Question

ok a nice one

Find all Pairs \((m,n)\) of positive integers such that\[\frac{n^2+1}{mn-1}\]is an integer too.

Answer 1

m=2, n=3 , ratio=2
n=2, m=3, ratio=1

Answer 2

m=1,n=2
m=1,n=3

Answer 3

still missing some of solutions

Answer 4

I got these:
m=1, n=2, ratio=5
m=1, n=3, ratio=5
m=2, n=1, ratio=2
m=2, n=3, ratio=2
m=3, n=1, ratio=1
m=3, n=2, ratio=1

Answer 5

quite right

Answer 6

I used a method where I assumed m = n + p and drew conclusions from the resulting equations

Answer 7

integer solutions could only exist if p=1, 2, -1 or -2

Answer 8

I'd be interested to know if there is another way of doing this

Answer 9

i'd like also to see ur reasoning @asnaseer

Answer 10

sure - let me type it up for you...

Answer 11

let m = n + p where p is some other integer (negative, zero or positive), then we have:\[\frac{n^2+1}{mn-1}=\frac{n^2+1}{(n+p)n-1}=\frac{n^2+1}{n^2+np-1}\]for this expression to yield an integer, we must satisfy at least this:\[n^2+1\ge n^2+np-1\]\[\therefore1\ge np-1\]\[\therefore2\ge np\]\[\therefore n\le\frac{2}{p}\]

Answer 12

if p is negative, then this inequality becomes:\[n\gt\frac{2}{p}\]which then gives us:\[p=2\implies n\le1\implies n=1\]\[p=1\implies n\le2\implies n=1\text{ or }2\]\[p=-1\implies n\gt-2\implies n=1,2,3...,\infty\]\[p=-2\implies n\gt-1\implies n=1,2,3,...,\infty\]

Answer 13

I then took each case and created a table to see which combinations gave a valid integer ratio

Answer 14

e.g. p=2, n=1, m=n+p=1+2=3, ratio=1

Answer 15

hope there isn't a flaw in my reasoning?

Answer 16

i enjoyed seeing ur solution...

Answer 17

thx :)

Answer 18

do you have an alternative method?

Answer 19

yes its a little bit longer than this.

Answer 20

ok - I won't push you to post it, but if you could (even a scan of paper written solution) then I would really appreciate it.

Answer 21

sure...i'll post it later. :)
Nice to see this group alive again.

Answer 22

thx - and yes - finally it awakes! :)

Answer 23

{m,n,ratio}
{{1, 2, 5}, {1, 3, 5}, {2, 1, 2}, {2, 3, 2}, {3, 1, 1}, {3, 2, 1}}

Answer 24

just a neat pint about this problem

if \[mn-1|n^2+1\]so\[mn-1|m^2n^2-1+n^2+1=n^2(1+m^2)\]\[mn-1|1+m^2\]so if \((m,n)\) is answer \((n,m)\) will be answer

Answer 25

*point

Answer 26

n=1 gives m=2,3 so 4 answers from here :
(1,2),(1,3),(2,1),(3,1)

no answer for m=n

suppose m>n\[kn-1=\frac{n^2+1}{mn-1}<\frac{n^2+1}{n^2-1}=1+\frac{2}{n^2-1}<2\]\[kn<3\]\[n=2 , k=1\]gives m=3 so 2 solution from here
(2,3),(3,2)

Answer 27

but why \[\frac{n^2+1}{mn-1}=kn-1\]\[\frac{n^2+1}{mn-1}=r\]its easy to show that\[r\equiv-1 \ \ \text{mod} \ n\]