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.

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