Find integer solutions to an equation with two variables

Question

I was wondering if it is possible to find integer solutions using some sort of effective method (not mindlessly substituting numbers). If it isn't possible please let me know, otherwise I would really appreciate it if you could show me how to solve this equation:

$\frac{8+n}{4n-1}=k$

for integer solutions.

Thanks.

Answer 1

So, $4n-1$ must divide $8+n$

As $4n-1$ is odd, iff $4n-1$ must divide $4(8+n)=4n-1+33$

So, $4n-1$ must divide $33$

So, $4n-1\in[\pm1,\pm3,\pm11\pm33]$.

Answer 2

HINT

We have that

$$\frac{8+n}{4n-1}=k \iff \frac14 \frac{4n-1+33}{4n-1}=k \iff \frac{33}{4n-1}=4k-1$$

Answer 3

$\dfrac{8+n}{4n-1}=k \overset{k\to x+y\\n\to x-y}{\implies} (4 x - 1)^2 - (4 y)^2 = 33$

Then $(x,y)=(-4,-4),(-4, 4),(2, -1),(2, 1)$,

and thus $(k,n)=(-8, 0), (0, -8), (1, 3), (3, 1)$.