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Pythagorean Triples

  1. Nov 3, 2004 #1
    Prove that every Pythagorean triple is of the form 3k, 4k, 5k. Could I say that 3k = x = 2st, 4k = y = t^2-s^2, and 5k = z = t^2 + s^2? those are the definitions of the pythagorean triple correct? can anyone say yea or nay? if nay, how can i make it correct?
  2. jcsd
  3. Nov 3, 2004 #2


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    You might want to check the exact wording of the problem, 5, 12, 13 is a Pythagorean triple that's not of the form you give.
  4. Nov 3, 2004 #3
    the wording is correct.
  5. Nov 3, 2004 #4

    matt grime

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    He/she didn't say that k was necessarily an integer....

    The general form of pythagorean triples is well known, try googling for them and I@m sure you'll find a nice proof.

    What you wrote certainly isn't true as the counter example shows. One counter example disproves it, so how are you going to amend your question?

    Where did the question appear?
  6. Nov 3, 2004 #5
    You have the general form of the triples correctly stated, but they are not of the form 3k, 4k, 5k, consider

    [tex]12^2+5^2=13^2, or 20^2+ 21^2 =29^2.[/tex]

    How could you argue about that? Is 29 a multiple of 5, is 13?
  7. Nov 4, 2004 #6


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    No it is not. Write down exactly what you've been asked.
  8. Nov 4, 2004 #7


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    IF you have copied the problem exactly then the problem is wrong:

    As has been pointed out, 5, 12, 13 is a Pythagorean triple (25+ 144= 169) but is NOT of the form 3k, 4k, 5k, even for k not an integer: if 3k= 5 then k= 5/3 but 4(5/3)= 20/3 is NOT equal to 12.

    It MAY be that the problem asks you to show that any triple of the form (3k, 4k, 5k) IS a Pythagorean triple. That exactly the opposite of what you wrote and is very easy to show.
  9. Nov 14, 2004 #8
    is the question demanding a counter-example here?
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