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Math Help - Infinite number of primitive Pythagorean triples

  1. #1
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    Lightbulb Infinite number of primitive Pythagorean triples

    Prove that there exist an infinite number of primitive Pythagorean triples x,y, and z with y even such that y is a perfect cube.

    I can write down the definition, but I have no idea how prove this...

    Any help is appreciated!

    [also under discussion in math links forum]
    Last edited by kingwinner; March 28th 2010 at 12:39 AM.
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  2. #2
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    Quote Originally Posted by kingwinner View Post
    Prove that there exist an infinite number of primitive Pythagorean triples x,y, and z with y even such that y is a perfect cube.

    I can write down the definition, but I have no idea how prove this...

    Any help is appreciated!

    Hint: if a,b,c\in\mathbb{Z}\,\,\,then\,\,\,a^2+b^2=c^2\Long  leftrightarrow a=m^2-n^2\,,\,b=2mn\,,\,c=m^2+n^2\,,\,\,m,n\in\mathbb{Z} .

    The triple (a,b,c) is called primitive if gcd(m,n)=1 and exactly one of them is even , so: can you find infinite pairs like  m,n as above? Well, there you have your infinite primitive pythagorean triples.

    Tonio
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  3. #3
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    Quote Originally Posted by kingwinner View Post
    Prove that there exist an infinite number of primitive Pythagorean triples x,y, and z with y even such that y is a perfect cube.

    I can write down the definition, but I have no idea how prove this...

    Any help is appreciated!
    WLOG We know for some  m>n

     x=m^2-n^2
     y=2mn
     z=m^2+n^2

    So let  m=4k^3 and  n=1 .

    Now observe  y=2\cdot4k^3 = (2k)^3 .
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    Quote Originally Posted by tonio View Post
    Hint: if a,b,c\in\mathbb{Z}\,\,\,then\,\,\,a^2+b^2=c^2\Long  leftrightarrow a=m^2-n^2\,,\,b=2mn\,,\,c=m^2+n^2\,,\,\,m,n\in\mathbb{Z} .

    The triple (a,b,c) is called primitive if gcd(m,n)=1 and exactly one of them is even , so: can you find infinite pairs like  m,n as above? Well, there you have your infinite primitive pythagorean triples.

    Tonio
    Yes, but the question requires also that y must be a perfect cube.
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    Quote Originally Posted by chiph588@ View Post
    WLOG We know for some  m>n

     x=m^2-n^2
     y=2mn
     z=m^2+n^2

    So let  m=4k^3 and  n=1 .

    Now observe  y=2\cdot4k^3 = (2k)^3 .
    Thanks, but can you explain how you came up with these answers and how can we prove that the resulting triples will all be "primitive" Pythag. triple and that y is a perfect cube?

    I have seen the theorem:
    The positive primitive solutions of x^2 + y^2 = z^2 with y even are x = r^2 - s^2, y=2rs, z = r^2 + s^2, where r and s are arbitrary integers of opposite parity with r>s>0 and (r,s)=1.

    Thank you!
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    Quote Originally Posted by kingwinner View Post
    Thanks, but can you explain how you came up with these answers and how can we prove that the resulting triples will all be "primitive" Pythag. triple and that y is a perfect cube?

    I have seen the theorem:
    The positive primitive solutions of x^2 + y^2 = z^2 with y even are x = r^2 - s^2, y=2rs, z = r^2 + s^2, where r and s are arbitrary integers of opposite parity with r>s>0 and (r,s)=1.

    Thank you!

    Well, you can choose ANY pair of coprime m,n\in\mathbb{N} with say even m, so he chooses m=4k^3\,,\,n=1...it is obvious these pairs are coprime, right? By the way, he could as well choose n=1\,,\,m=72k^3

    Tonio
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    Quote Originally Posted by tonio View Post
    Well, you can choose ANY pair of coprime m,n\in\mathbb{N} with say even m, so he chooses m=4k^3\,,\,n=1...it is obvious these pairs are coprime, right? By the way, he could as well choose n=1\,,\,m=72k^3

    Tonio
    Is there any restriction on k?

    Also, are you sure that m=72k^3 would work? 144 is not a perfect cube...
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    Quote Originally Posted by kingwinner View Post
    Is there any restriction on k?

    Also, are you sure that m=72k^3 would work? 144 is not a perfect cube...

    A mistake: it should have been m=2^2\cdot 3^3k=108k

    Tonio
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    In general, to ensure that the triple is "primitive", we need to ensure TWO things, i.e. (i) m and n must be of opposite parity AND (ii) gcd(m,n)=1, right?
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    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    In general, to ensure that the triple is "primitive", we need to ensure TWO things, i.e. (i) m and n must be of opposite parity AND (ii) gcd(m,n)=1, right?
    Right. Letting n=1 and m be a multiple of 4 always ensures this.
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    Quote Originally Posted by chiph588@ View Post
    Right. Letting n=1 and m be a multiple of 4 always ensures this.
    Any reason why gcd(m,n)=1 alone is not enough to guarantee that the triple is "primitive"?
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  12. #12
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    Quote Originally Posted by kingwinner View Post
    Any reason why gcd(m,n)=1 alone is not enough to guarantee that the triple is "primitive"?
    I think that's enough actually.
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    Quote Originally Posted by chiph588@ View Post
    I think that's enough actually.
    I just went through the proof of the theorem fairly carefully, and I think we do need BOTH conditions to be satisfied in order to ensure gcd(x,y,z)=1.
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    Quote Originally Posted by chiph588@ View Post
    I think that's enough actually.

    I don't think so: if both m,n are odd then their squares' difference/sum are both even, rendering even integers in the triple...

    Tonio
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