An example of how to solve equation over $\mathbb{Z}$

Question

I found an example of how to solve equation over $\mathbb{Z}$

Example

Solve the equation over $\mathbb{Z}$ : $$xy + 1 = 3x + y. $$ $$ xy = 1 + 3x + y \Longleftrightarrow (x-1) (y-3) = 2 $$ and $\textrm{Div} (2) = \{-2, -1,1,2\}$.

We can then describe the set of pairs $(x, y)$ solutions $$\textrm{S} = (1,3) + \{(-2, -1), (- 1, -2), (1,2), (2,1)\}$$

My thoughts:

From the equation $(1-x) (y-3) = 2$ I can see that $(x-1)\mid 2$ and $(y-3)\mid 2$ then $(x-1)\in \textrm{Div}(2) $ and $(y-3)\in \textrm{Div}(2) $ or $\textrm{Div} (2) = \{-2, -1,1,2\}$ thus

• $x-1=-2 \text{ and } y-3=-2 \Longrightarrow (x=-1,y=1)$
• $x-1=-1 \text{ and } y-3=-1 \Longrightarrow (x=-2,y=2)$
• $x-1=1 \text{ and } y-3=1 \Longrightarrow (x=2,y=4)$
• $x-1=2 \text{ and } y-3=2 \Longrightarrow (x=3,y=5)$

Thus the set of pairs $(x, y)$ solutions $\textrm{S}=\{ (x=-1,y=1),(x=-2,y=2),(x=2,y=4),(x=3,y=5)\}$

Am I right if not I really need help to understand the example

Thanks a lot.

Answer 1

Both $x-1$ and $y-3$ divide 2 but they also need to multiply to give $2$. So you should only consider $$ 2=2\times 1=1\times 2=(-2)\times(-1)=(-1)\times(-2). $$ These correspond to the 4 solutions given in the official solution.

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Edit to answer OP's question:

The $(1,3)$ is just a short-hand notation. From what I have written above (and similar comments by others), you can write $$ (x-1,y-3)\in\{(-2,-1),(-1,-2),(1,2),(2,1)\}. $$ This is equivalent to $$ (x,y)\in(1,3)+\{(-2,-1),(-1,-2),(1,2),(2,1)\}\equiv S. $$ Recall that for any set $A$, we define $a+A=\{a+x:x\in A\}$ (supposing that such addition makes sense).

Answer 2

There is a small mistake. The thing is that, if $x-1$ is a divisor of $2$, then $y-3$ must equal $\frac2{x-1}$. Only in this case, their product will evaluate to $2$.
What you did is assign the same divisor to $x-1$ and $y-3$, but this is not (necessarily) correct: if for example $x-1=1$ and $y-3=1$, their product can't be $2$.
The right approach would be to consider the following cases: $$\begin{array}{r|c|c}& x-\color{red}1&y-\color{blue}3\\\text{case 1:}&1&2\\\text{case 2:}&2&1\\\text{case 3:}&-1&-2\\\text{case 4:}&-2&-1\end{array}$$

The complete set of solutions can thus be described as $$(\color{red}1,\color{blue}3)+\{(1,2),(2,1),(-1,-2),(-2,-1)\},$$ which is nothing more than a shorthand for $$\{(\color{red}1+1,\color{blue}3+2),\;(\color{red}1+2,\color{blue}3+1),\;(\color{red}1+(-1),\color{blue}3+(-2)),\;(\color{red}1+(-2),\color{blue}3+(-1))\}.$$

But apart from this, your reasoning seems correct.

Answer 3

$(x-1)(y-3)=2$ combined with the fact that you are dealing with integers implies that:

$x-1=1$ and $y-3=2$, i.e. $(x,y)=(1,3)+(1,2)$

or:

$x-1=-1$ and $y-3=-2$, i.e. $(x,y)=(1,3)+(-1,-2)$

or:

$x-1=2$ and $y-3=1$, i.e. $(x,y)=(1,3)+(2,1)$

or:

$x-1=-2$ and $y-3=-1$, i.e. $(x,y)=(1,3)+(-1,-2)$

Answer 4

Another way of looking at this is to rearrange to give $$xy-3x=y-1=x(y-3)$$ so that $$x=\frac {y-1}{y-3}=1+\frac 2{y-3}$$

Then $y-3=-2, -1, 1, 2$ so that $y=1, 2, 4, 5$ and for each of these you can compute the value of $x$ as $0, -1, 3, 2$ respectively