Solve in integers the equation
(x+1)^2+(y+1)^2+xy(x+y+3)=2.
Proposed by Alessandro Ventullo, Milan, Italy
Solution:
Observe that \begin{array}{lll}(x+1)^2+(y+1)^2+xy(x+y+3)-1&=&x^2+y^2+2xy+2x+2y+1+xy(x+y+1)\\&=&(x+y+1)^2+xy(x+y+1)\\&=&(x+y+1)(x+1)(y+1), \end{array} so the equation can be rewritten as (x+y+1)(x+1)(y+1)=1. Therefore, at least one of the three factors is equal to 1. We have the three systems of equations
\begin{array}{rll} x+y+1&=&1 \\ (x+1)(y+1)&=&1, \end{array} \qquad \begin{array}{rll} x+1&=&1 \\ (x+y+1)(y+1)&=&1, \end{array} \qquad \begin{array}{rll} y+1&=&1 \\ (x+y+1)(x+1)&=&1. \end{array}
Solving each system we get that (x,y) \in \{(0,0),(-2,0),(0,-2)\}.
Observe that \begin{array}{lll}(x+1)^2+(y+1)^2+xy(x+y+3)-1&=&x^2+y^2+2xy+2x+2y+1+xy(x+y+1)\\&=&(x+y+1)^2+xy(x+y+1)\\&=&(x+y+1)(x+1)(y+1), \end{array} so the equation can be rewritten as (x+y+1)(x+1)(y+1)=1. Therefore, at least one of the three factors is equal to 1. We have the three systems of equations
\begin{array}{rll} x+y+1&=&1 \\ (x+1)(y+1)&=&1, \end{array} \qquad \begin{array}{rll} x+1&=&1 \\ (x+y+1)(y+1)&=&1, \end{array} \qquad \begin{array}{rll} y+1&=&1 \\ (x+y+1)(x+1)&=&1. \end{array}
Solving each system we get that (x,y) \in \{(0,0),(-2,0),(0,-2)\}.
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