Problem 73 Solution #3
Show that if p is a prime such that there is an integer b with p = b2 + 4, then Zp is not a unique factorization domain.
Proof: First notice that in Zp, the norm, $ N(a-b\sqrt{p})=a^2-b^2p $, is multiplicative and if N(x) = 1 for some x in Zp, x is a unit.
Now $ -N(b+\sqrt{p}) = -N(b-\sqrt{p}) = N(2) = 4 $, so if any of these are reducible, say xy = 2, then N(x)N(y) = 4 and since x and y not units, $ N(x)=N(y)=\pm 2 $.
However, we know p = b2 + 4, so p = b2 (mod 4). The only squares mod 4 are 0 and 1 and 4 does not divide p, so p = 1 (mod 4).
Thus we can see $ N(a+b\sqrt{p})=a^2-b^2p=a^2-b^2 \pmod{4} $. But since the only squares mod 4 are 0 and 1, we see there are no elements with $ N(x) = \pm 2 $. Thus $ (b+\sqrt{p})(b-\sqrt{p})=2*2 $ shows two different factorizations of 4 into irreducibles. So Zp is not a UFD.
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