## mukushla ok a nice one Find all Pairs $$(m,n)$$ of positive integers such that$\frac{n^2+1}{mn-1}$is an integer too. one year ago one year ago

1. eliassaab

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

2. sauravshakya

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

3. mukushla

still missing some of solutions

4. asnaseer

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

5. mukushla

quite right

6. asnaseer

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

7. asnaseer

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

8. asnaseer

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

9. mukushla

i'd like also to see ur reasoning @asnaseer

10. asnaseer

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

11. asnaseer

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}$

12. asnaseer

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$

13. asnaseer

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

14. asnaseer

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

15. asnaseer

hope there isn't a flaw in my reasoning?

16. mukushla

i enjoyed seeing ur solution...

17. asnaseer

thx :)

18. asnaseer

do you have an alternative method?

19. mukushla

yes its a little bit longer than this.

20. asnaseer

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.

21. mukushla

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

22. asnaseer

thx - and yes - finally it awakes! :)

23. eliassaab

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

24. mukushla

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

25. mukushla

*point

26. mukushla

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)

27. mukushla

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$