In this section we will demonstrate the converse. Thus x is irreducible iff |x| is prime.
For completeness, let's bring in the half quaternions. Thus we can use the previous theorem without disclaimers. In particular, the ring is a one-sided pid; every left or right ideal is principal.
Assume x is irreducible, and |x| = n, where p is a prime that divides n. If x consists of half integers, multiply by a half unit so that the coefficients of x are integers. This does not change |x|, nor the irreducibility of x.
Let H be the left ideal generated by x and p. Let w be an element of H, sx+tp for some s and t. What is |w| mod p? If t consists of half integers, multiply s t and w by a half unit on the left, so that t becomes an integer quaternion. This does not change |w|. Now |w| equals the sum of the squares of the coefficients of w mod p, so we can reduce the coefficients mod p first, then compute |w|. With this in mind, the term tp contributes nothing. We are left with sx. Therefore |w| = |s|×|x| mod p. Remember that |x| is 0 mod p, so every element in H has norm divisible by p. In particular, H does not include 1, and x and p do not span the entire ring.
Since H is a proper left ideal, and the ring is a one sided pid, let g generate H. In other words, g is the right gcd of p and x. Since x is irreducible, and g is not a unit, g is an associate of x. We may as well call it x, hence p is a left multiple of x. Write p = sx and take norms. The norm of x divides p2, and if it equals p2, then s is a unit. This makes x an associate of p. Otherwise |x| = p, and x times x conjugate = p.
If x is irreducible, its norm is prime, or x is an associate of p, and |x| = p2.
To complete the proof, we need to show that every prime splits. In other words, x = p, with norm p2, is not irreducible. That would prove x irreducible iff |x| irreducible. Let's see if p always splits.
Next let p be an odd prime, and restrict attention to the quaternion integers. Let H be the ideal generated by p, with index p4. Mod out by H to produce a quotient ring S with cardinality p4. The ring S is basically the quaternions reduced mod p.
Suppose S has no zero divisors. Then x times x conjugate, which is the norm of x, is a nonzero integer mod p. Multiply by another integer to get 1 mod p. Thus x is invertible. With no zero divisors, S is a division ring, and a finite division ring is a field.
Suddenly multiplication becomes commutative. Since ij = -ji, 2ij = 0, hence either i or j maps to 0. Yet i and j are units, and they must map to units. This is a contradiction.
Since S cannot be a finite field it contains zero divisors. Write xy = 0 in S, which means xy (in the quaternions) is divisible by p, even though x and y are not. Now p divides |x|×|y|; assume p divides |x|. In other words, p divides the norm of x, but does not divide all four coefficients of x.
If x is not irreducible, write x = ab. If p divides a or b it divides x, so p divides neither a nor b, yet it divides |ab|. It follows that p divides |a| or |b|. Assume p divides |a| and relabel a as x. Now the norm of x is smaller than it use to be. This process stops eventually, when x is irreducible.
We already showed the norm of x is either p or p2. But x is not divisible by p, hence it is not an associate of p. Therefore |x| = p, and p splits into x times x conjugate.
This completes the proof. If x is a quaternion or half quaternion, it is irreducible iff its norm is prime. Every integer, even a prime integer, splits in the quaternions.
Remember that each prime splits in the quaternion integers, without resorting to the half integers, so the same holds for our composite number n.