Question

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

Answer 1

All i can say is that m and n can't be equal to 1 simultaneously..

Answer 2

thats obvious.......

Answer 3

maybe like this:
\[n ^{2}+1=k(mn-1)\]
\[n^{2}-kmn+(1+k)=0\]
solving this you will get n like a function of m and constant k, so the pairs would be:
(m,f(m,k))
it won't look beatifull, but i think is ok...:)

Answer 4

nice try but i meant to find exact values for (m,n)
for example (m,n)=(2,1) is an answer

Answer 5

ok, thinking....:)

Answer 6

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

Answer 7

@eliassaab
thank u sir
exactly

Answer 8

@eliassaab how did you solve this?

Answer 9

ya, i also whant to know

Answer 10

I am interested in learning how to solve these types of problems - so any guidance from anyone would be greatly appreciated.

Answer 11

here is my solution


We start by checking n = 1, 2 . n^2 + 1 = 2, 5.
This gives the solutions (2, 1), (3, 1), (1, 2), (3, 2).

-------- So we assume hereafter that n > 2.

Let n^2 + 1 = (mn - 1)h. Then we must have h ≡ -1 (mod n). Put h = kn - 1.
Then n^2 + 1 = mkn^2 - (m + k)n + 1. Hence n = mkn - (m + k).
Hence n divides m + k. If m + k >2n,
then since n > 2 we have at least one of m, k >= n +1.
But then (mn - 1)(kn - 1) >= (n^2 + n - 1)(n - 1) > n^2 + 1.
So we must have m + k <=2n ---> m+k= n or 2n
Consider first m + k = n. Then mk=2 and we get (m,n)=(1,3), (2,3)
Finally, take m + k = 2n. Then mk=3. So There is no more solution from this.

Answer 12

thanks @mukushla - I /think/ I just about follow that reasoning :)

Answer 13

Other way:
I see it this way. Let the top be n and bottom m. Obviously n>m. Now little observation:
if m is divisor of n, n/m=k, notice that k is also divisor of n, since mk=n. So the divisores can be organized in so colled pairs (m,n/m), where m<n/m. So just chek all m<n and put it in corresponding pairs

Answer 14

@myko thank u
i'll think about it