# Pythagorean triples

• Mar 17th 2009, 04:35 PM
john_n82
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.
• Mar 17th 2009, 05:06 PM
PaulRS
• 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 $\displaystyle y$