We consider $\Bbb H (\Bbb Z)$ the ring of integer quaternions. We consider $H = \Bbb H (\Bbb Z)\cup \left(\Bbb H (\Bbb Z)+\dfrac{1+i+j+k}{2}\right)$ the eclidean ring called the Hurwitz quaternions.
Let be $x \in \Bbb H(\Bbb Z)$ a non-prime quaternion. Not being a prime in $H$ implies the existence of $a,b \in H$ such as $x = ab$ with $a,b$ non invertible.
Show that you can choose $a$ or $b$ in $\Bbb H (\Bbb Z)$.
I read this in a paper wich said it was obvious. But cannot figure out why...
EDIT: apparently, you can even choose $a$ and $b$ in $\Bbb H (\Bbb Z)$.
This result is for the proof that $x\in \Bbb H (\Bbb Z)$ is a prime iff $N(x)$ is a prime in $\Bbb Z$.