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From: M. M i c h a e l M u s a t o v on 27 Mar 2010 08:34
Np(f) for f = xn 􀀀 x 􀀀 1, n = 2; 3; 4; 5:
p n = 2 n = 3 n = 4 n = 5
2 0 0 0 0
3 0 0 0 0
5 1 1 0 0
7 0 1 1 0
11 2 1 1 0
13 0 0 1 0
17 0 1 2 2
19 2 1 0 1
23 0 2 1 1
              
59 2 3 1 0
              
83 0 1 4 0
5.2. The case n = 2. The discriminant of f = x2 􀀀 x 􀀀 1 is 5; the
polynomial f
has a double root mod 5; hence N5(f) = 1. For p 6= 5, we have
Np(f) =
(
2 if p  1 (mod5)
0 if p  2 (mod5):
If one de nes a power series F(q) =
P1
m=0 amqm by
F = q 􀀀 q2 􀀀 q3 + q4
1 􀀀 q5 = q 􀀀 q2 􀀀 q3 + q4 + q6 􀀀 q7 􀀀 q8 + q9 +    ;
the above formula can be restated as
Np(f) = ap + 1 for all primes p:
Note that the coecients of F are strongly multiplicative: one has
amm0 = amam0
for every m;m0  1. The corresponding Dirichlet series
P1
m=1 amm􀀀s is the
L-series
Q
p(1 􀀀 ( p
5 )p􀀀s)􀀀1.
434 JEAN-PIERRE SERRE
5.3. The case n = 3. The discriminant of f = x3 􀀀 x 􀀀 1 is 􀀀23; the
polynomial
f has a double root and a simple root mod 23; hence N23(f) = 2. For p
6= 23, one
has:
Np(f) =
(
0 or 3 if ( p
23) = 1
1 if(p
23) = 􀀀1:
Moreover, in the ambiguous case where ( p
23) = 1, p can be written either as x2 +
xy + 6y2 or as 2x2 + xy + 3y2 with x; y 2 Z; in the rst case, one has
Np(f) = 3;
in the second case, one has Np(f) = 0. (The smallest p of the form x2
+ xy + 6y2
is 59 = 52 + 5  2+ 6  22, hence N59(f) = 3; cf. table above.)
Let us de ne a power series F =
P1
m=0 amqm by the formula
F = q
1Y
k=1
(1 􀀀 qk)(1 􀀀 q23k)
=
1
2
 X
x;y2Z
qx2+xy+6y2 􀀀
X
x;y2Z
q2x2+xy+3y2

= q 􀀀 q2 􀀀 q3 + q6 + q8 􀀀 q13 􀀀 q16 + q23 􀀀 q24 +    :
The formula for Np(f) given above can be reformulated as:
Np(f) = ap + 1 for all primes p:
Note that the coecients of F are multiplicative: one has amm0 = amam0
if m and
m0 are relatively prime. The q-series F is a newform of weight 1 and
level 23. The
associated Dirichlet series is
1X
m=1
am
ms =
Y
p

1 􀀀 ap
ps +
 p
23
 1
p2s
􀀀1
:
5.4. The case n = 4. The discriminant of f = x4􀀀x􀀀1 is 􀀀283. The
polynomial
f has two simple roots and one double root mod 283, hence N283(f) = 3.
If p 6= 283,
one has
Np(f) =
8><
>:
0 or 4 if p can be written as x2 + xy + 71y2
1 ifp can be written as 7x2 + 5xy + 11y2
0 or 2 if
􀀀 p
283

= 􀀀1:
(These cases correspond to the Frobenius substitution of p being
conjugate in S4
to (12)(34) or 1, (123), (1234) or (12) respectively.)
A complete determination of Np(f) P can be obtained via a newform F =
1
m=0 amqm of weight 1 and level 283 given in [5, p. 80, example 2]:
F =q+
p
􀀀2q2 􀀀
p
􀀀2q3 􀀀q4 􀀀
p
􀀀2q5 +2q6 􀀀q7􀀀q9 +2q10 +q11 +
p
􀀀2q12 +   :
One has:
Np(f) = 1+(ap)2 􀀀
 p
283

for all primes p 6= 283:
I do not know any closed formula for F, but one can give one for its
reduction
mod 283; see Notes. This is more than enough to determine the integers
Np(f),
since they are equal to 0; 1;2 or 4.
ON A THEOREM OF JORDAN 435
5.5. The case n  5. Here the only known result seems to be that f =
xn 􀀀x􀀀1
is irreducible (Selmer [15]) and that its Galois group is the
symmetric group Sn.
No explicit connection with modular forms (or modular representations)
is known,
although some must exist because of the Langlands program.
Notes
1.1. Here is another interpretation of c0(f). Let K = Q[x]=(f) be the
number
eld de ned by f. We have [K : Q] = n  2. For every d  1, let ad(K)
be the number of the ideals a of the ring of integers of K with N(a) =
d.
The zeta function of K is the Dirichlet series
K(s) =
X
d1
ad(K)
ds :
Using standard recipes in analytic number theory, one can show that
Theorem
1 is equivalent to saying that K is lacunary: most of its coecients
are zero. More precisely, if we denote by NK(X) the number of d  X
with
ad(K) 6= 0, one has
NK(X)  cK
X
(logX)c0(f) for X !1;
where cK is a strictly positive constant (cf. Odoni [13] and Serre
[16, x3.5]).
As for Theorem 2, it can be reformulated as
NK(X) = O

X
(logX)1=n

for X !1;
with \O" replaced by \o" if n is not a power of a prime.
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