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Math Help - Pythagorean Triples!

  1. #1
    Newbie zenith20's Avatar
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    Pythagorean Triples!

    Dear all,
    i have recently noticed that all Pythagorean Triples include ONE and only one number that is a multiple of 5. but i have no proof for it.

    could anybody help me on this,
    thank you.


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  2. #2
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    Re: Pythagorean Triples!

    Hey zenith20.

    The formal way of stating this would be to prove that for:

    a^2 + b^2 = c^2 then a = 5x (x is an integer) and gcd(a,b) = gcd(a,c) = 1 since 5 is a prime number.
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  3. #3
    Lord of certain Rings
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    Re: Pythagorean Triples!

    (15,20,25) is a pythogorean triple with more than one multiple of 5. Hence I will assume you are talking about primitive pythogorean triples.

    We will prove your claim by contradiction. By definition, the primitive triples should not share a common factor. Suppose two of them shared a common factor of 5, the Pythogorus theorem would then imply that the third one should also contain a factor of 5. Then all of them would have a common factor of 5 contradicting primitiveness of the triple.
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    Newbie zenith20's Avatar
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    Re: Pythagorean Triples!

    many thanks to Isomorphism and Chiro,
    now i know that there could not be 'more than One'. but still i can not prove why 'there IS one'. i mean we can not have primitive Pythagorean triples 'without a multiple of 5' number.
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  5. #5
    Lord of certain Rings
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    Re: Pythagorean Triples!

    Quote Originally Posted by zenith20 View Post
    many thanks to Isomorphism and Chiro,
    now i know that there could not be 'more than One'. but still i can not prove why 'there IS one'. i mean we can not have primitive Pythagorean triples 'without a multiple of 5' number.
    That can be answered by simple modular arithmetic:

    Consider a pythogorean triple such that x^2 + y^2 = z^2. Let us supposed that neither x nor y has a factor of 5. Then observe that x^2, y^2 can leave a remainder of 1 or 4 when divided by 5.

    If x^2= y^2=1 \mod 5, then z^2 = 2 \mod 5. But a square of a number cannot be 2 mod 5.
    If x^2= y^2=4 \mod 5, then z^2 = 3 \mod 5. But a square of a number cannot be 3 mod 5.

    Thus one of the x^2, y^2 must be 1 mod 5 and the other one must be 4 mod 5. Adding we get that z is divisible by 5.

    So we have proved that if x and y both are not divisible by 5 then z is divisible by 5.
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: Pythagorean Triples!

    Quote Originally Posted by Isomorphism View Post
    That can be answered by simple modular arithmetic:

    ...So we have proved that if x and y both are not divisible by 5 then z is divisible by 5.
    Very nice. This same appoach can be used to prove that for Pyrthagorian primitives either x or y must be divisible by 3, and furthermore that either x or y must be even and z must be odd (using modular 4 arithmetic).
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