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Help needed with a proof

  1. Oct 1, 2006 #1
    hi, i have to prove that if n is a perfect square and n>11, then n2 - 19n + 89 is not a perfect square. i have came up with the following:

    n>11
    100 - 89 < 20n - 19n
    -20n + 100 < -19n + 89
    n2 - 20n + 100 < n2 - 19n + 89
    (n-10)2 < n2 - 19n + 89...........................(1)

    n>11
    92 - 81 < 19n - 18n
    -19n + 92 < -18n + 81
    n2 - 19n + 92 < n2 - 18n + 81
    n2 - 19n + 89 + 3 < (n-9)2
    n2 - 19n + 89 < (n-9)2............................(2)

    combining (1) and (2), we get,
    (n-10)2 < n2 - 19n + 89 < (n-9)2

    since n2 - 19n + 89 is between two consecutive perfect squares, it cannot be a perfect square itself. (QED)

    but my proof doesn't require n to be a perfect square (as stated in the problem). is the question wrong? or am i making some mistake in my proof?
     
    Last edited: Oct 1, 2006
  2. jcsd
  3. Oct 2, 2006 #2
    your proof seems correct to me
     
  4. Oct 2, 2006 #3

    HallsofIvy

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    Indeed, you don't need the n> 11 either. n2- 19n+ 89 is not a perfect square number for any n because n2- 19n+ 89 is not a perfect square polynomial.
     
  5. Oct 2, 2006 #4

    matt grime

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    I'm not sure I get that, Halls.

    x^2+9 is not a perfect square polynomial, yet when x=4, x^2+9=25. There are of course an infinite number of examples, I just wanted one where we evaluate x at a perfect square.
     
  6. Oct 2, 2006 #5
    I don't understand Halls either and I don't understand how the op got from
    n^2 - 19n + 89 + 3 < (n-9)^2
    to
    n^2 - 19n + 89 < (n-9)^2 ?
     
  7. Oct 2, 2006 #6

    matt grime

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    If a is less then b, is a-3 less than, more than, or equal to b?
     
  8. Oct 2, 2006 #7
    less than b

    by the way, please could you let me have a look at the maths questions you used to have ?
     
  9. Oct 2, 2006 #8

    shmoe

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    This polynomial is also a perfect square at n=11.

    The "n a perfect square" isn't necessary with the n>11 condition. That doesn't make the question wrong, just uneccesarily weaker than it could have been. You could actually remove either condition (n a square or n>11) and it would still be correct.
     
  10. Oct 2, 2006 #9

    matt grime

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    Now do you see how the above conlcusion was reached?
     
  11. Oct 2, 2006 #10
    but what did Halls mean by :n2- 19n+ 89 is not a perfect square number for any n because n2- 19n+ 89 is not a perfect square polynomial
     
  12. Oct 2, 2006 #11
    yes but how would it be proven explicitly without using proof by contradiction ?
     
  13. Oct 2, 2006 #12

    shmoe

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    You agree

    n^2 - 19n + 89 < n^2 - 19n + 89 + 3

    right? So

    n^2 - 19n + 89 + 3 < (n-9)^2

    implies

    n^2 - 19n + 89 < (n-9)^2

    "<" is transitive, a<b and b<c implies a<c
     
  14. Oct 2, 2006 #13
    i think if we remove the condition "n>11", then it wouldn't be correct.
    because, then n2 - 19n + 89 can be perfect square for n = 11.
    but we can remove the condition "n is a perfect square".
     
    Last edited: Oct 2, 2006
  15. Oct 2, 2006 #14

    HallsofIvy

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    I don't understand it myself! Another case of shooting from the hip.

    I'm tempted to go back and delete that post and pretend I never said any such thing!
     
  16. Oct 2, 2006 #15

    shmoe

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    I said as much above, but notice n=11 is not a perfect square. I said "either" condition, not "both". Maybe I should have specified that explicitly.
     
  17. Oct 2, 2006 #16
    sorry, my mistake. you are absolutely right.
     
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