The following diophantine equation came up in the past paper of a Mathematics competition that I am doing soon: $$ 2(x+y)=xy+9.$$

Although I know that the solution is $(1,7)$, I am unsure as of how to reach this result. Clearly, the product $xy$ must be odd since $2(x+y)$ must be even, however beyond that, I am unable to see anything else that I can do to solve the problem. I have also tried using the AM-GM inequality, however, it did not simplify the problem much:$$(x+y)+(x-xy+y)\le(\frac{(x+y)+(x+y-xy)}{2})^2.$$ Any help would be greatly be appreciated.


$2(x+y) = xy +9 \implies 2x - xy = 9 - 2y \implies x = \frac{9-2y}{2-y} = \frac{2y-9}{y-2} = 2 - \frac{5}{y-2}$.

If $x$ is an integer, then $\frac{5}{y-2}$ is also an integer. This will tell you what $y$ can be, then what $x$ can be. Trying these out will give you the solution $y=7, x=1$ and the solution $y=3,x=-3$, which then will give you four solutions, since you can switch $x,y$ (it doesn't change the equation) and get more solutions.

  • $\begingroup$ I would done in this way(+1) $\endgroup$ – lab bhattacharjee Apr 9 '17 at 13:11
  • $\begingroup$ @labbhattacharjee thank you $\endgroup$ – Teresa Lisbon Apr 9 '17 at 13:15
  • $\begingroup$ There are three other solutions... $\endgroup$ – Pedro A Apr 9 '17 at 14:35
  • 1
    $\begingroup$ @Hamsteriffic Answer edited. $\endgroup$ – Teresa Lisbon Apr 9 '17 at 17:38

We need to solve $$xy-2(x+y)+4=-5$$ or $$(x-2)(y-2)=-5$$ and it remains to solve the following systems.

$x-2=1$ and $y-2=-5$;

$x-2=-1$ and $y-2=5$;

$x-2=5$ and $y-2=-1$ and

$x-2=-5$ and $y-2=1$,

which gives the answer: $\{(1,7),(7,1), (3,-3),(-3,3)\}$.

  • $\begingroup$ Glad to see a more proper answer than the one favorited by the asker. Not to say that one is bad or anything, but yours most clearly and completely answers the question. $\endgroup$ – AmateurMathPirate Aug 24 '17 at 13:05
  • $\begingroup$ @AmateurMathGuy It happens often enough in our forum. Thank you for your interest! $\endgroup$ – Michael Rozenberg Aug 24 '17 at 13:10

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & 2\pars{x + y} = xy + 9 \implies y = {2x - 9 \over x - 2} = 2 + {5 \over 2 - x} \implies \left\{\begin{array}{ll} {\large\bullet} & \pars{2 - x} \mid 5 \\ {\large\bullet} & \pars{x \leq 2}\quad\mbox{or}\quad\pars{x \geq 5} \\ {\large\bullet} & x\ odd \end{array}\right. \end{align} If $\ds{x \geq 0}$ the only possibility is $\ds{\pars{x,y} = \pars{\color{red}{\large 1},\color{red}{\large 7}}}$ or $\ds{\pars{x,y} = \pars{\color{red}{\large 7},\color{red}{\large 1}}}$ because $\ds{\verts{2 - x} \leq 5}$.


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