Pythagorean triples

**john_n82** · March 17th 2009, 04:35 PM

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.

**PaulRS** · March 17th 2009, 05:06 PM

Case I : $z=16$ ( the case without solutions if you mean $ \mathbb{N}^* = \mathbb{N}/\left\{ 0 \right\} $ )

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

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

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

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

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

We have: $ 16^2 + y^2 = z^2 $ and factorise $ 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) : $ \left\{ \begin{gathered} z + y = 2^a \hfill \\ z - y = 2^b \hfill \\ \end{gathered} \right. $ for $ a,b \in \mathbb{N}/a + b = 8 $ and clearly $ a > b $ since $y>0$

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

Thus it follows: $ \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 $ a,b \in \mathbb{N}/a > b > 0 \wedge a + b = 8 $ and (*) will produce solutions (note that $ 2a > a + b = 8 \Rightarrow a > 4 $ 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 $y$

Thread: Pythagorean triples

LinkBack

Thread Tools

Display

Pythagorean triples

Similar Math Help Forum Discussions

Pythagorean triples

Pythagorean Triples II

pythagorean triples

Pythagorean Triples

pythagorean triples

Search Tags