Hellenic Mathematical Olympiad

Let $\frac{(a+b{)}^2+4a}{ab}=\kappa \in \mathbb{Z}$. The last one can be re-written as: $$(a+b{)}^2+4a=\kappa ab(1)$$ We will prove that $a$ is odd. Indeed, if $2|a$, then $2|(a+b{)}^2$, and thus $b$ is even, a contradiction.

We consider (1) as a quadratic with respect to $b$: $$b^2+b(2-\kappa )a+a^2+4a=0$$ The discriminant should be a perfect square. We have $$\Delta =a^2(\kappa -2{)}^2-4(a^2+4a)=a(a(\kappa -2{)}^2-4a-16).$$ It follows that the product of the numbers $a$, $a(\kappa -2{)}^2-4a-16$ is a perfect square, but since $a$ is odd, these two numbers are coprime and so each one of them is a perfect square, so $a$ is a perfect square.

Let $d=(a,b)$ and we write $a=dx$, $b=dy$, with $(x,y)=1$. Observe that since $d|b$ and $b$ is odd, $d$ is odd. The number $$\frac{(a+b{)}^2+4a}{ab}=\frac{d^2(x+y{)}^2+4dx}{d^{2}xy}=\frac{d(x+y{)}^2+4x}{dxy}$$ is an integer. So $x|d(x+y{)}^2+4x\Rightarrow x|d{y}^2$ and since $(x,y)=1$, we have $x|d$. Moreover, $d|d(x+y{)}^2+4x\Rightarrow d|4x$ and since $d$ is odd, $d|x$. From these two, we have that $d=x$, so $a=d^2$.