1
$\begingroup$

the original problem was to find all integer solutions for $xy+1=3(x+y)$.

$$xy+1=3(x+y)$$ $$xy+1=3x+3y$$ $$xy-3x=3y-1$$ $$x(y-3)=3y-1$$ $$x=\frac{3y-1}{y-3}$$

I was able to find all possible solutions using this method,

I was wondering if you can say that $3|xy+1$ or $x+y|xy+1$ and proceed to solve it in this fashion.

If it is possible please show me how.

2 Answers 2

2

We want to find a factorization if possible, so consider $xy-3x-3y+k = k-1$ as a re-writing of the condition. Then we can take any value for $k$ to make a factorization possible, and $(x-3)(y-3) = xy-3k-3y+9$, so take $k=9$ to get $$(x-3)(y-3) = 8$$

Then the integer ordered factor pairs of $8$ are $$\{(-1,-8), (-2,-4), (-4,-2), (-8,-1),(1,8), (2,4), (4,2), (8,1)\}$$ and the possible solutions $$(x,y)\in\{(2,-5), (1,-1), (-1,1), (-5,2),(4,11), (5,7), (7,5), (11,4)\}$$

Looking at $(x+y)\mid(xy+1)$, it seems like the best result from that probably takes us back on the same track, ie \begin{align} j(x+y)&=xy+1 \\ xy-jx-jy+1 &= 0 \\ xy-jx-jy + j^2 &= j^2-1 \\ (x-j)(y-j) & = j^2-1 \end{align}

... except here of course we already know what $j$ is.

In a different problem,
$\begin{align}(x+y) &\mid(xy+1) \\ \Rightarrow(x+y) &\mid(xy+1)-(x+y) \\ \Rightarrow(x+y) &\mid(xy-x-y+1) \\ \Rightarrow (x+y) &\mid(x-1)(y-1) \end{align}$
might be useful.

  • 0
    could you please explain ohw you made the step $(x+y)\mid(xy+1) \Rightarrow(x+y)\mid(xy-x-y+1)$ – 2017-01-29
  • 0
    @BasemFouda added a step to the answer, hope that improves it – 2017-01-29
4

$$x(3-y)-3(3-y)=1-9\iff8=(x-3)(y-3)$$

  • 0
    this is the first method but how do you use the second if possible – 2017-01-29
  • 1
    @BasemFouda: I do not understand your request. By this algebraic manipulation it turns out that the original problem is equivalent to factoring $8$. What else do you need, or why are you trying to make simple things difficult? – 2017-01-29
  • 0
    I was just wondering if you can apply the second method to the problem. I am not trying to solve it anymore, just exploring possible solutions. – 2017-01-29