Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
Using dynamical systems to prove that there
are infinitely many primes
Abstract. In this article we discuss proofs that there are infinitely many primes. We begin
with a guided survey of the many known proofs (including a few new minor variations). Then
we see how one of these ideas has been pushed in surprising new directions, involving some
beautiful themes from arithmetic dynamics.
1. A WIDE VARIETY OF PROOFS. There are many different proofs in the literature that there are infinitely many primes. Most of the proofs use the theorem that
Every integer q > 1 has a prime factor.
(This was first formally proved by Euclid on the way to establishing the Fundamental
Theorem of Arithmetic.) The idea, given any finite list P of primes, is to find an integer
q > 1 that is not divisible by any of the primes in P , so the prime factors of q are not
on the list P , and therefore P could not have been the complete list of primes to begin
with. This implies that no finite list contains all of the primes, and therefore there are
infinitely many primes.
Non-constructive proofs, developing Euclid’s idea. One can determine such an integer q , by establishing that
For any given integer N 6= 0 one can find
an integer q > 1 for which gcd(N, q) = 1.
If N = N (P), the product of all of the primes in P , then any such integer q has at
least one prime factor that does not belong to the list P . Euclid took N = N (P) and
q = N + 1, though there are many other possibilities for N and q :
• For any integer N > 2 that is divisible by N (P), we could take q = N + 1 or
q = N − 1. For example, q = 2N (P) − 1 or 2N (P) + 1, or perhaps N = n! where
n is the largest prime in P .1
• Partition P into two non-empty sets P = M ∪ N . Let m be the product of the
elements of M, and let n be the product of the elements of N , so that N = N (P) =
mn. Let q = m + n. We claim that gcd(q, N ) = 1, for if not some prime p divides
both q and N , so that p has to divide either m or n (as their product equals N ). Now
if p divides m then p divides q − m = n, and vice-versa, but m and n were selected
so as to have no common prime factor, a contradiction.
(In Euclid’s proof, one of M and N is the empty set.) See also [22].
A famous example comes from the number of home runs hit by Hank Aaron when
he overtook Babe Ruth’s record: 715 = 5 × 11 × 13 versus 714 = 2 × 3 × 7 × 17
1 [16] beautifully reformulates this idea: Let N = N (P) and p be a prime divisor of 2N + 1. If P contains
)π
all of the primes, then p divides N , so that 2N/p is an even integer. Therefore 0 = sin( (1+2N
) = sin( πp ),
p
π
π
which is impossible as p > 1, so that 0 < p < π and therefore sin( p ) > 0.
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
1
page 1
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
which together give the product of all the primes up to 17. Note that 714 + 715 =
1429 is prime.
• Modifying the last construction, one might instead take q = m − n or some other
linear combination am + bn in which the prime factors of a divide m and the prime
factors of b divides n. This works provided q 6= −1 or 1.
• For each prime p ∈ P , we let Np = N (P \ {p}) = N
P(P)/p, and let ap be any
integer whose prime factors all divide Np . Then take q = p∈P ap Np and again we
need to be careful that q 6= −1 or 1. There is an advantage to this construction: If P is
the set of all primes ≤ x, and if 1 < |q| ≤ x2 then q must itself be prime, and one can
show (see [1]) that every prime between x and x2 can be represented in this way.
• If f (x) is a polynomial with integer coefficients then p is a prime divisor of f
if there exists an integer m for which f (m) is divisible by p. We will now show that
any polynomial with integer coefficients, that is not of the form cxr , has infinitely
many distinct prime divisors: Suppose that our polynomial is xr f (x) where f (x) =
Pd
j
j=0 aj x is a polynomial with integer coefficients, for which a0 , ad 6= 0 and d ≥ 1.
Given a finite list P of primes for which N = N (P) > 2, we have
f (a0 N x) = a0 (1 + N xg(x)) where g(x) :=
d−1
X
ak+1 (a0 N )k xk .
k=0
If m is an integer for which mg(m) 6= 0 then |1 + N mg(m)| ≥ N − 1 > 1 and
so is divisible by some prime p, which evidently does not divide N . Therefore p is a
prime that is not in the list P , and divides f (a0 N m). See [8] for much more about
prime divisors of polynomials.
• We do not actually need to construct q to show that such an integer q exists: Euler
defined the totient function
φ(N ) := #{1 ≤ m ≤ N : gcd(m, N ) = 1},
and proved that
φ(N ) =
Y
pk−1 (p − 1)
pk kN
where the product is over all prime powers pk > 1 that divide N , such that pk+1 does
not divide N . This formula should be used by the reader to show that if N 6= 1, 2, 3, 4
or 6 then φ(N ) > 2, which implies that there exists an integer q in the range 1 < q <
N − 1 for which gcd(N, q) = 1.
• This gives the key idea behind the most beautiful of all proofs that there are
infinitely many primes, Furstenberg’s extraordinary proof using point set topology:
Define a topology on the set of integers Z in which a set S is open if it is empty or if for every
a ∈ S there is an arithmetic progression
Z(a, m) := {a + nm : n ∈ Z},
with m 6= 0, which is a subset of S. Evidently each Z(a, m) is open, and it is also closed since
[
Z(a, m) = Z \
Z(b, m).
b: 0≤b≤m−1, b6=a
2
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 2
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
If there are only finitely many primes p then A = ∪p Z(0, p) is also closed, and so Z \ A is
open. Now Z \ A is the set of integers without prime factors, and so Z \ A = {−1, 1}. However Z \ A = {−1, 1} is not open since {−1, 1} does not contain any arithmetic progression
Z(1, m). Hence there cannot be finitely many primes, so there are infinitely many primes.
These beautiful proofs only show that there cannot be finitely many primes and so
do not hint at how to construct infinitely many primes (other than just testing whether
each consecutive integer, 2, 3, 4, . . . is prime). In the next subsection we highlight
proofs that have the added bonus that they point us to where to look for our primes.
The construction of infinitely many primes. The idea is to
Construct an infinite sequence of distinct, pairwise coprime, integers a0 , a1 , . . .,
that is, a sequence for which gcd(am , an ) = 1 whenever m 6= n. We then obtain
infinitely many primes, prime factors of the an , as proved in the following lemma.
Proposition 1. Suppose that a0 , a1 , . . . is an infinite sequence of distinct, pairwise
coprime, integers. Let pn be a prime divisor of an whenever |an | > 1. Then the pn
form an infinite sequence of distinct primes.
Proof. The pn are distinct for if not then pm = pn for some m 6= n and so
pm = gcd(pm , pn ) divides gcd(am , an ) = 1,
a contradiction. As the an are distinct, any given value (in particular, −1, 0 and 1) can
be attained at most once, and therefore all but at most three of the an have absolute
value > 1, and so have a prime factor pn .
Here a few ways to construct such sequences:
• Modification of Euclid’s proof: Let a0 = 2, a1 = 3 and
an = a0 a1 · · · an−1 + 1 for each n ≥ 1.
If m < n then am divides a0 a1 . . . an−1 = an − 1 and so gcd(am , an ) divides
gcd(an − 1, an ) = 1, which implies that gcd(am , an ) = 1. Therefore if pn is a prime
divisor of an for each n ≥ 0, then p0 , p1 , . . . is an infinite sequence of distinct primes.
The recurrence for the an can be re-wriiten as
an+1 = a0 a1 · · · an−1 · an + 1 = (an − 1)an + 1 = f (an ),
where f (x) = x2 − x + 1.
n
• Fermat conjectured that the integers Fn = 22 + 1 are primes for all n ≥ 0.
His claim starts off correct: 3, 5, 17, 257, 65537 are all prime, but is false for F5 =
641 × 6700417, as Euler famously noted. It is an open question as to whether there
are infinitely many primes of the form Fn .2 Nonetheless we can prove that the Fn are
pairwise coprime, and so deduce that if pn is a prime divisor of Fn , then p0 , p1 , . . .
is an infinite sequence of distinct primes: Since the polynomial x + 1 divides x2L − 1
for any integer L ≥ 1, this also holds when we substitute in an integer in place of
2 The only Fermat numbers known to be primes, have n ≤ 4. We know that the F are composite for
n
5 ≤ n ≤ 30 and for many other n besides. It is always a significant moment when a Fermat number is factored
for the first time. It could be that all Fn , n > 4 are composite, or they might all be prime from some sufficiently
large n, onwards. Currently, we have no way of knowing what is the truth.
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
3
page 3
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
m
x. Taking x = 22 with L = 2n−1−m we deduce that Fm divides Fn − 2 and so
gcd(Fm , Fn ) = gcd(Fm , 2) = 1, as the Fermat numbers are obviously all odd.3
The Fn -values can be determined by a simple recurrence, as follows:
n
n
Fn+1 = (22 + 1)(22 − 1) + 2 = Fn (Fn − 2) + 1 = f (Fn ),
where f (x) = x2 − 2x + 2.
The (an )n≥0 and the (Fn )n≥0 are both examples of sequences (xn )n≥0 for which
xn+1 = f (xn )
for some polynomial f (x) ∈ Z[x]; the an with the polynomial x2 − x + 1, and the
Fn with the polynomial x2 − 2x + 2. The terms of the sequence are all given by the
recursive formula, so that
xn = f (f (. . . f (x0 ))) = fn (x0 )
where the notation fn means the polynomial obtained by composing f with itself n
times.
The key to the proof that the an ’s are pairwise coprime is that an ≡ 1 (mod am )
whenever n > m ≥ 0; and the key to the proof that the Fn ’s are pairwise coprime is
that Fn ≡ 2 (mod Fm ) whenever n > m ≥ 0. These congruences are not difficult
to prove using the polynomials that can define the sequences: Let f (x) = x2 − x + 1.
Then f (0) = 1 and f (1) = 1, so we deduce that f (f (. . . f (1))) = 1; that is, fn (0) =
1 for all n ≥ 1. To use this information we require the following lemma:
Lemma 1. Let g(x) be a polynomial with integer coefficients. If a, b and m are integers for which a ≡ b (mod m) then g(a) ≡ g(b) (mod m).
Proof. We first prove by induction that aj ≡ bj (mod m) for all integers j ≥ 1. It is
evidently true when j = 1, and then if it is true for j , we have aj+1 = aj · a ≡ bj · b =
Pd
bj+1 (mod m) using the induction hypothesis. Therefore if g(x) = g0 + j=1 gj xj
then
g(a) = g0 +
d
X
j=1
gj · aj ≡ g0 +
d
X
gj · bj = g(b)
(mod m),
j=1
as claimed.
Returning to iterations of the polynomial f (x) = x2 − x + 1: If n = m + k for
some k ≥ 1 then an = fk (fm (a0 )) = fk (am ) and so
an = fk (am ) ≡ fk (0) ≡ 1
(mod am ).
This implies that gcd(am , an ) divides gcd(am , 1) = 1, and so the an are pairwise
coprime.
This new proof also works for the Fermat numbers: Let f (x) = x2 − 2x + 2. Then
f (0) = 2 and f (2) = 2 and so fn (0) = 2 for all n ≥ 1. Therefore if n = m + k for
some k ≥ 1 then
Fn = fk (Fm ) ≡ fk (0) ≡ 2
(mod Fm ),
3 Goldbach was the first to use the prime divisors of the Fermat numbers to prove that there are infinitely
many primes, in a letter to Euler in late July 1730.
4
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 4
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
which implies that gcd(Fm , Fn ) divides gcd(Fm , 2) = 1, and so the Fn are pairwise
coprime.
This reformulation of two of the best-known proofs of the infinitude of primes
seems to open up a whole new vista of possibilities, which we will explore in the next
section. For now, though, we return to our survey of proofs that there are infinitely
many primes.
• Let a0 be a non-zero integers. Given a0 , . . . , an−1 , we define an by partitioning
the set {0, 1, . . . , n − 1} into two sets L ∪ M, and then letting
an =
Y
`∈L
a` +
Y
am .
(1)
m∈M
We claim that the aj are pairwise coprime, for if n is the smallest integer for which
there exists some k < n and
say k ∈ L,
Q some prime p which divides gcd(an , ak ), Q
then p divides both an and `∈L a` . Therefore p divides their difference, m∈M am ,
and so p divides some am with m ∈ M, which contradicts the assumption of n’s
minimality, as p divides gcd(am , ak ) where m 6= k < n.
In Somos and Haas [20], L ∪ M partitions the integers into the even and odd integers < n. Harris [12] begins with positive integers b0 , b1 , b2 with gcd(b0 , b2 ) = 1,
and then lets a0 = b0 , a1 = b0 b1 + 1, a2 = b0 b1 b2 + b0 + b2 . Thereafter, he uses
(1) to define the an , with M = {n − 2}. This ties in with continued fractions, for if
bn = a0 . . . an−3 for all n ≥ 3 then the continued fraction
[b0 , b1 , . . . , bn ] = an /dn for some integer dn , for every n ≥ 0.
Analytic proofs. The idea is to count the number of positive integers whose prime
factors only come from a given set of primes P = {p1 < p2 < . . . < pk }, up to some
large point x. These integers all take the form
e
pe11 pe22 · · · pkk for some integers ej , each ≥ 0.
(2)
• We are going to count the number of such integers up to x = 2m − 1, for an arbitrary
integer m ≥ 1, by studying this formula: As each pj ≥ 2, the smallest prime, we have
e
e
2ej ≤ pj j ≤ pe11 pe22 · · · pkk ≤ 2m − 1,
and so each ej ≤ m − 1, which implies that there are at most m possibilities for the
integer ej (namely the integers from 0 to m − 1). Therefore the number of integers of
the form (2), up to 2m − 1, equals
#{pe11 pe22
e
· · · pkk
m
≤ 2 − 1} ≤
k
Y
#{integers ej : 0 ≤ ej ≤ m − 1} = mk .
j=1
Now if P is the set of all primes then every positive integer is of the form (2), and
so the last equation implies that 2m − 1 ≤ mk for all integers m. We select m = 2k ,
so that 2k ≤ k 2 , which is false as k ≥ 5 (since we know the primes 2, 3, 5, 7, 11).
Therefore there cannot be only finitely many primes.
• We can modify the last proof to show that for every, not-too-fast-growing, infinite
sequence of integers, 1 ≤ a1 < a2 < . . ., there are infinitely many primes p which
each divide some aj (where j depends on p). By “not too fast growing”, we mean that,
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
5
page 5
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
for any fixed > 0 there exists infinitely many N for which aN ≤ 2N −1 . If not then
P := {p prime : p divides some aj } is finite. Taking m = [N ] for 0 < < k1 , so
that aN ≤ 2N −1 < 2m we have, by the above,
N = #{n : an ≤ aN } ≤ #{n : an ≤ 2m − 1} ≤ mk ≤ N k < N,
a contradiction.
e
• Euler noted that if the positive integer n = pe11 pe22 · · · pkk , then we may write
ek
e1
e2
1/n as 1/p1 · 1/p2 · · · 1/pk . Summing this over all integers n ≥ 1, we obtain
X
n≥1
n a positive integer
1
=
n a
1
e
. . . pkk
X
pe11 pe22
1 ,a2 ,...,ak ≥0
X 1
p e1
a ≥0 1
=
!
X 1
pe2
a ≥0 2
!
2
1


X 1

...
ek
p
k
a ≥0
k
k Y
1 −1
.
=
1−
pj
j=1
(3)
The last equality holds because each sum in the second-to-last line is over a geometric
progression. The right-hand side is a finite product of rational numbers, so is rational, whereas, as we will show, the left-hand side diverges to ∞, a contradiction, and
so there cannot be only finitely many primes. Now, to prove that the left-hand side
diverges, notice that
X
2k−1 ≤n<2k
n a positive integer
1
1
> k
n
2
X
1=
2k−1 ≤n<2k
n a positive integer
1
2k−1
=
k
2
2
as 1/n > 1/2k in this range for n. Therefore, partitioning the integers in [1, 2K ) into
intervals of the form [2k−1 , 2k ), we have


K
X
1≤n<2K
n a positive integer
1 X

=
n k=1 
X
2k−1 ≤n<2k
n a positive integer
1
 > K,
n
2
which diverges as 2K grows.
Even though there are infinitely many primes, a small modification of the proof of
(3), leads to Euler’s remarkable identity,
X
n≥1
n a positive integer
Y 1
1 −1
=
1− s
.
ns p prime
p
One does need to be a little careful about convergence issues. It is safe to write down
such a formula when both sides are “absolutely convergent”, which takes place when
s > 1; that is, when the sum of the absolute values of the terms converges. In fact they
are absolutely convergent even if s is a complex number so long as Re(s) > 1. This
6
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 6
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
function, discovered and developed by Euler, is known as the Riemann zeta-function,
in honour of the remarkable use Riemann made of it in his plan to give a very good
approximation for the number of primes up to any given large x.
Arithmetic proofs. Fermat’s little theorem implies that if p is an odd prime then
2p−1 ≡ 1
(mod p).
If P = {p1 , . . . , pk } is a finite set of odd primes, let L = L(P) := (p1 − 1) · · · (pk −
1). For each j we can write L = (pj − 1)Lj where Lj is an integer, which can be
given by the same product as L, excluding the factor pj − 1. Then
2L = (2pj −1 )Lj ≡ 1Lj ≡ 1
(mod pj ).
• The number q := 2L + 1 is odd, and q = 2L + 1 ≡ 1 + 1 = 2 (mod p) for
every odd prime p ∈ P , and so q is not divisible by any of the primes in P .
• The Mersenne number q := 2L+1 − 1 is odd, and q = 2 · 2L − 1 ≡ 2 · 1 − 1 =
1 (mod p) for every prime p ∈ P , and so q is not divisible by any of the primes in P .
• The number q := 2L+2 − 3 is odd, and q = 4 · 2L − 3 ≡ 4 − 3 = 1 (mod p)
for every odd prime p ∈ P , and so q is not divisible by any of the primes in P .
For the next two proofs (that there are infinitely many primes) we use the following
lemma:
Lemma 2. If 2m ≡ 1 (mod p) for odd prime p, then 2g ≡ 1 (mod p) where
g =gcd(m, p − 1).
Proof. Euclid’s algorithm, applied to m and p − 1, exhibits integers a and b for which
am + b(p − 1) = g . Therefore, by Fermat’s little theorem,
2g = 2am+b(p−1) = (2m )a · (2p−1 )b ≡ 1a · 1b ≡ 1
(mod p).
• Suppose that there are only finitely many primes and let q be the largest prime. If
p is a prime factor of the Mersenne number, 2q − 1, then 2q ≡ 1 (mod p). Therefore
2g ≡ 1 (mod p) where g = gcd(q, p − 1) by Lemma 2. Now g divides q , so g must
equal either 1 or q . However g cannot equal 1, else p divides 2g − 1 = 2 − 1 = 1.
Therefore q = g which divides p − 1. But then q ≤ p − 1 < p, so p is a larger prime
than q , contradicting its maximiality.
• Suppose that there are only finitely many primes p1 , . . . , pk andn let 2n be the
highest power of 2 dividing some pj − 1. The Fermat number q = 22 + 1 must be
n+1
n
divisible by some odd prime p, and so 22
− 1 = (22 )2 − 1 ≡ (−1)2 − 1 ≡ 0
(mod q), and therefore also mod p. By the lemma, 2g ≡ 1 (mod p) where g =
gcd(2n+1 , p − 1). Now g divides 2n+1 so g must be a power of 2, say g = 2m . Moreover m ≤ n as g = 2m divides p − 1, by the definition of n. Therefore
n
m
n−m
0 ≡ q = 22 + 1 = (22 )2
n−m
+ 1 ≡ 12
+1≡2
(mod p),
so that p divides 2, which is impossible as p is an odd prime.
This proof also yields that, for any integer N ≥ 1, there are infinitely many primes
≡ 1 (mod 2N ), and suitable modifications even allow one to prove that for any integer m ≥ 2, there are infinitely many primes ≡ 1 (mod m).
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
7
page 7
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
• Euler exhibited the inspiring identity
π
1 1 1
= 1 − + − + ...
4
3 5 7
The right-hand side can be re-written
where


1
χ(n) = −1

0
as the sum, over integers n ≥ 1, of χ(n)/n
if n ≡ 1 (mod 4),
if n ≡ −1 (mod 4),
if n is even
(One can write χ(n) = (−4/n), a Jacobi symbol). It is not difficult to see that χ(.) is
multiplicative, that is that
e
if n = pe11 pe22 · · · pkk then χ(n) = χ(p1 )e1 χ(p2 )e2 · · · χ(pk )ek .
Imitating the proof of (3), but replacing each term of the form 1/ms by χ(m)/ms , we
deduce that
π
=
4
X
n≥1
n a positive integer
−1
k χ(n) Y
χ(pj )
=
1−
.
n
pj
j=1
It is well-known that π (and so π/4) is irrational, but the right-hand side is a finite
product of rational numbers, so is rational, a contradiction.
Other types of proofs? There seem to be only a few that are genuinely different from
those that we have outlined above. My favourite is a new proof by Levent Alpoge [2],
based on van der Waerden’s Theorem, a deep result in combinatorics.
2. DYNAMICAL SYSTEMS AND THE INFINITUDE OF PRIMES. One can
model evolution by assuming that the future development of the subject of study follows from understanding its current state.4 This gives rise to dynamical systems, a rich
and bountiful area of study. One simple model is that the state of the object at time
n is denoted by xn , and given an initial state x0 one can find subsequent states via
a map xn → f (xn ) = xn+1 for some given function f (.). We call the set of points
x0 , x1 , . . . the orbit of x0 under the map x → f (x). In the complex plane, orbits of
linear polynomials are easy to understand,5 but quadratic polynomials can give rise to
evolution that is very far from what one might naively guess (the reader might look
into the extraordinary Mandelbrot set).
Here we are interested in the number theory of such orbits, when f is a polynomial
with integer coefficients. It helps to have a little terminology to describe this evolutory
process: Given an integer a0 we generate the orbit, (an )n≥0 , by taking
an → an+1 , where an+1 = f (an ).
This orbit is said to be eventually periodic if there exist integers N ≥ 0 and m ≥
1 such that an+m = an for all n ≥ N . If m is the minimal such integer then we
4 Not
5 One
8
taking into account any random mutations.
can verify: if f (t) = at + b then xn = x0 + nb if a = 1, and xn = an x0 + b/(1 − a) if a 6= 1.
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 8
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
say that the period length is m. If N is the minimal such integer then the numbers
a0 , a1 , . . . , aN −1 are pre-periodic points. For example if
9 → 2 → −3 → 5 → 11 → 13 → 11 → 13
then the orbit of 9 under this map is the set of numbers 9, 2, −3, 5, 11, 13, 11, 13, 11, 13 . . .
This is eventually periodic, with period 11 → 13 → 11 of length two (which could
also be written 13 → 11 → 13), and pre-period 9 → 2 → −3 → 5 of length four.
Notice that once two of the ai ’s are the same number then the sequence must be
periodic from that point onwards: For, if aN +m = aN then an+m = an for all n ≥ N ,
which follows from a suitable induction hypothesis as
an+m+1 = f (an+m ) = f (an ) = an+1 .
Generating infinite sequences of coprime integers. In section 1.2, we constructed
two infinite sequences of pairwise coprime integers which are each the orbit of an
integer a0 under a map x → f (x), for some polynomial f (x). In both cases fn (0) =
f (0) 6= 0 for all n ≥ 1, which we can write as
0 → a → a → ...
where a = f (0); so that 0 is a pre-periodic point for the map x → f (x). The key idea
in proving that the terms of the sequence a0 , a1 , . . . are pairwise coprime is encapsulated in the following lemma.
Lemma 3. Let f (x) ∈ Z[x]. In any integer orbit, (an )n≥0 , under the map x → f (x),
we have that
gcd(am , an ) divides fn−m (0),
whenever n > m ≥ 0. If 0 is a pre-periodic point for the map x → f (x) then
gcd(am , an ) divides the positive integer
L(f ) := lcm[fk (0) : k ≥ 1].
Proof. By Lemma 1 we have
an = fn−m (am ) ≡ fn−m (0)
(mod am ),
and so gcd(am , an ) divides gcd(am , fn−m (0)), which divides fn−m (0).
If 0 is pre-periodic then L(f ) is the lcm of a finite number of non-zero integers, and
so is itself a well-defined non-zero integer. The result follows since fn−m (0) divides
L(f ).
This shows that if 0 is pre-periodic then gcd(am , an ) belongs to a finite set of
possibilities, and we might hope to prove that gcd(an , L) = 1 for all n ≥ 1. If so then
the an are necessarily pairwise coprime, and therefore each an with |an | > 1, will
contain its own private prime factor (We say that pn is a private prime factor of an if
pn divides an but does not divide am for every m 6= n). We now run through various
examples to help us formulate what we will aim to prove.
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
9
page 9
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
— In our first example, f (x) = x2 − x + 1, we have L(f ) = 1, so the an are
pairwise coprime. Therefore if |an | > 1, then an has its own private prime factor pn .
If the orbit of a0 is not eventually periodic then the only possible exceptional an (that
is, for which |an | ≤ 1), is −1, since 0 and 1 belong to the orbit 0 → 1 → 1 → . . .
with a period. However there are no rational solutions to f (x) = −1, so if −1 appears
in an orbit, then it must be the first term of the orbit, that is a0 = −1. Therefore if the
orbit of a0 is not eventually periodic then, for every integer n ≥ 1, the integer an has
its own private prime factor.
— In our second example, f (x) = x2 − 2x + 2, we have L(f ) = 2, so if a0 is any
odd integer whose orbit is not eventually periodic then the an are pairwise coprime.
Again −1 is the only possible exception (since 1 → 1), and if it appears in an orbit
then a0 = −1. Therefore if the orbit of a0 is not eventually periodic then, for every
integer n ≥ 1, the integer an has its own private prime factor.
If a0 is even then each an is even, as the map 0 → 2 → 2 → . . . reduces to the map
0 → 0 → . . . (mod 2), and so gcd(am , an ) = gcd(am , 2) = 2. Moreover an ≡ 2
(mod 4) for all n ≥ 1 as 0, 2 → 2. Therefore the an /2 for n ≥ 1 are odd and pairwise coprime, and so an has its own private prime factor provided |an | > 2. Here
−2 is the only possible exception (since 2 → 2), and if it appears in an orbit then
a0 = −2,. Therefore if the orbit of a0 is not eventually periodic then, for every integer
n ≥ 1, the integer an has its own private, odd prime factor.
We have now seen that, whether or not a0 is even, every an with n ≥ 1, has its own
private prime factor which does not divide L(f ).6
— If f (x) = x2 − 6x + 6 then L(f ) = 6. In this case gcd(am , an ) = gcd(am , 6) =
gcd(a0 , 6) as gcd(t2 − 6t + 6, 6) = gcd(t2 , 6) = gcd(t, 6). In the example 3 →
−3 → 33 → . . . the an /3 are pairwise coprime, but a0 /3 and a1 /3 do not have their
own prime factors. In this example every an with n ≥ 2 has its own private prime
factor > 3, but not a0 or a1 .
Putting together what we have learned from these examples, we might hope for the
following theorem:
Theorem 1. Let f (x) ∈ Z[x] for which 0 is a pre-periodic point of the map x →
f (x). If the orbit of the integer a0 is not eventually periodic then the integer an has its
own private prime factor, which does not divide L(f ), provided n is sufficiently large.
How big is “sufficiently large”? We will show that this holds provided n ≥
2L(f ) + 6, a bound which depends only on f and not on the choice of a0 (as in
the above examples). It may be that the Theorem holds for all n ≥ n0 where n0 is
some number, perhaps even 2 or 3, which is independent of f as well as a0 , but we
have no idea how to prove such a result.
To prove Theorem 1 we will need to classify the possible orbits of 0 which are
eventually periodic, and to better understand the orbit of a0 reduced mod pk+1 where
pk kL(f ). We will discuss and sketch the proofs here, but full details can be found in
the arxiv version of this article.
3. POSSIBLE PERIODS, WITH 0 IN THE PRE-PERIOD We have already seen
the example f (x) = x2 − ax + a for which
0 → a → a → ...
6 For
10
n
this map, we always have the closed fomula an = (a0 − 1)2 + 1.
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 10
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
Now we exhibit several other examples. If f (x) = x2 − 2 then f (0) = −2, f (−2) =
2 and f (2) = 2, that is
0 → −2 → 2 → 2 → . . .
Although the period has length 1, as in our previous example, here the pre-period has
the longer length 2. We also deduce that gcd(am , an ) = 1 or 2 if m 6= n.
Next, for f (x) = x2 − x − 1, we have
0 → −1 → 1 → −1 → 1 → . . .
The iterates of 0 are again periodic, but now with a period of length 2 rather than 1.
We also deduce that gcd(am , an ) = 1 if m 6= n.
Finally for f (x) = 1 + x + x2 − x3 the pre-period and period both have length 2,
0 → 1 → 2 → −1 → 2 → −1 → . . .
We deduce that gcd(am , an ) = 1 or 2 if m 6= n.
The main result of this section is to show that this more-or-less accounts for all of
the possible types of periods with a pre-period involving 0.
Theorem 2. If 0 is pre-periodic for the map x → f (x) ∈ Z[x] then the orbit of 0 is
0 → a → a → . . . for some integer a 6= 0; or
0 → −a → a → a → . . . for a = −2, −1, 1 or 2; or
0 → 1 → a → 1 → . . . or 0 → −1 → a → −1 → . . . for some integer a 6= 0; or
0 → 1 → 2 → −1 → 2 → . . . or 0 → −1 → −2 → 1 → −2 → . . .
All periods have length one or two. This surprising result follows from Lemma 1:
Suppose that N is the smallest positive integer for which aN = a0 . We first prove
that aN +j = aj for all j ≥ 0, by induction: This is true for j = 0 by the hypothesis;
and if it is true for j then aN +j+1 = f (aN +j ) = f (aj ) = aj+1 , as desired.
Now assume that N > 1 so that a1 6= a0 . Lemma 1 implies that an+1 − an divides
f (an+1 ) − f (an ) = an+2 − an+1 for all n ≥ 0. Therefore
a1 − a0 divides a2 − a1 , which divides a3 − a2 , . . . , which divides
aN − aN −1 = a0 − aN −1 ; and this divides a1 − aN = a1 − a0 ,
the non-zero number we started with. We deduce that
|a1 − a0 | ≤ |a2 − a1 | ≤ . . . ≤ |aj+1 − aj | ≤ . . . ≤ |a1 − a0 |,
and so these are all equal. Now there must be some j ≥ 1 for which aj+1 − aj =
−(aj − aj−1 ), else each aj+1 − aj = aj − aj−1 = . . . = a1 − a0 in which case
0 = aN − a0 =
N
−1
X
j=0
January 2014]
(aj+1 − aj ) =
N
−1
X
(a1 − a0 ) = N (a1 − a0 ) 6= 0.
j=0
DYNAMICAL SYSTEMS AND PRIMES
11
page 11
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
Therefore aj+1 = aj−1 , and we deduce that
a2 = aN +2 = fN +1−j (aj+1 ) = fN +1−j (aj−1 ) = aN = a0 ,
as desired.
Lemma 1 can be used to understand the length two periods of a given map x →
f (x): If there is any period of length two then all periodic points are roots of f (x) +
x − T where T is the sum of the two elements in a period; moreover if there is a
periodic point of period one then it is unique and equals T /2. To prove this, let a → b
with a 6= b, be a period of length two, and let c → d be some other period. By Lemma
1, we have that
a − c divides f (a) − f (c) = b − d, which divides f (b) − f (d) = a − c,
and so a − c = ±(b − d); and, similarly a − d = ±(b − c). Therefore a + b = c +
d, for if not then a − c = b − d and a − d = b − c so that a = b which is false. So
we can let T = a + b, and if c = d then T = 2c.
Orbits and algebra. Before proving Theorem 2 we highlight linear transformations
linking different orbits: If (an )n≥0 is an orbit of the map x → f (x) then
(−an )n≥0 is an orbit of the map x → −f (−x)
(as −an+1 = −f (−(−an ))); and
(an − d)n≥0 is an orbit of the map x → gd (x), where gd (x) := f (x + d) − d
(as gd (an − d) = f (an ) − d = an+1 − d). This leads to the following result which
allows us to quickly identify eventually periodic orbits:
Corollary 1. If the orbit of a0 is eventually periodic then a2 = a4 .
Proof. If a0 is in the period then an+2 = an for all n ≥ 0 as the period length is either
one or two. If a0 is pre-periodic then 0 is pre-periodic for the map x → f (x + a0 ) −
a0 , with orbit 0 → a1 − a0 → a2 − a0 → . . .. By inspecting all the cases of Theorem
2 we see that a2 − a0 = a4 − a0 , and so the result follows.
Does a given map come from a polynomial with integer coefficients? Suppose that
f0 (x) ∈ Z[x] is given, and f0 (uj ) = vj for j = 1, 2, . . . , n, where the uj and vj are
given integers. We now prove that there exists a polynomial f (x) ∈ Z[x] for which
f (uj ) =
Qvnj for j = 1, 2, . . . , n, and f (u) = v for some given integers u and v , if and
only if j=1 (u − uj ) divides v − f0 (u):
If f (uj ) = vj for j = 1, 2, . . . , n then the polynomial f (x) − f0 (x) has roots
u1 , . . . , un , which happensQif and only if it is of the form m(x)g(x), for some
n
g(x) ∈ Z[x] with m(x) := j=1 (x − uj ). Therefore if f (u) = v then v − f0 (u) =
m(u)g(u), and so m(u) divides v − f0 (u), as g(u) is an integer. On the other
hand if m(u) divides v − f0 (u) then we can take f (x) = f0 (x) + Cm(x) where
C = (v − f0 (u))/m(u) ∈ Z.
The polynomial f0 (x) = 6 satisfies f0 (0) = f0 (6) = 6. Our result implies that
there is a polynomial f for which f (0) = f (6) = 6 and f (3) = v , if and only if 9 =
−u(u − 6) divides v − f0 (u) = v − 6. On the other hand Lemma 1 implies that 3 =
3 − 0 divides f (u) − f (0) = v − 6, and −3 = 3 − 6 also divides f (u) − f (6) =
v − 6, which is not as restrictive a requirement on v . This new result is therefore even
more powerful than Lemma 1.
We now proceed to a proof of Theorem 2.
12
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 12
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
When 0 is in the pre-period, and the period length is 1. We have the orbit 0 →
a → a for any f (x) of the form f (x) = a + x(x − a)g(x) for some g(x) ∈ Z[x].
If 0 → b → a → a with b 6= a is an orbit for x → f (x) then −b → 0 → a − b →
a − b for x → gb (x) = f (x + b) − b. Now 0 → a − b → a − b is an orbit of the map
x → a − b and so such a polynomial gb exists (and therefore f (x) = gb (x − b) + b
also exists) if and only if
ab = (0 − (−b))((a − b) − (−b)) divides (a − b) − 0 = a − b.
We deduce that b divides a − b and so divides a, and similarly a divides b; as b 6=
a, this implies that b = −a. Therefore a2 = −ab divides 2a = a − b, and so a =
−2, −1, 1 or 2. One can verify that we have such an orbit for f (x) = ±(2x2 − 1) if
a = ±1, and for f (x) = ±(x2 − 2) if a = ±2.
There is no orbit 0 → C → B → A → A else there is an orbit c → 0 → −a →
a → a (for x → gC (x)), by the previous paragraph: By Lemma 1 we then have c =
c − 0 divides f (c) − f (0) = 0 − b = a, but |c| 6= |a| else c would equal one of a or
−a. Therefore |a| = 2 and |c| = 1 and we have c(c2 − 4) = ±3 divides c2 − 2 =
−1, which is nonsense.
When 0 is in the pre-period, and the period length is 2. The period b → a → b is
an orbit of the map x → a + b − x, and so we can extend this to 0 → b → a → b if
and only if ab = (0 − a)(0 − b) divides (a + b − 0) − b = a, and so b = ±1.
If there is an orbit 0 → C → B → A → B then there is an orbit −C → 0 →
b → a → b with b = ±1, and we may assume b = 1 by replacing g(x) by −g(−x)
if b = −1. Now 0 → 1 → a → 1 is an orbit for the polynomial 1 + ax − x2 , so
can be extended by −C → 0 if and only if C(C + a)(C + 1) divides C 2 + aC − 1.
We see that C divides 1, so must equal 1 (as −C 6= b), and so 2(a + 1) divides a,
and therefore a = −2. Hence we have the orbit 0 → 1 → 2 → −1 → 2 (as well as
0 → −1 → −2 → 1 → −2). This is an orbit for the polynomial 1 + x + x2 − x3 .
This cannot be extended by −r → 0 else r = 0 − (−r) divides 1 − 0 = 1, so
equals ±1, which is impossible, as −1, 1 → −2 not 0. This completes the proof of
Theorem 2.
4. PRIVATE PRIME FACTORS. PROOF SKETCH OF THEOREM 1. When
0 → a → a under the map x → f (x), we now sketch a proof that for any integer
a0 , there exists a divisor D = D(a0 ) of a such that the integers an /D are pairwise
coprime, and coprime with a, for all n ≥ n1 (f ), which is the maximum of p − 1 +
k over all prime powers pk dividing a: By Lemma 3 we know that if m < n then
gcd(am , an ) =gcd(am , a), so we need only worry about the prime divisors p of a.
One begins by showing that if m is the smallest integer ≥ 0 for which p|am then
m ≤ p − 1 by the pigeonhole principle. Suppose that pk ka and pj kan . As an (an − L)
divides an+1 − L, we deduce that p2j |an+1 if 2j ≤ k , and pk kan+1 if 2j > k . This
implies that pk kan whenever n ≥ n1 (f ), and the claim follows.
Define a sequence (bj )j≥0 of integers by bj := an1 +j /D for all j ≥ 0, so that
bj+1 = g(bj ) for all j ≥ 0, where g(x) = f (Dx)/D. Here g(x) ∈ Z[x] since
g(0) = a/D, so that 0 → g(0) → g(0). By the same methods as above one can
show that for any map x → g(x) in which the orbit of 0 looks like 0 → b → b, if
−1 or 1 appears in an orbit, then it must be one of the first three terms. Therefore
|an1 +j /D| = |bj | ≥ 2 if j ≥ 3. We deduce that an has its own private prime factor,
not dividing a, for each n ≥ n1 (f ) + 3.
If 0 → −a → a → a with a = ±1 or ±2, then by Lemma 3, either the an are
pairwise coprime or, when a = ±2, the an are even for all n ≥ 1 and the an /2 are
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
13
page 13
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
pairwise coprime and odd for all n ≥ 2. These numbers are then all > 1 in absolute
value for n ≥ 5, and so each an contains its own private prime factor once n ≥ 5.
In every other case in Theorem 2, the orbit of 0 eventually has period 2. The orbit
of 0 under the map x → f2 (x) is of the form 0 → a → a, of period 1. Moreover
the orbit of a0 under x → f (x) is partitioned into the two orbits a0 , a2 , a4 , . . . and
a1 , a3 , a5 , . . . under the map x → f2 (x), and we have gcd(am , an ) = 1 whenever m
and n have different parity (that is, belong to different orbits), since fk (0) = 1 or −1
whenever k is odd. We apply the results of the first paragraph to the two orbits under
f2 (.), which are coprime to each other, and so deduce that each an has its own private
prime factor, which does not divide L(f ) = a, once n ≥ 2n1 (f ) + 6.
Finally observe that p + k − 1 ≤ pk ≤ L(f ) whenever pk kL(f ), so that n1 (f ) ≤
L(f ), and the result follows.
5. OTHER DYNAMICAL SYSTEMS TO FIND AN INFINITY OF PRIMES.
In Theorem 1 we saw that if 0 is a pre-periodic point for the map x → f (x) then
any given integer orbit is either eventually periodic, or every an has its own private
prime factor once n is sufficiently large. Do we expect something like this be true for
other polynomials f (x) ∈ Z[x]? If f (x) has degree d > 1 then one can show that if
(an )n≥0 is not eventually periodic then there exist real numbers 1 ≥ α > 0 and β ,
which depend only on f , and τ > 1 which depends on f and a0 , such that
n
If n is sufficiently large then |an | is the integer nearest to ατ d + β.
(4)
The an grow so fast that the product of the gcd(am , an ) with m < n is far smaller
than an , so it seems very likely that an has its own new prime factors. However this
argument cannot be made into a proof since some of the primes that divide the am with
m < n, might divide an to a very high power. We will discuss this possibility further
in section 5.
Before this more general discussion, we highlight another class of polynomials f
that have been used to construct infinitely long sequences of coprime integers, in an
elementary way.
Another iteration to describe Euclid’s construction. Let A0 = 2, and An+1 =
An (An + 1) for all n ≥ 0, so that
A1 = 2 × 3, A2 = 2 × 3 × 7, A3 = 2 × 3 × 7 × 43, etc.
Taking a0 = A0 with an = An /An−1 for all n ≥ 1, we again obtain Euclid’s sequence (see, e.g. [21, 18])! Whereas before we described the iteration as an+1 =
f (an ) with f (x) = x2 − x + 1, now we have An+1 = F (An ) with F (x) = x(x +
1). For the proof with the Fermat numbers we may similarly let F (x) = x(x + 2).
Does this generalize? Do we obtain all the same sequences as before?
The idea is to take G(x) ∈ Z[x] with G(0) 6= 0, and let F (x) = xG(x). Then, for
given integer A0 = a0 , we let
An+1 = F (An ) = An G(An ), and then
an+1 = An+1 /An = G(An ) for all integers n ≥ 0.
We deduce, by induction, that
An = a0 a1 · · · an .
14
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 14
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
Theorem 3. Let F (x) = xG(x) with G(x) = G(0) + xh H(x) where H(x) ∈ Z[x],
and G(0), H(0) 6= 0. If (an )n≥0 is not eventually periodic then the integers an , with
n ≥ max{|G(0)| + 4, |H(0)| + 2}, each have their own private prime factor.
Proof sketch. If the (an )n≥0 are eventually periodic then the Am all belong to the
finite set {` ∈ Z : G(`) = an for some n ≥ 1}. Therefore Am+d = Am for some
m + d > m, so either some Am = 0, or 1 = Am+d /Am = am+d am+d−1 . . . am+1 ,
so either am+1 = 1 or am+2 = 1 or am+1 = am+2 = −1. In the first case, if it is
the smallest such m then am = 0 and each subsequent an = G(0). If an = 1 then
An = An−1 and therefore an+1 = G(An ) = G(An−1 ) = an , so an = 1 for all n ≥
m + 2. Finally if am+1 = am+2 = −1 then the An have period two by an analogous
induction argument and an = −1 for all n ≥ m + 1.
In this proof we saw that if |aN aN +1 | ≤ 1 then an is periodic for all n ≥ N + 2.
Hence if (an )n≥0 is not eventually periodic then |aN aN +1 | ≥ 2 for all N . Therefore if
N ≥ 2k(F ) + 2 then |AN −1 | > max{2|G(0)|, |H(0)|}, where k(F ) is the largest
integer k for which 2k ≤ max{2|G(0)|, |H(0)|}.
We now deduce if |an | > 1 for all n ≤ 2k(F ) + 1 then (an )n≥0 is not eventually
periodic. For if (an )n≥0 is eventually periodic then aN = −1, 0 or 1 for some N ≥
2k(F ) + 2. Then G(AN −1 ) = aN , and so AN −1 = AN −1 − 0 divides G(AN −1 ) −
G(0) = aN − G(0). Now aN = G(0) else |AN −1 | ≤ |G(0)| + 1 ≤ 2|G(0)|, a contradiction. Therefore AN −1 is a root of G(x) − aN = xh H(x), and so is a root of
H(x) as AN −1 6= 0. Therefore AN −1 |H(0) and so |AN −1 | ≤ |H(0)|, another contradiction.
Now, if m < n then am |Am |An−1 and so an = G(An−1 ) ≡ G(0) (mod am ),
which implies that gcd(am , an ) = gcd(am , G(0)) which divides gcd(an , G(0)).
Therefore if prime p - G(0) then p can divide at most one of the an .
Suppose that pk kG(0). As before (for those f for which 0 is pre-periodic), if m
is the smallest integer for which p divides Am then m ≤ p − 1. Then p|am , and
p|am |An for all n ≥ m, and therefore an+1 = G(An ) ≡ G(0) ≡ 0 (mod p). which
implies that pj+1 |am am+1 · am+j |Am+j for all j ≥ 0. Therefore if n ≥ m + k +
1(≥ p + k ) then an = G(An−1 ) ≡ G(0) (mod pk+1 ) and so pk kan .
Therefore, if D is the product of those prime powers pk kG(0) for which p divides
some an , then D|an and the an /D are pairwise coprime, and coprime with G(0),
when n ≥ |G(0)| + 1. These an /D will have their own private prime factor unless
G(An−1 ) = an = −D or D. That is An−1 is a root of G(x) − d for some positive
or negative divisor d of G(0). If d 6= G(0) then An−1 divides G(0) − d, so that
|An−1 | ≤ |G(0)| + |d| ≤ 2|G(0)|; if d = G(0) then An−1 is a root of H(x), and
so divides H(0), so that |An−1 | ≤ |H(0)|. Therefore n ≤ 2k(F ) + 1, and the result
follows.
The “taking the pth power” example. Let F (x) = (x + 1)p − 1 and A0 = a − 1
n
so that An = ap − 1 for all n ≥ 0 and
n
ap − 1
an := pn−1
.
a
−1
Now G(0) = p and, by Fermat’s Little Theorem, F (b − 1) = bp − 1 ≡ b − 1
(mod p) and so An ≡ a − 1 = A0 (mod p) by induction. Therefore if a 6≡ 1
(mod p) then gcd(an , p) divides (An , p) = (A0 , p) = 1. This implies that if |a| > 1
and a 6≡ 1 (mod p) then the integers (an )n≥0 are pairwise coprime, and so have their
own private prime factor if n ≥ 1.
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
15
page 15
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
n
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
n−1
Now if prime qn divides an then ap ≡ 1 (mod qn ), but ap
6≡ 1 (mod qn ) as
qn - an−1 (as an and an−1 are coprime). The order of a (mod qn ) is the least integer
r > 0 for which ar ≡ 1 (mod qn ). Lemma 2 implies that r =gcd(r, pn ), so r divides
pn , but r does not divide pn−1 and therefore r = pn . Fermat’s Little Theorem, together
with Lemma 2, then implies that r = pn divides qn − 1; that is,
qn ≡ 1
(mod pn ) for all n ≥ 1.
In particular qn ≡ 1 (mod pm ) whenever n ≥ m; that is, we obtain an infinite sequence of distinct primes ≡ 1 (mod pm ).
n
If a ≡ 1 (mod p) then ap ≡ 1 (mod p) for all n ≥ 0 and so
an =
p−1
X
n
(ap )j ≡
j=0
p−1
X
1j = p ≡ 0
(mod p).
j=0
Therefore gcd(am , an ) = gcd(am , p) = p; and so the integers (an /p)n≥0 are pairwise coprime. This implies that if |a| > 1 then the integers (an )n≥0 each have their
own private prime factor, not equal to p, for all n ≥ 1.
n−1
If p = 2 then an = a2
+ 1 and so an+1 = f (an ) where f (x) = x2 − 2x + 2,
which we recognize from earlier. However the sequence an is new when p > 2; in
fact, it could not have been generated by a map an → f (an ), where f (x) ∈ Z[x]. For
if it is then, as an+1 /apn → 1 as n → ∞, we deduce that f (x) = xp + terms of lower
degree, and the limit of (f (an ) − apn )/ap−1
n , as n → ∞, will yield the coefficient of
xp−1 . However (f (an ) − apn )/anp−1 = (an+1 − apn )/ap−1
→ ∞ as n → ∞, so an+1
n
cannot equal f (an ) for some fixed f (x) ∈ Z[x].
Constructing F from f . Suppose that 0 → a → a under the map x → f (x). Let
F (x) = f (x + a) − a = x(x + a)H(x) for some H(x) ∈ Z[x]. Now, given a0 and
then an+1 = f (an ) for all n ≥ 0, let A0 = a0 − a and define An+1 = F (An ) for
all n ≥ 0. We claim that An = an − a for all n ≥ 0: This is true for n = 0 and
then An+1 = F (An ) = f (An + a) − a = f (an ) − a = an+1 − a. Now An+1 =
an An H(An ) so all of the prime factors of the an divide the An . On the other hand, if
prime p|H(An ) but p - aH(0)H(−a) then p|An+1 and p - An (as p - H(0)) so p a0 a1 · · · an−1 , and p - an = An + a (as p - H(−a)), and therefore am ≡ a (mod p)
as p|Am so p - am for all m ≥ n + 1. (The same argument works here if F (x) =
xr (x + a)s H(x).)
Other polynomials? What of the sequences (an )n≥0 generated by x → f (x) where
0 is neither pre-periodic, nor a root of f (x)? At the start of this section we noted that,
if the an are not eventually periodic, and
n
If n is sufficiently large then an is the integer nearest to ατ d + β .
for some τ = τ (a0 ) > 1, and 0 < α ≤ 1. Now if n = m + k then gcd(am , an ) =
gcd(am , fk (0)) and so is bounded by min{|am |, |fk (0)|}, which implies that
Y
m≤n−1
gcd(am , an ) ≤
Y
0≤m≤n/2
|am | ·
Y
n/2
|fk (0)| ≤ C d
0≤k≤n/2
for some constant C = C(a0 ) ≥ (τ (a0 )τ (0))2 . This is evidently far, far smaller than
an for n large, so only a small part of an comes from its gcds with the am for m < n,
16
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 16
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
and so we believe that an is likely to have a prime divisor that the am , with m < n,
do not. However it is feasible that some of the primes dividing the am might divide
an to extraordinarily high powers. We know of no unconditional way to show that this
is impossible, but we can recover such a result if we assume the abc-conjecture,7 so
called because it limits how big is the contribution of prime powers to any solution of
the equation
a+b=c
in pairwise coprime integers. Applied to this equation with b = 1, so that c = a + 1,
one gets the following consequence of the abc-conjecture: For any fixed > 0 there
exists a constant κ > 0 such that for any integer a we have
Y
p ≥ κ |a|1−
p prime
p|a(a+1)
The product here is over all primes p which divide the polynomial f (x) = x2 + x
evaluated at x = a. There is a far-reaching generalization which is a consequence
of the abc-conjecture. It applies to the prime divisors of any given polynomial (see
[15, 6, 9]):
Consequence of the abc-conjecture: Suppose that f (x) ∈ Z[x] has at least two distinct roots (in other words, f (x) is not of the form a(x + c)d ). For any fixed > 0
there exists a constant κ,f > 0 such that for any integer a we have
Y
p ≥ κ,f |a|1−
p prime
p|f (a)
We say that pn is a primitive prime factor of an if pn divides an but does not divide
am for any m < n. It is possible that a primitive prime factor, pn of an , divides aM
for some M > n, so pn is not as unique as a “private” prime factor. We now show,
assuming the abc-conjecture, that if n is sufficiently large then an has a primitive
prime factor: If not then every prime factor of an must divide am for some m < n,
and so
Y
p prime
p|an
Y
p=
p|
p prime
Qn−1
m=0 gcd(am ,an )
p≤
n−1
Y
n/2
gcd(am , an ) ≤ C d
,
m=0
as we proved above. On the other hand, applying (4), and then taking = 31 in the
abc-conjecture (so that κ = κ1/3,f > 0), we deduce that, for sufficiently large n,
Y
n−1
n/2
τ d /2 ≤ κ|an−1 |2/3 ≤
p ≤ Cd ,
p prime
p|an =f (an−1 )
which is impossible, as dn−1 goes to infinity, as n grows, much faster than dn/2 . Therefore we have proved:
7 This has recently been claimed to have been proved by Mochizuki, but the experts have as yet struggled
to agree, definitively, that Mochizuki’s extraordinary ideas are all correct.
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
17
page 17
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
Theorem 4. Assume the abc-conjecture. Suppose that f (x) ∈ Z[x] has at least two
distinct roots. If (an )n≥0 is generated by x → f (x) and is not eventually periodic,
then an has a primitive prime factor for all sufficiently large n.
Primitive prime factors of rational functions. Now suppose that φ(t) = f (t)/g(t),
with f (t), g(t) ∈ Z[t]. Beginning with a rational number r0 , we define rn+1 = φ(rn )
for all n ≥ 0, and write rn = an /bn where gcd(an , bn ) = 1 and bn ≥ 1.
Suppose that f (0) = 0 but f (t) is not of the form atd . Ingram and Silverman [13]
showed that if the orbit of r0 is not eventually periodic then an has a primitive prime
factor for all sufficiently large n. It is believed that some result of this sort should
hold whether or not f (0) = 0, and indeed this has been confirmed in [11] under the
assumption of the abc-conjecture.
Faber and Granville [7] showed that for any rational function φ of degree ≥ 2, if
the orbit of r0 is not eventually periodic then either there is a primitive prime factor in
the numerator of an+1 − an for all sufficiently large n, or there is a primitive prime
factor in the denominator of an+1 − an for all sufficiently large n.
All these cases give rise to new proofs that there are infinitely many primes!
6. PRIMITIVE PRIME FACTORS, FOR DEGREE ONE F , AND BEYOND.
The Mersenne numbers, Mn = 2n − 1 can be generated by taking M0 = 0 and then
Mn = f (Mn−1 ) for all n ≥ 1, with f (x) = 2x + 1. The growth of the numbers
involved, now that f is a linear polynomial, is simply exponential (rather than doubly
exponential in n, as we saw in (4) when the degree of f is ≥ 2), provided that the
leading coefficient of f has absolute value > 1.8 If n ≡ r (mod m) then we can
write n = qm + r to obtain
Mn = 2qm+r − 1 = (2m )q · 2r − 1 ≡ 1q · 2r − 1 = 2r − 1
(mod Mm )
and so gcd(Mn , Mm ) = Mr . We deduce, by induction, that gcd(Mn , Mm ) =
Mgcd(m,n) . In this case the gcd’s grow large, but we can still hope to find a primitive prime factor of Mn when n is sufficiently large. For example, for n = 2, 3, 4, 5,
the Mn have primitive prime factors 3, 7, 5, 31 then M6 = 63 does not, and subsequently we have 127, 17, 73, 11, . . .; indeed Mn has a primitive prime divisor for all
n
−bn
for all
n > 6. More than that, the Bang-Zsimondy theorem states that if xn = a a−b
n ≥ 0, with a > b ≥ 1 then xn has a primitive prime divisor for all n > 1, other than
the example xn = M6 above. These provide yet more infinite sequences of distinct
prime numbers.
A Lucas sequence (an )n≥1 begins a0 = 0, a1 = 1 and then satisfies the secondorder linear recurrence
an = pan−1 + qan−2 for all n ≥ 2,
for arbitrary integers p and q , in which the discriminant ∆ := p2 + 4q is non-zero.9 In
1913, Carmichael showed that if ∆ > 0 then an has a primitive prime factor for each
n > 12, the most exceptional case being the Fibonacci sequence Fn = Fn−1 + Fn−2
in which F2 = 1, F6 = F33 and F12 = F34 F42 have no primitive prime factors. In
2001, Bilu, Hanrot and Voutier [4] proved, following up on groundbreaking work of
Schinzel [19], that if ∆ < 0 then an has a primitive prime factor for each n > 30, the
8 If
f (x) = x + b then fn (a) = a + nb. If f = −x + b then fn (a) = a or b − a as n is even or odd.
n
−bn
sequence (xn )n≥0 with xn = a a−b
, is the special case with p = a + b and q = −ab.
9 The
18
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 18
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
most exceptional case being an = an−1 − 2an−2 in which a2 = 1,
a3 = a5 = a13 = −1, a8 = a4 , a12 = a24 a6 , a18 = −a6 a9 and a30 = −a26 a10 a15 .
Calculations reveal that a primitive prime factor usually divides an to the power
one; that is, p divides an but p2 does not. One might conjecture that if n is sufficiently
large then an has a primitive prime factor p for which pkan , though there has been
no success with this question. Recently, inspired by remarks of Lenstra, Granville was
able to show [10] that whenever 2n − 1 has a primitive prime factor then it has a
primitive prime factor which divides it to an odd power. More generally, if an is a
Lucas sequence with gcd(p, q) = 1 and ∆ > 0, where 2kq , then an has a primitive
prime factor which divides it to an odd power, provided n 6= 1, 2 or 6.
It is much more difficult to work with the prime factors of recurrence sequences
that do not start with 0. For example, we expect that xn = 2n − 3 has primitive prime
divisors for all n > 7, but we only know how to prove that this is true for some infinite
sequence of values of n (see the third arithmetic proof that there are infinitely many
primes, above).
REFERENCES
1. Takashi Agoh, Paul Erdős, and Andrew Granville, Primes at a (somewhat lengthy) glance. Amer. Math.
Monthly 104 (1997), 943-945.
2. Levent Alpoge, van der Waerden and the primes, Amer. Math. Monthly 122 (2015), 784–785.
3. Richard Bellman, A note on relatively prime sequences, Bull. Amer. Math. Soc. 53 (1947), 778–779.
4. Y. Bilu, G. Hanrot and P. Voutier, Existence of primitive divisors of Lucas and Lehmer numbers, J. Reine
angew. Math 539 (2001), 75-122
5. A. W. F. Edwards Infinite Coprime Sequences, The Mathematical Gazette, 48 (1964), 416-422
6. Noam Elkies, ABC implies Mordell, Internat. Math. Res. Notices (1991), 99-109.
7. Xander Faber and Andrew Granville, Prime factors of dynamical sequences, J. Reine angew. Math 661
(2011), 189-214.
8. Irving Gerst and John Brillhart, On the Prime Divisors of Polynomials, Amer. Math. Monthly 78 (1971),
250-266.
9. Andrew Granville, ABC allows us to count squarefrees, Internat. Math. Res. Notices 19 (1998), 991-1009.
10. Andrew Granville, Primitive prime factors in second-order linear recurrence sequences, Acta Arithmetica, 155 (2012), 431-452.
11. C. Gratton, K. Nguyen, and T. J. Tucker, ABC implies primitive prime divisors in arithmetic dynamics,
Bull. Lond. Math. Soc. 45 (2013), 1194-1208.
12. V.C. Harris, Another proof of the infinitude of primes, Amer. Math. Monthly 63 (1956), 711.
13. Patrick Ingram and Joseph H. Silverman, Primitive divisors in arithmetic dynamics, Math. Proc. Cambridge Philos. Soc 146 (2009), 289-302.
14. Holly Krieger, Primitive prime divisors in the critical orbit of z d + c, Int. Math. Res. Not. 23 (2013),
5498-5525.
15. Michel Langevin, Partie sans facteur carr de F (a, b) (modulo la conjecture abc, in Sminaire de Thorie
des Nombres (1993/94), Publ. Math. Univ. Caen
16. Sam Northshield, A one-line proof of the infinitude of primes, Amer. Math. Monthly 122 (2015), 466.
17. Brian Rice, Primitive prime divisors in polynomial arithmetic dynamics, Integers (electronic), 7:A26
(2007), 16 pages.
18. Filip Saidak, A new proof of Euclid’s theorem, Amer. Math. Monthly 113 (2006), 937-938.
19. A. Schinzel, Primitive divisors of the expression An − B n in algebraic number fields, J. Reine
angew. Math 268/269 (1974), 27-33.
20. Michael Somos and Robert Haas, A linked pair of sequences implies the primes are infinite, Amer. Math.
Monthly 110 (2003), 539–540.
21. M.V. Subbarao, On relatively prime sequences, Amer. Math. Monthly 73 (1966), 1099–1102.
22. John Thompson, A method for finding primes, . Amer. Math. Monthly 60 (1953), 175.
For a blog keeping track of proofs of the infinitude of primes, please see the website
http://www.cut-the-knot.org/proofs/Furstenberg.shtml
January 2014]
DYNAMICAL SYSTEMS AND PRIMES
19
page 19
Mathematical Assoc. of America
American Mathematical Monthly 121:1
September 23, 2016
3:44 p.m.
MonthlyArticleCorrectSubmittedVersion.tex
and the links at the end of that proof
20
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
page 20