text{ or }(x,y)=(5,5)\text{ or }(x,y)=(3,8)\text{ or }(x,y)=(7,4). \end{gathered}$$

Alternatively, from the equation $(y-3)x^2+(y-3)(y-7)x=2(y^2-8y+30)$ we get: $$(y-3)[x^2+(y-7)x]=2y^2-16y+60(1)$$ From which, since $x$, $y$ are positive integers, arises that: $(y-3)|2y^2-16y+60$. Since $2y^2-16y+60=(y-3)(2y-10)+30$ arises that $y-3|30$. Therefore $y-3\in \{\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30\}$, and since $y-3\ge -2$ we get: $y-3\in \{\pm 1,\pm 2,3,5,6,10,15,30\}$. For each of these values, by substituting to (1) we obtain a trinomial of $x$.