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

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eliassaab
 3 years ago
Best ResponseYou've already chosen the best response.0m=2, n=3 , ratio=2 n=2, m=3, ratio=1

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1still missing some of solutions

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2I 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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2integer solutions could only exist if p=1, 2, 1 or 2

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2I'd be interested to know if there is another way of doing this

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1i'd like also to see ur reasoning @asnaseer

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2sure  let me type it up for you...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2let 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:\[n^2+1\ge n^2+np1\]\[\therefore1\ge np1\]\[\therefore2\ge np\]\[\therefore n\le\frac{2}{p}\]

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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2e.g. p=2, n=1, m=n+p=1+2=3, ratio=1

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2hope there isn't a flaw in my reasoning?

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1i enjoyed seeing ur solution...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2do you have an alternative method?

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1yes its a little bit longer than this.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2ok  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.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1sure...i'll post it later. :) Nice to see this group alive again.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2thx  and yes  finally it awakes! :)

eliassaab
 3 years ago
Best 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}}

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1just 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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1n=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)

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