Spring Mathematical Tournament

Let $F(x,y)=x^3-x^{2}y+y^2+x-y$. Note that $F(1,1)=1$ and $F(1,2)=2$. We shall prove that the equation $F(x,y)=3$ has no solution in positive integers. Write this equation in the form $$y^2-(1+x^2)y+x^3+x-3=0.$$ Its discriminant with respect to $y$ equals $$D=(1+x^2{)}^2-4(x^3+x-3)=x^4-4x^3+2x^2-4x+13.$$ Since $D<(x^2-2x-1{)}^2$ for any integer $x\ge 2$ and $D>(x^2-2x-2{)}^2$ for any integer $x\ge 6$, the equation $F(x,y)=3$ has no positive integer solutions if $x\ge 6$. Direct verifications show the same if $x\in \{1,2,3,4,5\}$. Hence the wanted integer is 3.