<s>[INST] Positive integer solutions to $xy=4x+7y$ I think I have found all the positive integer solutions to $xy=4x+7y$. I did first was to make $y$ the subject of the expression, and then I
$$xy-7y=4x$$
$$y=\frac{4x}{x-7}$$
$$y=\frac{4}{x-7} * x$$
I started finding the divisors of $4$: $1,2,4$, and setting $x-7$ to be equal to these divisors. I then got the solution pairs: $(8, 32), (9, 18), (11, 11)$. 
I also realised that if $x-7$ and $x$ were divisible by $7$, then $7$ could be 'cancelled out' from the expression. I substituted $x=14, 21, 35$, and got $3$ more solution pairs: $(14, 8)$, $(21, 6)$, $(35, 5)$. After this, I concluded that there might not be any more solutions, because $$\lim_{x\to \infty} (4*\frac{x}{x-7}) = 4*1=4,$$
and that means that when $x$ is another multiple of $7$, $y$ will never equal $4$.
However, how can I prove that there are no other solutions rigorously, without checking each number case by case? Is there a way to find the total number of integer solutions without knowing what they are?
 [/INST] Let $\gcd(x,y) = k$ and let $x = ak; y = bk$ and therefore $\gcd(a,b) = 1$.
$abk^2 = 4ak + 7bk$
$abk = 4a + 7b$  
So $a|7b$ so $a|7$ so $a = 7$ or $a=1$.
$b|4a$ so $b|4$ so $b=1,2$ or $4$.
$k = \frac 4b + \frac 7a$.
So there are $6$ solutions.
$a=1; b=1; k=11$ and $x=y = 11$.  $121 = 44 + 77$
$a=1; b=2; k=9$ and $x=9;y=18$.  $9*18 = 4*9 + 7*18$.
$a=1; b=4; k=8$ and $x=8;y=36$.  $8*36 = 4*8 + 7*36$
$a=7; b=1; k=5$ and $x=35; y=5$.  $5*35= 4*35+7*5$
$a=7; b=2; k=3$ and $x=21; y=6$.  $6*21=4*21 + 7*6$
$a=7; b=4; k= 2$ and $x=14; y=8$.  $8*14 = 4*24 + 7*8$.
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