## Factor in the Gaussian integers

Factor the following elements in $\mathbb{Z}[i]$: $7+70i$, $1 - 7i$, $17 + i$, $219 + 219i$.

Note that $7 + 70i = 7(1+10i)$. Since $N(1+10i) = 101$ is prime, $1+10i$ is irreducible. Note that no Gaussian integer has norm 7 since $a^2+b^2 = 7$ has no integer solutions. Thus 7 is irreducible, as any factorization of 7 must have a factor with norm 1.

Note that $1-7i = (2+i)(1-2i)(1-i)$. We saw previously that each of these factors is irreducible.

Note that $17+i = (2-5i)(1-2i)(i-1)$. We saw previously that these factors are irreducible.

Note that $219 + 219i = 3 \cdot (1+4i) \cdot (1-4i) \cdot (1+i)$. We saw previously that these factors are irreducible.