How to solve $xy=5x+5y$ for integer solutions [closed]

Question

I did a bit of research and as I understand it would probably involve Diophantine equations. Unfortunately I have no idea where to start. Any help would be greatly appreciated. Thanks ^^

Answer 1

$(x-5)(y-5)=xy-5x-5y+25=25$

Now we have proved $(x-5)(y-5)=25$. Because both $x$ and $y$ are integers, both $x-5$ and $y-5$ are integers. Factorizing 25 gives the following possibilities:

$25=1 \times 25$

$25=5 \times 5$

$25=25 \times1$

$25=(-1) \times(-25)$

$25=(-25) \times(-1)$

$25=(-5) \times (-5)$

All possible values of $x-5$:

$1,5,25,-1,-25,-5$

The corresponding possible values of $y-5$:

$25,5,1,-25,-1,-5$

Now all integer solutions appear:

$x=6,y=30$

$x=10,y=10$

$x=30,y=6$

$x=4,y=-20$

$x=-20,y=4$

$x=0,y=0$

Answer 2

Hint $\ $ This type of diophantine equation is solvable by a generalization of completing the square. Namely, completing a square generalizes to completing a product, using the AC-method, viz.

$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\large \times\,a}\iff\ &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

Now we need only test when $\,ad+bc\,$ can be split into two factors of the above form. A nonzero integer has only finitely many splittings into two factors, so only finitely many cases need testing.

Answer 3

You can, if you like, begin in straightforward algebraic fashion, namely by first solving for $y$ in terms of $x$:

$$xy=5x+5y\implies xy-5y=5x\implies y={5x\over x-5}={5(x-5)+25\over x-5}=5+{25\over x-5}$$

It's at this point that the number theory comes in: In order for $y$ to be an integer, we need $x-5$ to divide $25$, which is to say $x-5\in\{\pm1,\pm5,\pm25\}$, for a total of six pairs $(x,y)$ that solve the equation.

The real key step here consists of not stopping at $y={5x\over x-5}$, but realizing that $5+{25\over x-5}$ is a more useful expression for the problem at hand.