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Formaula for pythagorian triples watch

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    ive been thinking bout this recently and i belive there are a infinate number, and also ive been seeing similarities in them as they get bigger,

    any1 else have ny thoughts on this
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    It can be shown that, the three sides of a right angled triangle can be written in the form

    p^2-q^2
    p^2+q^2
    2pq

    And yes, there are similarities. For example, you have a 3,4,5 triangle, along with a 6,8,10 one

    so if a^2+b^2=c^2
    and you double a and b, and let the sum of their squares be x
    you get 4a^2+4b^2=x^2
    a^2+b^2=c^2, x^2/4=c^2
    so x=2c

    There is also quite obvious stuff, such that, they are either all even or only one is even. But that is obvious from the fact that an odd+odd=even, and even+odd=odd and even+even=even
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    http://www.nrich.maths.org/discus/me...tml?1102785766
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    The product of a pythagorean triple is a mutltiple of 60.
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    if the proof difficult? ^
    I tried showing that the square of the product [(a^4)(b^2) + (a^2)(b^4)] is a multiple of 3600. But I'm not sure cancelling c is a good idea because the set of numbers have to follow the rule and can't just be any 3 integers.
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    (Original post by lgs98jonee)
    It can be shown that, the three sides of a right angled triangle can be written in the form

    p^2-q^2
    p^2+q^2
    2pq

    There is also quite obvious stuff, such that, they are either all even or only one is even. But that is obvious from the fact that an odd+odd=even, and even+odd=odd and even+even=even
    Yes, one must be even and one must be odd, but thats not where the p^2 - q^2, p^2 + q^2, and 2pq come from. In fact, this parameterization requires quite a bit of work. Its interesting, however, because its literally half the proof that Fermat used in his infinite descent argument for n=4.
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    (Original post by mik1a)
    if the proof difficult? ^
    I tried showing that the square of the product [(a^4)(b^2) + (a^2)(b^4)] is a multiple of 3600. But I'm not sure cancelling c is a good idea because the set of numbers have to follow the rule and can't just be any 3 integers.
    If you know a bit of modular arithmetic then it's not too bad.

    Square numbers can be 0 or 1 mod 3, can be 0 or 1 mod 4, and can be 0, 1 or 4 mod 5.

    So looking at a^2 + b^2 = c^2 we see that:

    mod 3 this eqn has to be 0 + 1 = 1 or 0 + 0 = 0;
    mod 4 this eqn has to be 0 + 1 = 1 or 0 + 0 = 0;
    mod 5 this eqn has to be 0 + 0 = 0 or 0 + 1 = 1 or 1 + 4 = 0 or 0+4=4.

    Any which way there has to be factors of 3,4 and 5 in abc and so the product is divisible by 60.
 
 
 
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