8. Euler’s prime-producing polynomial 19

π1

ρ1

Figure 2. (Based on [Zau83].)

We can assume that N (π1) ≤ N (ρ1). (If this does not hold initially,

interchange the two factorizations.) For ξ, γ ∈ Z[η] still to be chosen, define

(1.10) α := (ρ1ξ − π1γ)ρ2 · · · ρj.

Then

α = αξ − π1

α

ρ1

γ

= π1(π2 · · · πkξ − ρ2 · · · ρjγ).

Factoring the parenthetical expression, we deduce that α has a factorization

into irreducibles where one of the irreducibles is π1. We will choose ξ and γ

so that π1 ρ1ξ. Then π1 ρ1ξ−π1γ, and so we may deduce from (1.10) that

α has a factorization into irreducibles, none of which is a unit multiple of

π1. So α possesses two distinct factorizations into irreducibles. If further,

γ and ξ satisfy

N (ρ1ξ − π1γ) N (ρ1),

then N (α ) is smaller than N (α), and so we have a contradiction to our

choice of α.

So it remains to show that it is possible to choose ξ, γ ∈ Z[η] with the

following two properties:

(P1) π1 ρ1ξ,

(P2) N (ρ1ξ − π1γ) N (ρ1), or equivalently, ξ −

π1

ρ1

γ 1.

Since N (π1) ≤ N (ρ1), the complex number π1/ρ1 lies on or inside the unit

circle. Suppose first that π1/ρ1 lies outside the shaded region indicated in