Math Help - Pythagorean triples

1. 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.

• 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 $y$