SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Since $x^2-px+q=0$ has integral solution, then we may assume that $$\{\begin{array}{l}p=a+b\\ q=ab\end{array}\text{ with }a,b\in {\mathbb{Z}}^{+}.$$ Note that $q-p+1=ab-(a+b)+1=(a-1)(b-1)\ge 0$ then $q+1\ge p$. By the same way on the second equation $x^2-qx+p=0$, then we have $p+1\ge q$. Thus $p=q$ or $p+1=q$ or $q+1=p$. We consider three cases:

1. If $p=q$ then $x^2-px+p=0$ then we have $$\Delta =p^2-4p=k^2\iff (p-2{)}^2-k^2=4.$$ Hence, $(p-2-k)(p-2+k)=4$. Note that $(p-2-k)+(p-2+k)=2(p-2)$ then these numbers have the same parity, so we just need to consider $p-2-k=p-2+k=2$ or $p=4,k=0$. In this case, we have one pair satisfies the given condition $(p,q)=(4,4)$.

2. If $p+1=q$ then $x^2-px+p+1=0$, we have $$\Delta =p^2-4p-4=k^2\iff (p-2{)}^2-k^2=8.$$ Hence $(p-2-k)(p-2+k)=8$. Similarly, these numbers have the same parity then $p-2-k=2$, $p-2+k=4$ or $p=5,k=5$. In this case, we have one pair satisfies the given condition $(p,q)=(5,6)$.

3. If $q+1=p$, similarly, we have $(p,q)=(6,5)$.

Therefore, $(p,q)=(4,4),(5,6),(6,5)$. $\square$