Results 1 to 2 of 2

Thread: Pythagorean triples

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    28

    Pythagorean triples

    Find all Pythagorean triples (not necessary primitive) with one of the three integers equal to 16. (i.e. x, y, z in N* with x^2 + y^2 = z^2 and 16 in {x, y, z}).

    I know how to find the answer for this problem but don't know how to show my answer in a proper way. The answer is {(16, 12, 20), (12, 16, 20), (16, 30, 34), (30, 16, 34), (16, 63, 65), (63, 16, 65)}. Please help me to show all the steps. Thank you so much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    • Case I : $\displaystyle z=16$ ( the case without solutions if you mean $\displaystyle
      \mathbb{N}^* = \mathbb{N}/\left\{ 0 \right\}
      $)

    Note that: $\displaystyle
    x^2 + y^2 \equiv 0\left( {\bmod .4} \right) \Leftrightarrow x \equiv y \equiv 0\left( {\bmod .2} \right)
    $ Just because $\displaystyle
    x^2 \equiv 0,1\left( {\bmod .4} \right)\forall x \in \mathbb{Z}
    $

    So: $\displaystyle
    x^2 + y^2 = 16^2
    $ implies that $\displaystyle
    x = 2x_1
    $ and $\displaystyle y=2y_1$ with $\displaystyle x_1,x_2 \in \mathbb{Z}^ +
    $

    So: $\displaystyle
    x_1 ^2 + y_1 ^2 = 8^2
    $

    And repeat the process until we get: $\displaystyle
    x_4 ^2 + y_4 ^2 = 1
    $ which is a contradiction since $\displaystyle
    x_4 ^2 + y_4 ^2 \geqslant 2
    $!


    • Case II : Without loss of generality ( since we can exchange $\displaystyle x$ for $\displaystyle y$ freely) take $\displaystyle x=16$

    We have: $\displaystyle
    16^2 + y^2 = z^2
    $ and factorise $\displaystyle
    16^2 =2^8 = z^2 - y^2 = \left( {z + y} \right) \cdot \left( {z - y} \right)
    $

    So we must have (since 2 is prime) : $\displaystyle
    \left\{ \begin{gathered}
    z + y = 2^a \hfill \\
    z - y = 2^b \hfill \\
    \end{gathered} \right.
    $ for $\displaystyle
    a,b \in \mathbb{N}/a + b = 8
    $ and clearly $\displaystyle
    a > b
    $ since $\displaystyle y>0$

    Sum the 2 equations to get: $\displaystyle
    2 \cdot z = 2^a + 2^b
    $, $\displaystyle
    a > 0
    $ hence $\displaystyle
    2z - 2^a = 2^b
    $ is even, and so $\displaystyle b>0$

    Thus it follows: $\displaystyle
    \begin{gathered}
    z = 2^{a - 1} + 2^{b - 1} \hfill \\
    y = 2^{a - 1} - 2^{b - 1} \hfill \\
    \end{gathered}
    $ (*) (solve the system)

    Now take $\displaystyle
    a,b \in \mathbb{N}/a > b > 0 \wedge a + b = 8
    $ and (*) will produce solutions (note that $\displaystyle
    2a > a + b = 8 \Rightarrow a > 4
    $ hence $\displaystyle a$ can take 3 values $\displaystyle a=5,6,7$)

    Remember to consider the case $\displaystyle y=16$ which is totally analogous, all you have to do is to exchange $\displaystyle x$ for $\displaystyle y$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pythagorean triples
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: Jun 20th 2011, 03:58 AM
  2. Pythagorean Triples II
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Nov 2nd 2010, 07:13 AM
  3. pythagorean triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Nov 23rd 2009, 08:15 PM
  4. Pythagorean Triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Dec 8th 2008, 10:18 PM
  5. pythagorean triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Nov 2nd 2008, 04:28 AM

Search Tags


/mathhelpforum @mathhelpforum