Write $5$ and $6+8i$ as products of irreducible elements in $\mathbb{Z}[i]$.

Question

Write $5$ and $6+8i$ as products of irreducible elements in $\mathbb{Z}[i]$.

I am not sure how to do $5$. Do I assume $5=\alpha \beta$, and so $v(\alpha\beta)=v(\alpha)v(\beta)=5$. Then since $5\equiv 1$ (mod $4$), how do I find two non-associate irreducible elements in $\mathbb{Z}[i]$ that equal $5$?

For $6+8i$, what I did was factor out $2$. Then $2(3+4i)=(1+i)(1-i)\alpha\beta$. Then $v(\alpha\beta) = 25$ or $\alpha = \beta = 5$. However, since $5\equiv 1$ (mod $4$), I am again stuck on the dilemma where I cannot find two non-associate irreducible elements that equal $5$.

Answer 1

Ah, here's the mistake: $v(5)\neq 5$, $v(5)=25$. Now, $5=1+4 = 1^2+2^2 = (1+2i)(1-2i)$. Similarly $2 = 1^2 + 1^2 = (1+i)(1-i)$. Can you finish now?

Answer 2

Note that $N(\alpha\beta)$, which I assume is $v$ in your case, is 25 not 5 as $N(\alpha\beta) = N(5) = 5^2$.

Then, $25 = N(\alpha)N(\beta)$ and the only way to factor is when both norms equal 5. Can you go on from here?