Solve in integers the equation $$(2x+y)(2y+x)=9\min(x,y).$$
Proposed by Titu Andreescu.
Solution:
Assume without loss of generality that $x \leq y$. Then, we have to find the integer solutions of the equation
$$2y^2+5xy+(2x^2-9x)=0.$$ To this aim, the discriminant of this equation in $y$ must be a perfect square, so there exists $t \in \mathbb{Z}$ such that
$$\Delta_y=9x(x+8)=t^2 \implies x(x+8)-\dfrac{t^2}{9}=0,$$ i.e. $$\left(x+4-\dfrac{t}{3}\right)\left(x+4+\dfrac{t}{3}\right)=16.$$
If both the two factors are equal to $\pm 4$, it must be $t=0$, so $x=0$ or $x=-8$, which give $y=0$ and $y=10$ respectively. If the two factors are distinct, since they have the same parity, it must be $$\begin{array}{lll}x+4-t/3&=&\pm 2 \\ x+4+t/3&=&\pm 8, \end{array} \qquad \begin{array}{lll}x+4-t/3&=&\pm 8 \\ x+4+t/3&=&\pm 2. \end{array}$$
Since $x(x+8) \geq 0$, an easy check shows that $x=1$, and for this value we get $y=1$. Therefore, all the integer solutions are $$(0,0), (1,1), (-8,10), (10,-8).$$
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