Mathematical Reflections 2016, Issue 1 - Problem J361

Problem:
Solve in positive integers the equation $$\dfrac{x^2-y}{8x-y^2}=\dfrac{y}{x}.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:

The given equation can be retwritten as $$x^3+y^3+27-9xy=27.$$ Using the well-known identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$ we obtain $$(x+y+3)(x^2+y^2+9-xy-3x-3y)=27.$$ Since $x,y$ are positive integers, then $x+y+3 \geq 5$ and we have two cases.

(i) $x+y+3=9$ and $x^2+y^2+9-xy-3(x+y)=3$. From the first equation we have $x+y=6$, which gives $x^2+y^2=36-2xy$. Substituting these values into the second equation, we have $(36-2xy)+9-xy-18=3$, which gives $xy=8$. So, we obtain the system of equations
$$\begin{array}{rcl} x+y&=&6 \\ xy&=&8, \end{array}$$ which gives $(x,y) \in \{(2,4),(4,2)\}$. An easy check shows that the only solution to the given equation is $(x,y)=(4,2)$.

(ii) $x+y+3=27$ and $x^2+y^2+9-xy-3x-3y=1$.  From the first equation we have $x+y=24$, which gives $x^2+y^2=576-2xy$. Substituting these values into the second equation, we have $(576-2xy)+9-xy-72=3$, which gives $xy=170$. So, we obtain the system of equations $$\begin{array}{rcl} x+y&=&24 \\ xy&=&170, \end{array}$$ which yields no solutions in positive integers.

In conclusion, $(x,y)=(4,2)$.