China Western Mathematical Olympiad

choose any $(a,b)$ such that $b$ is the smallest. Then the quadratic equation $$x^2+(1-kb)x+b^2+b=0$$ has an integral root $x=a$. Let $x=a^{\prime }$ be the second root, it follows from $a+a^{\prime }=kb-1$ that $a^{\prime }\in \mathbb{Z}$, and from $$a\cdot a^{\prime }=b(b+1)$$ that $a^{\prime }>0$. Hence, we have $$\frac{b+1}{a^{\prime }}+\frac{a^{\prime }+1}{b}=k.$$ And it follows from the assumption on $b$ that $a\ge b,a^{\prime }\ge b.$ So one of $a$ and $a^{\prime }$ is equal to $b$. Without loss of generality, we may assume $a=b$, so $k=2+\frac{2}{b}$, and so $b\mid 2$ i.e., $b=1$ or $2$, and $k=3$ or $4$, respectively. If $a=b=1$, then $k=4$; if $a=b=2$, then $k=3$. Consequently, $k=3$ or $4$ is the only solution. $\square$