Exhibit some irreducible Gaussian integers

Find all the irreducible elements a + bi \in \mathbb{Z}[i], where |a|, |b| \leq 5.


Note that if \alpha is irreducible, then all the associates of \alpha are also irreducible, as well as their conjugates. Thus from each irreducible Gaussian integer we can easily construct seven others. We claim that if \alpha is a Gaussian integer, then one of these eight is of the form a+bi where 0 \leq a \leq b. To see this, let \alpha = a+bi and note the following.

  • If 0 \leq a \leq b, then a+bi has this form.
  • If 0 \leq b \leq a, then b + ai has this form.
  • If a \leq 0 \leq b and |a| \leq |b|, then (-a) + bi has this form.
  • If a \leq 0 \leq b and |a| \geq |b|, then b + (-a)i has this form.
  • If b \leq 0 \leq a and |a| \leq |b|, then a + (-b) has this form.
  • If b \leq 0 \leq a and |a| \geq |b|, then (-b) + ai has this form.
  • If a \leq b \leq 0, then (-b) + (-a)i has this form.
  • If b \leq a \leq 0, then (-a) + (-b)i has this form.

So every Gaussian integer is conjugate-associate to one of the form a+bi where 0 \leq a \leq b. Suppose a = b \neq 1; then a+bi = a(1+i) is reducible. Thus every irreducible Gaussian integer is conjugate-associate to either 1+i or to a+bi where 0 \leq a < b.

To find the irreducible Gaussian integers a+bi satisfying |a|,|b| \leq 5, it suffices to consider the following 16 cases.

  1. i: This is a unit, and so not irreducible.
  2. 2i: Note that 2i = (1+i)^2 has a nontrivial factorization, and so is not irreducible.
  3. 3i: Note that N(3i) = 9. Suppose N(x+yi) = 3, so that x^2+y^2 = 3. This equation has no solutions in the integers. Thus in any factorization of 3i, some factor must have norm 1, and thus must be a unit. Hence 3i is irreducible.
  4. 4i: Note that 4i = (2)(2i) has a nontrivial factorization, and so is not irreducible.
  5. 5i: Note that 5i = (1+2i)(2+i) has a nontrivial factorization, and so is not irreducible.
  6. 1+i: Note that N(1+i) = 2 is prime, so that this element is irreducible.
  7. 1+2i: Note that N(1+2i) = 5 is prime, so that this element is irreducible.
  8. 1+3i: Note that 1+3i = (1+i)(2+i) has a nontrivial factorization, and so is not irreducible.
  9. 1+4i: Note that N(1+4i) = 17 is prime, so that this element is irreducible.
  10. 1+5i: Note that 1+5i = (i-1)(2-3i) has a nontrivial factorization, and so is not irreducible.
  11. 2+3i: Note that N(2+3i) = 13 is prime, so that this element is irreducible.
  12. 2+4i: Note that 2+4i = 2(1+2i) has a nontrivial factorization, and so is not irreducible.
  13. 2+5i: Note that N(2+5i) = 29 is prime, so that this element is irreducible.
  14. 3+4i: Note that 3+4i = (2i-1)(1-2i) has a nontrivial factorization, and so is not irreducible.
  15. 3+5i: Note that 3+5i = (4i-1)(1-i) has a nontrivial factorization, and so is not irreducible.
  16. 4+5i: Note that N(4+5i) = 41 is prime, so that this element is irreducible.
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