démonstration

n∈Nσ(n)
S={n|3|n3-σ(n)−n}S
x2+xy +y2
x, y ∈N
j= exp(2iπ
3)Z[j]
α=a+bj ∈Z[j]α N(α) := |α|2|α|
α n α ∈Z[j]
n=N(α)
Z[j]
α, β ∈Z[j]N(αβ) = N(α)N(β)
α∈Z[j]β∈Z[j]N(α) = N(β)β=x+jy
x, y ≥0
Z[j] 1
N(a)N(b) = |a|2|b|2=|ab|2=N(ab)
N(a)N(b)N(ab)
N(j) = |j|2= 1
α, jα, j2α, −α, −jα, −j2α β
u=a+bj N(u) = 1 1 = N(a+bj) = (a+bj)(a+j2b)
u|1u u v uv = 1
1 = N(u)N(v)N(u)=1
p3
p⇔p≡1[3].
f=X2+X+ 1 K:= Q[X]
<f>
K
Z[j]Z[i]Z[j]
p∈Z Z[j]

pZ[j]
f f
p deg(f) = 2 f
(disc(f)
p) = 1
disc(f) = 12−4·1 = −3 (disc(f)
p)=(−1 p)(3 p) = (−1 p)(−1)p−1
2(p
3)
(−1
p)=(−1)p−1
2(disc(f)
p)=(p 3)
1p1 3 p
Z[j] 1 3
p∈Z Z[j]p=αβ
α, β ∈Z[j]
p2=N(α)N(β)α β N(α), N(β)>1
N(a) = N(b) = p p
p a, b ∈ZN(a+bj) = p
(a+bj)·(a+bj2) = p N (a+bj) = N(a+bj2) = p > 1a+bj
a+bj2p p
Z[j]
n
p2 3 valp(n)
n2 3
n= 3val3(n)n0m2m∈Nn0
p≡1[3]
n0m2=m2+m·0 + 02
m23 = 1 + 1 + 1 n
q≡2[3] n
2 3 n=N(a+bj)q
n q ≡2[3] q
n0=n
qn0q= (a+bj)(a+j2b)a, b ∈Z
qZ[j]q
a+bj a +bj2j2j
q|(a+bj)x, y ∈Zq·(x+jy) = a+bj
q2N(x+jy) = n n1
n
2 3 n1
n

n3|n3-σ(n)−n x2+xy +y2
x, y ∈N
n3|n3-σ(n)n= 3vΠk
i=1pti
iπl
j=1qsj
j
n σ(n)6≡ 0[3] i
σ(pti
i)6≡ 0[3] j σ(qsj j)6≡ 0[3] i≤k σ(pti i) =
1 + pi+. . . +pti
i≡1 + . . . + 1[3] ≡ti+ 1[3] σ(pti i)≡0[3]
ti≡2[3] j≤l σ(qsj
j) = 1+qj+. . .+qsj j≡1−1+1−1+. . .+(−1)sj[3]
sj≡1[2]
3-σ(n)sjn
2. n n =x2+xy +y2
x, y ∈N