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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 ??

vonbrand
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Educ
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2 Answers2

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$(0,2)$ 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 $x,y$?
  • what about others solutions am i right – Educ Jun 20 '14 at 10:45
  • @Educ :

    see the edit...

     –  Jun 20 '14 at 11:02
  • how could you choice exactly this possibilities and if we took 1st possibilty gives $x= 2$ and $y = 5 \implies 2.5 \neq 2$ – Educ Jun 20 '14 at 11:05
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    @Educ: like you said... my $(x-1)$ has to be a factor of $2$ and at the same time $y-3$ has to be a factor of $2$ and at the same time their product has to be $2$.... I can not take $x-1=2$ and $y-3=2$ as this would give me $(x-1)(y-3)=4$ which is not what i wanted.. –  Jun 20 '14 at 11:07
  • as you said the all solutions are : ${ (3,4);(-1,2);(2,5),(0,1)}$ – Educ Jun 20 '14 at 11:13
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    @Educ : yes.. those are all solutions.. Very well done :) –  Jun 20 '14 at 11:14
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$(x-1)(y-3)=2\implies(x-1,y-3)\in\{(1, 2), ((2,1),(-1,-2),(-2,-1)\}.$