A Lipschitz integer is a Quaternion with integer coefficients. The norm is defined as $N(a+ib+jc+kd)=a^2+b^2+c^2+d^2$ which is a multiplicative function
$N:\mathbb H\to\mathbb R$, $N(\alpha\beta)=N(\alpha)N(\beta)$
Now since any natural number can be written as the sum of four squares of integers, all prime numbers in $\mathbb N$ are composite Lipschitz integers
$a^2+b^2+c^2+d^2=(a+ib+jc+kd)(a-ib-jc-kd)$
Obviously, if $\alpha$ is a composite, then so is $N(\alpha)$.
In the ring of Gaussian integers it also holds that if both the real and the imaginary terms are non-zero, then the norm is prime if the Gaussian number is prime. My question is
If a Lipschitz integer is irreducible and all four terms are non-zero, is then the norm a prime ?
Due to Wikipedia a Hurwitz number is irrecducible iff it's norm is a prime number.
A Hurwitz number is either a Lipschitz number or a quaternion number where all components are of the form odd number divided with $2$.
For Hurwitz numbers Euclidian division is possible.
For complex or Gaussian integers $u= a + ib$ and $v= c + id$, with the norm $N(v) > 0$, there always exist $z = e + if$ and $z_1 = r + is$ such that
$u = vz + z_1$, where $N(z_1) < N(v)$.
However, for Lipschitz integers $m = a+ib+jc+kd$ and $n =e+if+jg+kh$ it can happen that $N(z_1) = N(n)$. This motivated a switch to Hurwitz integers, for which the condition $N(z_1) < N(n)$ is guaranteed.