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Math Help - Pythagorean triples

  1. #1
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    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.
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  2. #2
    Super Member PaulRS's Avatar
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    • Case I : z=16 ( the case without solutions if you mean <br />
\mathbb{N}^*  = \mathbb{N}/\left\{ 0 \right\}<br />
)

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

    So: <br />
x^2  + y^2  = 16^2 <br />
implies that <br />
x = 2x_1<br />
and y=2y_1 with x_1,x_2 \in \mathbb{Z}^ +  <br />

    So: <br />
x_1 ^2  + y_1 ^2  = 8^2 <br />

    And repeat the process until we get: <br />
x_4 ^2  + y_4 ^2  = 1<br />
which is a contradiction since <br />
x_4 ^2  + y_4 ^2  \geqslant 2<br />
!


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

    We have: <br />
16^2  + y^2  = z^2 <br />
and factorise <br />
16^2 =2^8 = z^2  - y^2  = \left( {z + y} \right) \cdot \left( {z - y} \right)<br />

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

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

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

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

    Remember to consider the case y=16 which is totally analogous, all you have to do is to exchange x for y
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