Equation over Z

Question

Solve the equation $xy+1=3x+y$ over $\mathbb{Z}^2$

Indeed, $$ xy+1=3x+y \Longleftrightarrow (x-1)(y-3)=2 $$

or $ \textrm{Div}(2)=\{k \in \mathbb{Z}/ k|2 \}=\{-1;1;-2;2\}$ Then

$(x-1)/2 \implies x-1 \in \textrm{Div}(2) \implies x\in \{0,2,-1,3 \} $ $(y-3)/2 \implies y-3 \in \textrm{Div}(2) \implies y\in \{2,4,1,5 \} $ $$S=\{(0,2),(2,4),(-1,1),(2,5)\}$$ AM i right ??

Answer 1

$(0,2)$ is not a solution because $(0-1)(2-3)\neq 2$

What you have observed $xy+1=3x+y \Longleftrightarrow (x-1)(y-3)=2$ is appreciable..

For it to hold in $\mathbb{Z}^2$ only possibilities are..

• $(x-1)=2$ and $(y-3)=1$.. What does this say about $x,y$?
• $(x-1)=-2$ and $(y-3)=-1$.. What does this say about $x,y$?
• $(x-1)=1$ and $(y-3)=2$.. What does this say about $x,y$?
• $(x-1)=-1$ and $(y-3)=-2$.. What does this say about $x,y$?

Answer 2

$(x-1)(y-3)=2\implies(x-1,y-3)\in\{(1, 2), ((2,1),(-1,-2),(-2,-1)\}.$