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Thread: Pythagorean Help!

  1. #1
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    Pythagorean Help!

    Question:

    Show that there exists infinitely many primitive Pythagorean triples x, y, z whose even member x is a perfect square.

    I think I can use the fact that n is an arbitrary odd integer and then consider the triple 4n^2, n^4-4, and n^4+4. However, I am stuck on how to put it all together.
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  2. #2
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    Seeing Triples

    I'm sorry, I'm not that familiar with methods of generating Pythagorean Triples.

    You are saying that if n is odd, then $\displaystyle (4n^2)^2+(n^4-4)^2 = (n^4+4)^2 $ . Okay, I verified that, simple enough. Since this works for all odd $\displaystyle n=3,5,7,9,...$ there are an infinite number of triples of this form. For $\displaystyle n$ odd, $\displaystyle 4n^2$ is even and the other two values are odd. Since $\displaystyle 4n^2=(2n)^2$ it is an even perfect square.

    Now you have to show that there exist an infinite number of n's for which $\displaystyle gcd(4n^2,n^4-4,n^4+4)=1$ , thus making the triple primitive.

    Well, since $\displaystyle (n^4+4)-(n^4-4)=8$, $\displaystyle gcd(n^4-4,n^4+4) \leq 8$ . But since they are both odd, their $\displaystyle gcd \in \{1,3,5,7\}$. But for $\displaystyle p \in \{3,5,7\}$ if $\displaystyle p|n^4-4$, then adding 8 will certainly mean $\displaystyle p\not| n^4+4$ . Therefore, for all n odd, $\displaystyle gcd(n^4-4,n^4+4) =1$

    Suppose $\displaystyle p|4n^2$ for some odd prime $\displaystyle p$. Then $\displaystyle p|n$. So $\displaystyle p|n^4$, so $\displaystyle p\not| n^4\pm 4$. So $\displaystyle gcd(4n^2,n^4-4)=gcd(4n^2,n^4+4)=1$.

    Ergo, for n odd, this triple is primitive.
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  3. #3
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    Hello, cathwelch!

    Show that there exists infinitely many primitive Pythagorean triples $\displaystyle x, y, z$
    whose even member $\displaystyle x$ is a perfect square.
    I was given this set of generating equations: .$\displaystyle \begin{array}{ccc}x &=& 2mn \\ y &=& m^2-n^2 \\ z &=& m^2+n^2\end{array}$

    . . where the triple is primitive if $\displaystyle \text{GCD}(m,n) = 1.$



    If $\displaystyle x = 2mn$ is a square, let: .$\displaystyle m = 2p^2,\;n = q^2$, where $\displaystyle \text{GCD}(p,q) = 1$ and $\displaystyle q$ is odd.

    Then we have: .$\displaystyle \begin{Bmatrix}x &=& 4p^2q^2 \\ y &=& 4p^4 - q^4 \\ z &=& 4p^4+q^4 \end{Bmatrix} \quad\text{ where GCD }\!(p,q) = 1\,\text{ and }\,q \text{ is odd.}$


    Examples: .$\displaystyle \begin{array}{ccccccc}
    (p,q) = (1,3) & \Rightarrow & (m,n) = (2,9) & \Rightarrow & (x,y,z) = (36,77,86) \\
    (p,q) = (2,3) & \Rightarrow & (m,n) = (8,9) & \Rightarrow & (x,y,z) = (144,17,145) \\
    (p,q) = (3,5) & \Rightarrow & (m,n) = (18,25) & \Rightarrow & (x,y,z) = (900,301,949)
    \end{array}$

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