Equivalently, $(x-y).(3-x-y) = (n-1).(x^2 + xy + 3y)\implies (x-y).(3-x-y)=(n-1).x.(x+y)+3y.(n-1)$. now assume , $n\geq 2$, and $(x+y)>3$, which is certainly impossible since, $(n-1).x.(x+y)>(x-y).(3-x-y)\implies (x-y).(3-x-y) < (n-1).(x^2 + xy + 3y)$, so if, $(x+y)\leq 3\implies (x,y)=(1,1),(2,1),(1,2)$, considering all of these pairs indicates $n=1$, secondly, if, $n<2$ i.e if $n=1$ then certainly the equation has at least one solution, for example, $x=y$. so the only value is $\boxed{n=1}$.

sayanjoddar wrote: Equivalently, $(x-y).(3-x-y) = (n-1).(x^2 + xy + 3y)\implies (x-y).(3-x-y)=(n-1).x.(x+y)+3y.(n-1)$. now assume , $n\geq 2$, and $(x+y)>3$, which is certainly impossible since, $(n-1).x.(x+y)>(x-y).(3-x-y)\implies (x-y).(3-x-y) < (n-1).(x^2 + xy + 3y)$, so if, $(x+y)\leq 3\implies (x,y)=(1,1),(2,1),(1,2)$, considering all of these pairs indicates $n=1$, secondly, if, $n<2$ i.e if $n=1$ then certainly the equation has at least one solution, for example, $x=y$. so the only value is $\boxed{n=1}$. Can you please explain why is x+y>3 case impossible? I don't understand why such inequality stands. x can be far less than y so (x-y)×(3-x-y) can be large product of two negative numbers.

In fact, the answer is $\displaystyle n\in \{1,3,4,9\}$: Put $\displaystyle x=y$, and you get $\displaystyle n=1$. Suppose that $\displaystyle n >1$. From the given equation, we have $\displaystyle y^{3} -(( n-1) x+3n) y+3x-nx^{2} =0.$ Let $\displaystyle D=(( n-1) x+3n))^{2} -4\left( 3x-nx^{2}\right) .$ For $\displaystyle ( x,n) \in \mathbb{N}^{2} ,\ D$ is a perfect square $\displaystyle \Leftrightarrow $ there exists $\displaystyle y\in \mathbb{N}$ satisfying the equation. (Note that $\displaystyle ( n-1) x+3n >0$) So it's sufficient to find all $\displaystyle ( x,n) \in \mathbb{N}^{2}$ such that $\displaystyle D$ is a perfect square. We can verify the following inequality holds: $\displaystyle (( n+1) x+3n-6)^{2} < D< (( n+1) x+3n)^{2}$ (Note that we can also write $\displaystyle D=(( n+1) x+3n)^{2} -12( n+1) x$) So we can take an integer $\displaystyle a\in [ 1,5]$ such that $\displaystyle D=(( n+1) x+3n-a)^{2}$. Then we get $ \displaystyle 2( 6-a)( n+1) x=a( 6n-a)$. If $\displaystyle a$ is odd, then LHS is even and RHS is odd. So we want to check the following two cases: Case $\displaystyle a=2$ $\displaystyle ( n+1)( 3-2x) =4\ \Leftrightarrow \ ( x,n) =( 1,3)$ Case $\displaystyle a=4$ $\displaystyle ( n+1)( 6-x) =10\ \Leftrightarrow \ ( x,n) =( 4,4) ,( 5,9)$ Therefore, $\displaystyle \{1,3,4,9\}$ gives all solutions indeed.