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?
 A: If $xy = 4x+7y$, then $xy-4x-7y=0$. Complete the factorization:
$$
  (x-7)(y-4) = xy-4x-7y+(\text{something}) = (\text{the **same** something})
$$
So $x-7$ and $y-4$ must be a factorization of the (something). Note: Just because $x$ and $y$ are positive integers doesn't mean that $x-7$ and $y-4$ are necessarily positive.
A: Your approach can also be modified to work:
Lemma: if $a\mid bc$ and $a$ and $b$ are relatively prime, then $a\mid c$.
Proof: Exercise. (If you assume the fundamental theorem of arithmetic i.e. unique factorization, then this is easy.  If not, then this is harder, and is one of the steps in one standard proof of the fundamental theorem of arithmetic.)
If we have the lemma, we can now split into 2 cases:
Case 1: $7\mid x$
Let $x=7a$. Then we can cancel 7 as you said to get
$$y=4*\frac{a}{a-1}$$
which has to be an integer, so $a-1\mid 4$.
Case 2: $7\not\mid x$
Then $x-7\mid4$, as you showed.
A: This is how I answered (using @Zach Teitler's hint):
Starting from $xy-4x-7y+28=28$, and so we have $(x-7)(y-4) = 28$. For the $(x-7)$ bracket, the factors of $28$ which are greater than $7$ are $14$ and $28$, which give $(21,6)$ and $(35, 5)$.
For the $(y-4)$ bracket, the factors which are greater than $4$ are $7,14,28$, which give the solutions $(11,11), (9,18), (8, 32)$.
A: 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 $x=14; y=8$.  $8*14 = 4*24 + 7*8$.
A: Hint:
$$4x=4(x-7)+28$$
So, $x-7$ must divide $28$
