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ok a nice one
Find all Pairs \((m,n)\) of positive integers such that\[\frac{n^2+1}{mn1}\]is an integer too.
 one year ago
 one year ago
ok a nice one Find all Pairs \((m,n)\) of positive integers such that\[\frac{n^2+1}{mn1}\]is an integer too.
 one year ago
 one year ago

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eliassaabBest ResponseYou've already chosen the best response.0
m=2, n=3 , ratio=2 n=2, m=3, ratio=1
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
still missing some of solutions
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
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
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
I used a method where I assumed m = n + p and drew conclusions from the resulting equations
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
integer solutions could only exist if p=1, 2, 1 or 2
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
I'd be interested to know if there is another way of doing this
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
i'd like also to see ur reasoning @asnaseer
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
sure  let me type it up for you...
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
let m = n + p where p is some other integer (negative, zero or positive), then we have:\[\frac{n^2+1}{mn1}=\frac{n^2+1}{(n+p)n1}=\frac{n^2+1}{n^2+np1}\]for this expression to yield an integer, we must satisfy at least this:\[n^2+1\ge n^2+np1\]\[\therefore1\ge np1\]\[\therefore2\ge np\]\[\therefore n\le\frac{2}{p}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
if p is negative, then this inequality becomes:\[n\gt\frac{2}{p}\]which then gives us:\[p=2\implies n\le1\implies n=1\]\[p=1\implies n\le2\implies n=1\text{ or }2\]\[p=1\implies n\gt2\implies n=1,2,3...,\infty\]\[p=2\implies n\gt1\implies n=1,2,3,...,\infty\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
I then took each case and created a table to see which combinations gave a valid integer ratio
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
e.g. p=2, n=1, m=n+p=1+2=3, ratio=1
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
hope there isn't a flaw in my reasoning?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
i enjoyed seeing ur solution...
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
do you have an alternative method?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
yes its a little bit longer than this.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
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.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
sure...i'll post it later. :) Nice to see this group alive again.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
thx  and yes  finally it awakes! :)
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
{m,n,ratio} {{1, 2, 5}, {1, 3, 5}, {2, 1, 2}, {2, 3, 2}, {3, 1, 1}, {3, 2, 1}}
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
just a neat pint about this problem if \[mn1n^2+1\]so\[mn1m^2n^21+n^2+1=n^2(1+m^2)\]\[mn11+m^2\]so if \((m,n)\) is answer \((n,m)\) will be answer
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
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\[kn1=\frac{n^2+1}{mn1}<\frac{n^2+1}{n^21}=1+\frac{2}{n^21}<2\]\[kn<3\]\[n=2 , k=1\]gives m=3 so 2 solution from here (2,3),(3,2)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
but why \[\frac{n^2+1}{mn1}=kn1\]\[\frac{n^2+1}{mn1}=r\]its easy to show that\[r\equiv1 \ \ \text{mod} \ n\]
 one year ago
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