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Primitive Pyth. triples: Solutions of x^2 + y^2 = 2 z^2

  1. Mar 29, 2010 #1
    Theorem: The positive primitive solutions of [tex]x^2 + y^2 = z^2[/tex] with y even are [tex]x = r^2 - s^2, y = 2rs, z = r^2 + s^2[/tex], where r and s are arbitrary integers of opposite parity with r>s>0 and gcd(r,s)=1.

    Using this theorem, find all solutions of the equation [tex]x^2 + y^2 = 2z^2[/tex]
    (hint: write the equation in the form [tex](x+y)^2 + (x-y)^2 = (2z)^2[/tex])
    ==================================

    The above theorem characterizes all "PRIMITIVE Pythagorean triples", but what is the statement of the theorem that characterizes ALL "Pythagorean triples" (not necessarily primitive)?

    (i) The positive solutions of [tex]x^2 + y^2 = z^2[/tex] with y even are precisely [tex]x = (r^2 - s^2)d, y = (2rs)d, z = (r^2 + s^2)d[/tex], where d is any natural number, r and s are arbitrary integers of opposite parity with r>s>0 and gcd(r,s)=1.

    (ii) The positive solutions of [tex]x^2 + y^2 = z^2[/tex] with y even are precisely [tex]x = r^2 - s^2, y = 2rs, z = r^2 + s^2[/tex], where r and s are arbitrary integers with r>s>0.

    Which one is correct?

    I hope someone can help me out. Thank you!
     
  2. jcsd
  3. Mar 29, 2010 #2
    The only difference is a d, which, as it says, represents any natural number.
     
  4. Mar 29, 2010 #3
    So I think (i) correctly characterizes ALL Pythagorean triples.

    Is (ii) wrong?
    i.e. not all Pythagorean triple is of the form [tex]x = r^2 - s^2, y = 2rs, z = r^2 + s^2[/tex], r>s>0?
    For example, 6,8,10 is a Pythagorean triple, but I think it cannot be written in the above form?
     
  5. Mar 29, 2010 #4
    The most important triples are the fundamental or primitative ones. We can always add a constant to all elements in any triple.

    If you can read through the proof of (2), which starts from scratch, you will get a better understanding of this.
     
  6. Mar 30, 2010 #5
    But can EVERY Pythagorean triple (primitive or imprimitive) be written in the form [tex]x = r^2 - s^2, y = 2rs, z = r^2 + s^2[/tex] for some r,s E Z, r>s>0?
     
  7. Mar 30, 2010 #6
    No. For example, [itex]9^2 + 12^2 = 15^2[/itex], but 15 cannot be written as the sum of two squares.

    Petek
     
  8. Mar 30, 2010 #7
    OK, so (i) is correct and (ii) is wrong.

    Back to our original problem...

    [tex]x^2+y^2=2z^2\iff(x+y)^2+(x-y)^2=(2z)^2[/tex]
    <=> [tex]x+y,x-y,2z[/tex] is a Pythagorean triple (but it's not necessarily primitive)
    [tex] \begin{cases} x+y=(m^2-n^2)d\\x-y=2mnd\\2z=(m^2+n^2)d \end{cases} [/tex]
    where d is natural number, m and n are integers of opposite parity with m>n>0 and gcd(m,n)=1.

    Now how can we characterize ALL solutions of [tex]x^2 + y^2 = 2z^2[/tex]? Is there any restriction on "d"?

    Thank you!
     
  9. Mar 30, 2010 #8

    jbunniii

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    If [itex]d[/itex] is any natural number, then [itex]x+y[/itex], [itex]x-y[/itex], and [itex]2z[/itex] form a Pythagorean triple. Now just follow your "if and only ifs" in reverse to conclude that [itex]x^2 + y^2 = 2z^2[/itex].
     
  10. Mar 30, 2010 #9
    But I believe that when we're asked to find all solutions of the equation [tex]x^2 + y^2 = 2z^2[/tex], the answer should be of the form
    x=...
    y=...
    z=...

    How can we get this?
     
  11. Mar 30, 2010 #10

    jbunniii

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    Well, you can solve these three equations for x, y, and z:

    [tex]
    \begin{cases} x+y=(m^2-n^2)d\\x-y=2mnd\\2z=(m^2+n^2)d \end{cases}
    [/tex]

    The first two are a linear system of two equations with two unknowns (x and y) so should be easy to solve. The third one is even easier - just divide both sides by 2.

    Is that what you meant?

    [Edit]: Hmm, I guess you will have to impose constraint(s) to ensure that the solutions are integers.

    For example:

    [tex]z = \frac{1}{2}(m^2 + n^2)d[/tex]

    We're still assuming that [itex]m[/itex] and [itex]n[/itex] have opposite parity, right? That means [itex]m^2 + n^2[/itex] is odd, so [itex]d[/itex] had better be even. You will have to check [itex]x[/itex] and [itex]y[/itex] as well once you have expressions for them.
     
    Last edited: Mar 30, 2010
  12. Mar 30, 2010 #11
    Sovling for x,y,z, I get:
    [tex]x = d(m^2 - n^2 + 2mn) / 2[/tex]
    [tex]y = d(m^2 - n^2 - 2mn) / 2[/tex]
    [tex]z = d(m^2 + n^2) / 2[/tex]

    So d must be even, is this the ONLY restriction?
     
    Last edited: Mar 30, 2010
  13. Mar 30, 2010 #12

    jbunniii

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    Yes, that looks like the only restriction.
     
  14. Mar 30, 2010 #13
    So I think the final answer is
    [tex]x = d(r^2 - s^2 + 2rs) / 2[/tex]
    [tex]y = d(r^2 - s^2 - 2rs) / 2[/tex]
    [tex]z = d(r^2 + s^2) / 2[/tex]
    where d is any EVEN natural number, r and s are any integers of opposite parity with r>s>0 and gcd(r,s)=1?
    Does this include ALL solutions of [tex]x^2 + y^2 = 2z^2[/tex]????
     
    Last edited: Mar 31, 2010
  15. Mar 31, 2010 #14
    OK, now I've thought about it more, and I think that are serious problems in my formula above. We're actually missing A LOT of solutions. For example, x=y=z is a solution for any integer (e.g. x=y=z=6 is a solution), but it's definitely not a Pythagorean triple, the formula above misses all of those. Also, all the negative solutions are missing from the above formula because by definition a Pythagorean triple is positive while solutions to the equation can be positive or negative.

    So the above formula certainly does not include all solutions, how can we fix this problem? How can we correctly write down a formula that includes all the solutions?

    Can someone kindly explain how we can summarize our final answer that includes all solutions into a compact formula?
    Thanks a million!
     
    Last edited: Mar 31, 2010
  16. Mar 31, 2010 #15
    x = y = z isn't a solution to [itex]x^2 + y^2 = z^2[/itex] but it is a solution to [itex](x +y)^2 + (x - y)^2 = (2z)^2[/itex]. That is,

    [tex](x + x)^2 + (x - x)^2 = (2x)^2[/tex]

    Petek
     
  17. Apr 1, 2010 #16
    If I want to solve [tex]x^2 + y^2 = 2z^2[/tex] in the integers (rather than in the natural numbers), i.e. x,y,z E Z, what would be the answer? How can we fully describe the set of ALL integer solutions? Right now, I'm having a lot of trouble trying to describe ALL integer solutions without missing any. Does anyone happen to know the correct answer?

    Thanks!!
     
  18. Apr 1, 2010 #17

    CRGreathouse

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    The solutions in integers are just
    (x, y, z)
    (-x, y, z)
    (x, -y, z)
    (-x, -y, z)
    (x, y, -z)
    (-x, y, -z)
    (x, -y, -z)
    (-x, -y, -z)
    for every solution (x, y, z) in positive integers, together with (0, 0, 0).
     
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