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arXiv:2411.04213v1 [math.NT] 6 Nov 2024
CYCLOTOMIC PRIMES
CARL POMERANCE
Abstract. Mersenne primes and Fermat primes may be thought
of as primes of the form Φm(2), where Φm(x) is the mth cyclotomic
polynomial. This paper discusses the more general problem of
primes and composites of this form.
1. Introduction
Studied since antiquity, we have the Mersenne primes. These are
prime numbers 1 less than a power of 2, so of the form 2n−1. To
be prime it is necessary that n=pis prime, but this is not sufficient,
e.g., p= 11. The first 4 of these primes were known to Euclid and
they played a key role in his work on perfect numbers. We now know
more than 50 Mersenne primes, the largest at present being 2p−1 with
p= 136,279,841, see [9]. Evidently it takes some doing to check the
primality of numbers this large!
It is widely believed that there are infinitely many Mersenne primes,
and also infinitely many primes pwith 2p−1 composite. Though both
assertions are still unsolved, there is a conditional proof of the second
one based on the prime k-tuples hypothesis: If p≡3 (mod 4) is prime
with p > 3 and q= 2p+ 1 is prime, then 2p−1 is composite. Indeed,
the conditions imply that q≡7 (mod 8) so that (2/q) = 1. This
implies that q|2(q−1)/2−1 = 2p−1, and the condition p > 3 implies
that q < 2p−1. Thus qis a proper divisor of 2p−1 implying the
latter is composite. For example, 23 |211 −1. It remains to note that
the prime k-tuples hypothesis implies there are infinitely many primes
p≡3 (mod 4) with 2p+ 1 prime.
Also studied for centuries are the Fermat primes. These are primes
that are 1 more than a power of 2, so of the form 2n+ 1. To be prime
(and >2) it is necessary that nitself is a power of 2. Again, this is not
sufficient. Fermat knew that 22k+ 1 is prime for k= 0,1,2,3,4 and
Mathematics Department, Dartmouth College, Hanover, NH 03755, USA. email:
carlp@math.dartmouth.edu
Date: November 8, 2024.
2010 Mathematics Subject Classification. 11N32, 11N25.
Key words and phrases. Mersenne prime, Fermat prime, cyclotomic polynomial,
abc conjecture.
1
2 CARL POMERANCE
he conjectured that it is always prime. However, Euler showed that
641 |225+ 1. It is now known that 22k+ 1 is composite for all larger
values of kup to 32, and also some sporadic larger values as well. It is
conjectured that all but finitely many are composite and that perhaps
224+ 1 is the largest Fermat prime. Nothing has been proved here,
even conditionally.
What Mersenne primes and Fermat primes have in common is that
they are cyclotomic primes. These are primes of the form Φm(2), where
Φmis the mth cyclotomic polynomial. This is the minimal polynomial
in Q[x] for e2πi/m and it has degree ϕ(m), Euler’s function. We have
the twin identities:
xm−1 = Y
d|m
Φd(x),Φm(x) = Y
d|m
(xd−1)µ(m/d).
Note that if pis prime, then Φp(x) = (xp−1)/(x−1), so that Φp(2) =
2p−1. Further Φ2k+1(x) = (x2k+1 −1)/(x2k−1) = x2k+ 1, so that
Φ2k+1 (2) = 22k+ 1. We also have the Wagstaff primes, see [16], which
are primes of the form Φ2p(2) = (2p+ 1)/3, where pis an odd prime.
So, a cyclotomic prime is a prime of the form Φm(2). We can ask if
there are infinitely many of them and also if there are infinitely many
numbers of this form that are composite. It turns out that there are
infinitely many composites for fairly trivial reasons. The substance of
this paper is to show that there are infinitely many nontrivial compos-
ites. We make this precise in the next section.
2. Basics and statement of results
Let
φm:= Φm(2).
We say a prime factor pof φmis primitive if it does not divide any φk
for k < m. Otherwise we say pis intrinsic. For an odd prime plet ℓ(p)
denote the mutlplicative order of 2 in (Z/pZ)×. We have that pis a
primitive prime factor of φmif and only if ℓ(p) = m. Further, φmhas
an intrinsic prime factor pif and only if m=pjℓ(p) for some positive
integer j, in which case pis the largest prime factor of mand pkφm.
If mis of this form, let δm=p, and otherwise let δm= 1. Thus, every
prime factor of
ψm:= φm/δm
is primitive.
We know that for each m /∈ {1,6}, there is at least one primitive
prime factor of φm; this is Bang’s theorem. The numbers ψmare pair-
wise coprime and except for m= 1 or 6, they are all >1 (cf. [18]).
CYCLOTOMIC PRIMES 3
Due to the factorization
4x4+ 1 = (2x2+ 2x+ 1)(2x2−2x+ 1),
there is a further generic factorization of φmbeyond δmψmwhen m≡4
(mod 8). Note that
24k+2 + 1 = 4(2k)4+ 1 = (22k+1 + 2k+1 + 1)(22k+1 −2k+1 + 1),
which leads to the factorization
φ8k+4 = gcd(φ8k+4,22k+1 + 2k+1 + 1) gcd(φ8k+4,22k+1 −2k+1 + 1)(1)
=: φ+
8k+4φ−
8k+4.
By dividing out an intrinsic prime factor if it exists, we have
ψ8k+4 =: ψ+
8k+4ψ−
8k+4.
Further, this factorization is nontrivial for k≥3, a result due to
Schinzel [19]. The factorization (1) is known under the name Au-
rifeuille, see [3].
Note that
(2) φm∈[2ϕ(m)−1,2ϕ(m)+1),
see [10, Theorem 3.6], [18, Theorem 4.3]. Also, using [3, eqs. (13), (14)]
it is not hard to show that
(3) φ+
8k+4, φ−
8k+4 ≍2ϕ(8k+4)/2,
where the notation indicates the 2 items on the left side are of the same
magnitude as the item on the right side.
To state our results, consider the sets
C1={ψm:m6≡ 4 (mod 8)}, C2={ψm:m≡4 (mod 8)}.
Theorem 1. For sufficiently large values of x, the set C1contains more
than x3/5composite numbers ψmwith m≤xand the set C2contains
more than x3/5numbers ψmwith m≤xthat are not the product of two
primes.
The exponent 3/5 in the theorem is not optimal, this is discussed
below. The proof uses the deep result that for some θwith 1/2< θ < 1,
there are infinitely many primes psuch that p−1 has a large prime
factor q > pθ. We can also prove a slightly stronger result conditional
on the abc conjecture.
Theorem 2. Assume the abc conjecture. The set C1contains infinitely
many numbers divisible by at least 2distinct primes and the set C2
contains infinitely many numbers divisible by at least 3distinct primes.
4 CARL POMERANCE
We remark that the abc conjecture has been used for similar purposes
in [20] and [2, Theorem 3]. We also remark that this theorem gives
an abc-conjecture-conditional solution of a problem of Schinzel [19, p.
561].
Throughout the letters p, q will always denote prime numbers. We
also let P+(n) denote the largest prime factor of n > 1, and we let
P+(1) = 1.
3. An elementary approach
Here we prove a somewhat weaker version of Theorem 1 where the
exponent 3/5 is replaced with 1/2. Let xbe large and consider primes
p≤x. It follows from Erd˝os–Murty [6, Theorem 1] that the number of
such primes pwith ℓ(p)>5x1/2(log x)2is ∼x/ log x. For any positive
integer dlet Ld(x) denote the set of primes p∈(x/2, x] with
p≡3 (mod 4),
ℓ(p) = (p−1)/d,
ℓ(p)>5x1/2(log x)2.
It follows that
(4) X
d≤1
5x1/2/(log x)2
#Ld(x)∼x
4 log x.
Thus, for large xthere is some number d0≤1
5x1/2/(log x)2with
#Ld0(x)> x1/2.
Consider now the values of m=ℓ(p) = (p−1)/d0for p∈Ld0(x). They
are all distinct, bounded by x, and 6≡ 4 (mod 8). For large x,ψmis
easily seen to be > x (using m > x1/2and (2)) and p|ψm. It follows
that ψmis composite. Thus, C1contains more than x1/2composite
numbers ψmwith m≤x.
By changing 3 (mod 4) above to 5 (mod 8) and using (3), the anal-
ogous argument shows that at least one of ψ+
m, ψ−
mis composite, so C2
contains more than x1/2numbers ψmwith m≤xwhich are not the
product of two primes.
4. Proof of Theorem 1
Denote by π(x;d, a) the number of primes p≤xwith p≡a(mod d).
Our principal tool is the following theorem. Let θ= 3/5.
CYCLOTOMIC PRIMES 5
Proposition 1. We have
X
q>xθ
π(x; 4q, 2q+ 1) log q≫xand X
q>xθ
π(x; 8q, 4q+ 1) log q≫x.
The analogous result for π(x;q, 1) is well known with varying values
of “θ” in the literature. The current champions are Baker and Harman
[1], who essentially have θ= 0.677, though they do not state their
result in the same way. Probably the techniques of their paper would
allow the same value of θin Proposition 1, but we do not pursue the
optimal value at this point. Other results in their paper have been
recently strengthened (see [13]); conjecturally any value of θ < 1 may
be used.
We now sketch a proof of Proposition 1. With Λ the von Mangoldt
function, we have
X
d≤x
π(x; 4d, 2d+ 1)Λ(d) = X
p≡3 (mod 4)
p≤x
X
d|(p−1)/2
Λ(d) + O(x/ log x)
=X
p≡3 (mod 4)
p≤x
log(p−1) + O(x/ log x)
=1
2x+O(x/ log x).
Further, by the Bombieri–Vinogradov theorem plus a small additional
argument using the Brun–Titchmarsh inequality (see [15]) to clean up
the boundary cases, we have
X
d≤x1/2
π(x; 4d, 2d+ 1)Λ(d)∼1
4x, x → ∞.
Thus,
X
d>x1/2
π(x; 4d, 2d+ 1)Λ(d)∼1
4x, x → ∞.
The contribution to this last sum when dis composite is o(x), so we
have
X
q>x1/2
π(x; 4q, 2q+ 1) log q∼1
4x, x → ∞.
By the Brun–Titchmarsh inequality,
X
x1/2<q≤xθ
π(x; 4q, 2q+ 1) log q≤X
x1/2<q≤xθ
2xlog q
ϕ(4q) log(x/4q)
∼xlog(5/4) <0.23x.
6 CARL POMERANCE
Thus, with the prior display, we have
X
q>xθ
π(x; 4q, 2q+ 1) log q≫x,
which shows the first assertion in Proposition 1. The second assertion
follows in a similar manner.
To achieve a cosmetically more appealing version of our results, note
that since
X
xθ<q≤xθ(log x)2
1
q≪log log x
log x=o(1),
the Brun–Titchmarsh theorem implies that in both parts of Proposi-
tion 1 we may replace q > xθwith q > xθ(log x)2.
Since no prime p≤xis divisible by 2 different primes q > xθ, we
have the following result.
Corollary 1. We have
X
p≡3 (mod 4)
P+(p−1)>xθ(log x)2
p≤x
1≫x/ log xand X
p≡5 (mod 8)
P+(p−1)>xθ(log x)2
p≤x
1≫x/ log x.
An elementary argument shows that the number of primes pwith
ℓ(p) = kis ≪k/ log k, so it follows that the number of primes p≤x
with ℓ(p)≤x1−θ/(log x)2is ≪x2(1−θ)=o(π(x)). (Note that there is a
similar argument in [8]. Also here one could appeal to [6, Theorem 1].)
However, a prime pcounted in either part of Corollary 1 either has
ℓ(p)≤x1−θ/(log x)2or ℓ(p)> xθ(log x)2. Hence we have
X
p≡3 (mod 4)
ℓ(p)>xθ(log x)2
p≤x
1≫x/ log xand X
p≡5 (mod 8)
ℓ(p)>xθ(log x)2
p≤x
1≫x/ log x.
Thus, using the notation of the previous section we have the improve-
ment on (4):
X
d≤x1−θ/(log x)2
#Ld(x)≫x
log x,
and an analogous result holds for primes p≡5 (mod 8). Thus, by the
same argument as in the previous section, we have Theorem 1.
5. Conditional results
For a positive integer nlet rad(n) denote the largest squarefree di-
visor of n. The abc conjecture asserts that for each fixed ǫ > 0, there
are at most finitely many coprime positive integer triples a, b, c with
CYCLOTOMIC PRIMES 7
a+b=cand rad(abc)< c1−ǫ. In this section we will prove Theorem 2,
which is conditional on the abc conjecture, and also discuss some other
conditional results.
Proof of Theorem 2. First suppose that m6≡ 4 (mod 8). We know
from Theorem 1 that there are infinitely many such mwith ψmcom-
posite, and in fact, there are ≥xθsuch m≤xwhen xis large. Further,
each such mis of the form kqqwhere q > xθand ψkqqis divisible by
a prime p=pq≤xwith ℓ(p) = kqq. The only way for a composite
number not to be divisible by at least 2 distinct primes is if it is a prime
power, namely piwhere i≥2, so suppose that ψkqq=pi. Consider the
abc equation 1 + (2kqq−1) = 2kqq. Since ψkqq≥φkqq/q, we have
rad(abc)≤2pq(2kqq−1)/φkqq.
Assuming the abc conjecture this would be impossible for large qif
there is some fixed ǫ > 0 such that φkqq>2ǫkqq(since 2pq =O(x2) =
2o(kqq)). Using (2) this would follow if ϕ(kq)>2ǫkq. We now show that
we may assume this is indeed the case.
Recalling that ℓ(p) = kqq, write j= (p−1)/q, so that kq|j. And
so if ϕ(kq)/kq≤2ǫ, then ϕ(j)/j ≤2ǫ. For a given value of j≤x1−θ
we consider primes q≤x/j such that jq + 1 is prime. By Brun’s or
Selberg’s sieve, the number of such qis ≪x/(ϕ(j)(log x)2). In the
proof of Proposition 1 we showed there are ≥(.02 + o(1))x/ log xpairs
q, p. Let
J={j≤x1−θ:ϕ(j)/j ≤2ǫ}.
We will show that Pj∈J1/ϕ(j)≪ǫlog xwith ǫsmall enough this will
be negligible in comparison with (.02 + o(1)) log x. This would follow
from Erd˝os [5, Theorem 1] (also see [12, Theorem B]), but we prefer to
use the simpler approach in [11, Section 3].
We have (j/ϕ(j))2=Pd|jh(d), where his multiplicative, supported
on the squarefrees, and has h(p) = (2p−1)/(p−1)2. Then
X
n≤z
n
ϕ(n)2=X
d≤z
h(d)jz
dk< z Y
p1 + h(p)
p<4.5z.
Thus, for any δ > 0,
X
n≤z
ϕ(n)/n≤δ
1<4.5δ2z.
A partial summation argument then shows that
X
n≤z
ϕ(n)/n≤δ
1
n<4.5δ2log(ez),
8 CARL POMERANCE
so that writing 1/ϕ(n) = 1/n ·n/ϕ(n),
X
n≤z
1
2δ<ϕ(n)/n≤δ
1
ϕ(n)<2
δ4.5δ2log(ez) = 9δlog(ez).
We apply this with z=x1−θand at δ= 2ǫ, ǫ, 1
2ǫ, . . . getting
X
j∈J
1
ϕ(j)<36ǫlog(ex).
Thus if ǫ=ǫ0is small enough, we would have the number of q, p
pairs with ϕ(j)/j ≤ǫ0being < .01x/ log x. But we have seen in the
previous section that the total number of q, p pairs generated is ≥
(.02 + o(1))x/ log x. Thus, we can discard those with ϕ(j)/j ≤ǫ0and
still be left with ≫x/ log xpairs. So, we will have ϕ(j)/j > ǫ0, which
implies that ϕ(kq)/kq> ǫ0. Thus, the abc conjecture is in play to show
that the equation 1 + (2kqq−1) = 2kqqwith ψkqqa power of pcannot
occur when xis large.
The situation for kq ≡4 (mod 8) is completely analogous; we sup-
press the details.
By being more careful with the estimates one can get counts in The-
orem 2 like the ones we found in Theorem 1. We further remark that a
variant of our proof can show that for asymptotically all m,ψmis not
square-full, and the same goes for ψ+
8k+4 and ψ−
8k+4.
We mentioned in the introduction that the prime k-tuples conjecture
can be used to show that there are infinitely many primes pwith ψp=
2p−1 composite. We add here a couple of thoughts. First, since we
know that 8 and 9 form the only pair of consecutive numbers which
are nontrivial powers (a result of Mih˘ailescu [14]), it follows that 2p−1
cannot be a nontrivial power, so in this case, the abc conjecture is not
necessary. Second, using the Hardy–Littlewood version of the k-tuples
conjecture, we have the number of primes p≤xwith 2p−1 divisible
by at least 2 different primes is ≫x/(log x)2.
We can prove there are more cyclotomic composites assuming Artin’s
primitive root conjecture. If 2 is a primitive root for p, we have pa
prime factor of ψp−1and so ψp−1is composite for plarge. By Hoo-
ley’s GRH conditional proof of Artin’s conjecture, we have ≫x/ log x
primes p≤xwhich have 2 as a primitive root. Further, the proof
is amenable to insisting that p≡3 (mod 4) and also the same holds
when p≡5 (mod 8). So the GRH implies there are quite a few cy-
clotomic composites. And as above, the abc conjecture can be used to
show these composites are usually not prime powers. This result can be
CYCLOTOMIC PRIMES 9
improved a little by considering primes p≤kx with ℓ(p) = (p−1)/k
for various small values of kand using sieve methods to show that
(p−1)/k = (p′−1)/k′has few solutions when k6=k′are small. Thus,
with a little work it may be possible to show, assuming the GRH, that
there are ≫xlog log x/ log xintegers l≤xof the form ℓ(p) for some
prime p≪xlog x.
6. Statistics and surmises
Concerning Table 1, Gallot [7] previously enumerated the cases where
φmis prime for m≤6500. Our calculations agree with his. In our work
we used Mathematica and in particular their PrimeQ function, which
we understand is a probabilistic test for primality. So, it is possible that
some of the prime declarations made are false, but this seems unlikely,
given that there are not very many of them. One of the larger primes
unearthed here is φ60,287 which has 17,090 decimal digits. Note that
when PrimeQ declares a number is not prime, this conclusion is not
in doubt. Since PrimeQ is notably slower than checking if the Fermat
congruence 3n≡3 (mod n) holds, we first used that and confirmed the
few primality assertions with PrimeQ. (We used the base 3 since every
ψmis either a prime or a base 2 pseudoprime. See [17] where these
thoughts are developed.) Many of the large primes uncovered here
have indeed been certified (including φ60,287) in the ongoing project
factordb.com. (Thanks are due to Yves Gallot for informing me about
this.)
Heuristically there are at most finitely many examples where ψpiℓ(p)
is prime. Is ψ127·7the largest such example? It is a prime of 226 decimal
digits. There are several examples where both ψ+
mand ψ−
mare prime.
The largest that we found in our calculations to 216 is m= 1132, where
the two primes each have 85 decimal digits. Probably there are at most
finitely many of these “twin cyclotomic primes”.
The counts in Table 1 look to be proportional to k2, and this is
supported heuristically as well. Indeed, one can model ψmas a random
number near 2ϕ(m)which has all prime factors larger than m. So the
“probability” that it is prime (given that m6≡ 4 (mod 8)) is about
eγlog m/ϕ(m) log 2. The sum of these quantities up to 215 is about
223.4, and up to 216 is about 254.4, which are not bad matches with
the table.
One can enlarge further the realm of cyclotomic primes to look at
the primitive parts of an−1, where a > 2. Also one can look at the
Fibonacci sequence, as well as other Lucas sequences, for example see
10 CARL POMERANCE
Table 1. Counts for m≤2kwith φmprime, ψmprime,
ψ+
mprime, ψ−
mprime
k#mwith φmprime ψmprime ψ+
mprime ψ−
mprime
1 1 1 0 0
2 3 3 1 0
3 7 6 1 0
4 14 13 2 0
5 23 25 4 1
6 33 36 7 5
7 49 52 13 8
8 64 68 20 16
9 81 86 24 25
10 99 106 30 33
11 122 129 34 43
12 140 147 44 54
13 167 174 50 59
14 195 202 61 64
15 221 228 72 74
16 255 262 85 83
Drobot [4]. We suspect our methods carry over, but we leave this topic
for another day, and perhaps another person.
Acknowledgments
I thank Michael Filaseta, Yves Gallot, Florian Luca, Mits Kobayashi,
Pieter Moree, Paul Pollack, and Sam Wagstaff for their valuable com-
ments and interest.
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CYCLOTOMIC PRIMES 11
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Appendix
Here we list the values of mcorresponding to the counts in Table 1.
Values of mwith φmprime:
2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 19, 22, 24, 26, 27, 30, 31,
32, 33, 34, 38, 40, 42, 46, 49, 56, 61, 62, 65, 69, 77, 78, 80, 85, 86, 89, 90,
93, 98, 107, 120, 122, 126, 127, 129, 133, 145, 150, 158, 165, 170, 174,
184, 192, 195, 202, 208, 234, 254, 261, 280, 296, 312, 322, 334, 345, 366,
374, 382, 398, 410, 414, 425, 447, 471, 507, 521, 550, 567, 579, 590, 600,
607, 626, 690, 694, 712, 745, 795, 816, 897, 909, 954, 990, 1106, 1192,
1224, 1230, 1279, 1384, 1386, 1402, 1464, 1512, 1554, 1562, 1600, 1670,
1683, 1727, 1781, 1834, 1904, 1990, 1992, 2008, 2037, 2203, 2281, 2298,
2353, 2406, 2456, 2499, 2536, 2838, 3006, 3074, 3217, 3415, 3418, 3481,
3766, 3817, 3927, 8370, 9583, 9689, 9822, 9941, 10192, 10967, 11080,
11213, 11226, 11581, 11614, 11682, 11742, 11766, 12231, 12365, 12450,
12561, 13045, 13489, 14166, 14263, 14952, 14971, 15400, 15782, 15998,
16941, 17088, 17917, 18046, 19600, 19937, 20214, 20678, 21002, 21382,
21701, 22245, 22327, 22558, 23209, 23318, 23605, 23770, 24222, 24782,
27797, 28958, 28973, 29256, 31656, 31923, 33816, 34585, 35565, 35737,
12 CARL POMERANCE
36960, 39710, 40411, 40520, 42679, 42991, 43830, 43848, 44497, 45882,
46203, 47435, 48387, 48617, 49312, 49962, 49986, 50414, 51603, 51945,
53977, 55495, 56166, 56898, 56955, 57177, 58315, 58534, 58882, 60287
Values of mwith ψm< φmand ψmprime:
18, 20, 21, 54, 147, 342, 602, 889
Values of mwith ψ+
mprime:
4, 12, 20, 28, 36, 44, 60, 68, 76, 84, 100, 108, 116, 132, 140, 180, 204,
220, 228, 252, 276, 340, 356, 484, 588, 628, 652, 700, 756, 924, 1132,
1292, 1452, 1516, 2300, 2484, 2604, 2964, 3116, 3276, 3420, 3540, 3940,
3988, 4892, 5100, 5268, 5908, 6620, 7812, 8964, 9084, 9324, 9468, 10308,
11980, 12188, 12204, 13724, 13860, 15252, 17052, 18476, 20676, 21916,
24252, 25004, 25508, 28692, 29460, 29492, 31692, 34236, 34380, 35700,
38428, 40564, 41316, 45028, 46076, 50332, 51148, 51204, 56588, 58796
Values of mwith ψ−
mprime:
28, 36, 44, 52, 60, 84, 108, 116, 132, 140, 172, 188, 196, 212, 220, 252,
260, 276, 292, 316, 348, 372, 420, 444, 452, 516, 604, 668, 812, 868, 924,
956, 964, 1044, 1132, 1204, 1276, 1412, 1468, 1500, 1540, 1564, 1828,
2124, 2172, 2228, 2252, 2452, 2532, 2716, 2764, 2868, 3484, 3852, 4844,
5316, 5468, 6164, 7828, 9516, 9684, 10924, 12164, 15860, 19516, 20588,
21292, 24180, 25100, 25212, 28612, 30988, 31460, 32340, 34404, 38132,
42660, 43084, 46292, 46980, 52740, 56668, 60676
Mathematics Department, Dartmouth College, Hanover, NH 03755,
USA
Email address:carlp@math.dartmouth.edu
