prove or disprove in $\mathbb{Z}[i]$ decomposition of real integer to 2 elements with the same norm is unique up to multiplying by unit

Question

let $a,b,c\in\mathbb{Z}[i]$such that a=bc and say $|b|=B$ and $|c|=C$ . prove or disprove, there is no other decomposition such: $a=de ,|d|=B,|e|=C$ up to multiplying b and c by unity.

Edit: this claim is not true for complex number, but is it for reals? prove it!

Answer 1

No it doesn't hold for real number either: counterexample: $25=(4+3i)(4-3i)=5*5$ yet $(4+3i)$ has the same norm as $5$, as well $(4-3i)$ and $5$.

Answer 2

We can easily find a counterexample by finding two non-associates with the same norm, such as $1+7i$ and $5+5i$. Then, all we need is $$(1+7i)(x) = (5+5i)(y)$$ This yields $$\frac{y}{x} =\frac{1+7i}{5+5i} = \frac{4+3i}{5}$$ so by choosing $x = 5$ and $y = 4+3i$, we see that $$(1+7i)(5) = (5+5i)(4+3i)$$ which serves as a counterexample, since $|1+7i| = |5+5i|$ and $|5| = |4+3i|$, yet neither are associates.