If it is, is it trivial?
**Theorem**
There does not exist a pythagorean triple $a^2 + b^2 = c^2$ $\{a,b,c \in \Bbb N\}$ where $b \ge a$ and $a|b$
**PROOF**
Suppose $\exists$ a constant k {$k | k \ge 1, k \in \Bbb N$} and $k \cdot a$ forms a pythagorean triple with a.
I.e $a^2 + (k \cdot a)^2 = c^2$
$a^2 + k^2 \cdot a^2 =c^2 $
$a^2(k^2 + 1) = c^2$
$c = \sqrt{a^2(k^2 + 1)}$
$c = a \sqrt{k^2 + 1} $
$ \exists $ $ k: \sqrt{k^2 + 1} \in \Bbb Q$
However $\forall$ $ a \in \Bbb N$, $ a \ge 1$ $ \sqrt{a^2 + 1} \notin \Bbb Q$*
Where $ \Bbb Q$ is the set of rational numbers.
*: The proof of this theorem is left as an exercise to the reader.