How to solve the diophantine equation $2(x+y)=xy+9$?

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.

• I would done in this way(+1) – lab bhattacharjee Apr 9 '17 at 13:11
• @labbhattacharjee thank you – Teresa Lisbon Apr 9 '17 at 13:15
• There are three other solutions... – Pedro A Apr 9 '17 at 14:35
• @Hamsteriffic Answer edited. – 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: $\{(1,7),(7,1), (3,-3),(-3,3)\}$.

• 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. – AmateurMathPirate Aug 24 '17 at 13:05
• @AmateurMathGuy It happens often enough in our forum. Thank you for your interest! – Michael Rozenberg Aug 24 '17 at 13:10
