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.

Multiply by $4$ and divide the two polynomials on the left. You get $4k=\frac{4n+32}{4n-1}=1+\frac{33}{4n-1}$. Then $4n-1$ must divide $33$. — conditionalMethod, Dec 06 '19 at 18:10

score 3 · Answer 1 · edited Dec 06 '19 at 18:37

3

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

edited Dec 06 '19 at 18:37

Matt Samuel

58,164

answered Dec 06 '19 at 18:10

lab bhattacharjee

274,582

+1. Hence $n=0,1,3$ of -8. – almagest Dec 06 '19 at 18:16
@almagest ...or $0$. – Franklin Pezzuti Dyer Dec 06 '19 at 18:17
Sorry for the downvote, was a fat finger error on my phone screen. – Matt Samuel Dec 06 '19 at 18:38

score 2 · Answer 2 · answered Dec 06 '19 at 18:12

2

HINT

We have that

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

answered Dec 06 '19 at 18:12

user

154,566

score 2 · Answer 3 · answered Dec 06 '19 at 18:47

2

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

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

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

answered Dec 06 '19 at 18:47

Dmitry Ezhov

1,653

Wolfram Alpha agrees with the result here that. $$k\in{-8,0,1,1}\land n\in{0,-8,3,1}$$ – poetasis Jul 30 '20 at 20:10

Find integer solutions to an equation with two variables

3 Answers3