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mukushla Group Title

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

  • 2 years ago
  • 2 years ago

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  1. eliassaab Group Title
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    m=2, n=3 , ratio=2 n=2, m=3, ratio=1

    • 2 years ago
  2. sauravshakya Group Title
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    m=1,n=2 m=1,n=3

    • 2 years ago
  3. mukushla Group Title
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    still missing some of solutions

    • 2 years ago
  4. asnaseer Group Title
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    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

    • 2 years ago
  5. mukushla Group Title
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    quite right

    • 2 years ago
  6. asnaseer Group Title
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    I used a method where I assumed m = n + p and drew conclusions from the resulting equations

    • 2 years ago
  7. asnaseer Group Title
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    integer solutions could only exist if p=1, 2, -1 or -2

    • 2 years ago
  8. asnaseer Group Title
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    I'd be interested to know if there is another way of doing this

    • 2 years ago
  9. mukushla Group Title
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    i'd like also to see ur reasoning @asnaseer

    • 2 years ago
  10. asnaseer Group Title
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    sure - let me type it up for you...

    • 2 years ago
  11. asnaseer Group Title
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    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:\[n^2+1\ge n^2+np-1\]\[\therefore1\ge np-1\]\[\therefore2\ge np\]\[\therefore n\le\frac{2}{p}\]

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

    • 2 years ago
  13. asnaseer Group Title
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    I then took each case and created a table to see which combinations gave a valid integer ratio

    • 2 years ago
  14. asnaseer Group Title
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    e.g. p=2, n=1, m=n+p=1+2=3, ratio=1

    • 2 years ago
  15. asnaseer Group Title
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    hope there isn't a flaw in my reasoning?

    • 2 years ago
  16. mukushla Group Title
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    i enjoyed seeing ur solution...

    • 2 years ago
  17. asnaseer Group Title
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    thx :)

    • 2 years ago
  18. asnaseer Group Title
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    do you have an alternative method?

    • 2 years ago
  19. mukushla Group Title
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    yes its a little bit longer than this.

    • 2 years ago
  20. asnaseer Group Title
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    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.

    • 2 years ago
  21. mukushla Group Title
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    sure...i'll post it later. :) Nice to see this group alive again.

    • 2 years ago
  22. asnaseer Group Title
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    thx - and yes - finally it awakes! :)

    • 2 years ago
  23. eliassaab Group Title
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    {m,n,ratio} {{1, 2, 5}, {1, 3, 5}, {2, 1, 2}, {2, 3, 2}, {3, 1, 1}, {3, 2, 1}}

    • 2 years ago
  24. mukushla Group Title
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    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

    • 2 years ago
  25. mukushla Group Title
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    *point

    • 2 years ago
  26. mukushla Group Title
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    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)

    • 2 years ago
  27. mukushla Group Title
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    but why \[\frac{n^2+1}{mn-1}=kn-1\]\[\frac{n^2+1}{mn-1}=r\]its easy to show that\[r\equiv-1 \ \ \text{mod} \ n\]

    • 2 years ago
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