8. Euler’s prime-producing polynomial 21

complex number corresponding to the intersection of e and f. The angle

determined by e1 and e2 cuts f into a segment of length |D|/3 1, and

so there is a point of Z + Zη on f within this angle. We choose such a point

ξ for which the distance from ξ to µ is as small as possible. See Figure 3.

We claim that the distance from ξ to e is strictly smaller than

√

3/2.

This is clear if both ξ + 1 and ξ − 1 fall within the angle determined by e1

and e2, since in that case, the distance from ξ to µ must be at most 1/2. So

suppose that ξ + 1 falls outside this angle; the case when ξ − 1 falls outside

is analogous. Then ξ − 1 must lie within the given angle. Now, if ξ is to the

right of µ, then in order that ξ be at least as close to µ as ξ − 1, it must be

that the distance from ξ to µ is at most 1/2. So we can assume that ξ falls

to the left of µ. This is the scenario depicted in Figure 3. In this case we

we use the following argument: Let ν represent the intersection of e1 and

f; then the distance between ξ and ν is smaller than 1. Since e1 makes an

angle of

60◦

with f, elementary trigonometry shows that the distance from

ξ to e1 is strictly smaller than

√

3/2. But the perpendicular line segment

from ξ to e1 meets e. So the distance from ξ to e is also strictly smaller than

√

3/2.

It follows that the unit disc centered at ξ intersects e in a segment of

total length 1. (Indeed, let τ be the point on e for which the line from ξ

to τ is perpendicular to e, so that the distance from ξ to τ is strictly smaller

than

√

3/2. Then by the Pythagorean theorem, τ divides the segment in

question into two parts, each of length 1/2.) Since |π1/ρ1| ≤ 1, it follows

that we can choose a rational integer γ so that γπ1/ρ1 lies within the open

unit disc centered at ξ.

We claim that with the above choices of ξ and γ, both (P1) and (P2)

hold. Condition (P2) is guaranteed by the choice of γ, so it remains only to

verify (P1). For this it is enough to prove that ξ is prime. Indeed, suppose

that ξ is prime but (P1) fails. Then

ρ1ξ = π1κ

for some κ. Since ξ is prime, it must divide either π1 or κ. But ξ cannot

divide π1; if it did, then since π1 is irreducible, we would have that π1 is

a unit multiple of ξ. But then π1 would be prime since ξ is prime. This

contradicts the observation made above that none of the πi are prime. So

ξ must divide κ; but then dividing through by ξ we find that π1 divides ρ1.

That implies that π1 and ρ1 are unit multiples of each other, which again

contradicts our initial observations.

Why should ξ be prime? Since ξ is a point of the lattice Z + Zη lying

on f, we have ξ = n + η for some integer n. Moreover, since ξ belongs to