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Problem from GCSE Paired Pilot

Discussion in 'Mathematics' started by GoldMaths, Jun 22, 2011.

  1. GoldMaths

    GoldMaths New commenter

    Two square numbers have a sum of 50005.

    I have found two solutions:

    98^2 + 201^2 = 50005
    102^2 + 199^2 = 50005

    Question is are there any more?
    Is there a logical way to approach this? I have used excel to solve so far....
     
  2. GoldMaths

    GoldMaths New commenter

    54^2 + 217^2 = 50005
    87^2 + 206^2 = 50005
     
  3. Which paper and which board? Presumably calculator allowed. Seems like T&I at first glance - is that on the syllabus?
     
  4. GoldMaths

    GoldMaths New commenter

    Think it was Methods 2 AQA but cant be 100%.

    Ridiculously hard if your supposed to do it using trial and improvement.
     
  5. Was there no lead-in part and where was it in the paper ie early on or at the end?
     
  6. GoldMaths

    GoldMaths New commenter

    (a) Show that (mn ? 1)^2
    + (m + n)^2
    = (m^2
    + 1)(n^2
    + 1)
    [3]

    The two shorter sides of a right-angled triangle are 2m and m^2
    ? 1 where m is a
    positive integer.

    (b) Show that the hypotenuse is always a positive integer.
    [2]


    (c) Find two square numbers which have a sum equal to 50005
    [2]
     
  7. Can do the first two parts but struggling to see the link with part (c) - if there is one.
     
  8. bombaysapphire

    bombaysapphire Star commenter

    50005 = 5 x 10001
    5 = 2^2+1 so m=2
    10001 = 100^2+1 so n=100
    Square numbers 199 and 102.
    So yep - definitely a link from (a) to (c).
     
  9. bombaysapphire

    bombaysapphire Star commenter

    [​IMG]
     
  10. GoldMaths

    GoldMaths New commenter

    Is there anyway to solve this without a trial and error approach?
     
  11. bombaysapphire

    bombaysapphire Star commenter

    See post 8. 50005 clearly factorises into 5 and 10001.
    These appear to have been chosen because it is easy to spot that they are both 1 bigger than a square number.
     
  12. DM

    DM New commenter

  13. If you want to use part (a), then the only starting points are
    50005 = 5 X 10001 or 137 X 365 or 73 X 885.
    I don't know if there are solutions not using part (a), but I'll tell you later.
     
  14. Using Maple 13, I got the following complete set of solutions:
    (54, 217); (87, 206); (98, 201); (102, 199).
    I assume this is what DM found using the Mathematica/Alpha machine.
    So yes, I suppose the method is brute force.
     
  15. mmmmmaths

    mmmmmaths New commenter

    Not brute force, use part a) see post 8.
     
  16. DM

    DM New commenter

    Poley knew how to answer the question using the result provided but was answering the wider question about finding the other solutions.
     
  17. Nice one. Who said GCSEs were too easy?

     
  18. Really nice question - will be using that as a starter in the next week.
     

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