8. Euler’s prime-producing polynomial 17
Lemma 1.11. An element α Z[η] is a unit precisely when N (α) = 1.
The only units in Z[η] are ±1.
Proof. If α is a unit, then N (α) · N
(α−1)
= 1. Moreover, both N (α) and
N (α−1) are positive integers, so that N (α) = N (α−1) = 1. Conversely, if
N (α) = 1, then αα = 1, and so α is a unit. Finally, notice that if y = 0,
then
N (x + yη) =
x2
xy +
Ay2
= (x
y/2)2
+
1
4
(4A
1)y2

4A 1
4
7
4
1.
So x + can be a unit only when y = 0. In this case we must have
N (x) =
x2
= 1, and this occurs exactly when x = ±1.
Lemma 1.12. If α is a nonzero, nonunit element of Z[η], then α can be
written as a product of irreducible elements of Z[η].
Proof. If the claim fails, there is a nonzero, nonunit α of smallest norm
for which it fails. Clearly α is not irreducible, and so we can write α = βγ,
where β and γ are nonzero nonunits. Hence N (α) = N (β)N (γ). Since N (β)
and N (γ) are each larger than 1, both N (β) and N (γ) must be smaller than
N (α). So by the choice of α, both β and γ factor as products of irreducibles,
and thus α does as well. This contradicts the choice of α.
We can now prove one of the two outstanding implications:
Proof that (iii) (i). Let η = (−1 +

D)/2. Suppose 0 n A 1.
We have
(1.9)
n2
+ n + A = (n η)(n ¯) η = (n η)(n + 1 + η).
Let p be a prime dividing
n2
+ n + A. We claim that p is not irreducible in
Z[η]. Indeed, since Z[η] is a unique factorization domain by hypothesis, if p
were irreducible, then p would be prime. So from (1.9), we would have that
p divides n η or n + 1 + η. But this is impossible, since neither n/p η/p
nor (n + 1)/p + η/p belongs to Z[η] = Z + Zη.
Hence we can write p = αβ, where α, β Z[η] and neither α nor β is a
unit. Taking norms, we deduce that
p2
= N (p) = N (α)N (β). Since α and
β are not units, we must have N (α) = N (β) = p.
Write α = x + for integers x, y. Then y = 0 (since p is a rational
prime), and so
p = N(α) =
x2
xy +
Ay2
= (x
y/2)2
+ (A
1/4)y2
A 1/4.
Thus (since p is an integer) p A. Moreover, since 0 n A 1,
n2
+ n + A (A
1)2
+ (A 1) + A = (A 1)A + A =
A2.
Previous Page Next Page