Using Dynamical Systems to Construct Infinitely Many Primes
aa r X i v : . [ m a t h . N T ] A ug Mathematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 1
Using Dynamical Systems to ConstructInfinitely Many Primes
Andrew Granville
Abstract.
Euclid’s proof can be reworked to construct infinitely many primes, in many differ-ent ways, using ideas from arithmetic dynamics.
1. CONSTRUCTING INFINITELY MANY PRIMES.
The first known proof thatthere are infinitely many primes, which appears in Euclid’s
Elements , is a proof bycontradiction. This proof does not indicate how to find primes, despite establishingthat there are infinitely many of them. However a minor variant does do so. The key toeither proof is the theorem that
Every integer q > has a prime factor. (This was first formally proved by Euclid on the way to establishing the fundamentaltheorem of arithmetic.) The idea is to Construct an infinite sequence of distinct, pairwise coprime, integers a , a , . . . ,that is, a sequence for which gcd ( a m , a n ) = 1 whenever m = n . We then obtaininfinitely many primes, the prime factors of the a n , as proved in the following result. Proposition 1.
Suppose that a , a , . . . is an infinite sequence of distinct, pairwisecoprime, integers. Let p n be a prime divisor of a n whenever | a n | > . Then the p n form an infinite sequence of distinct primes.Proof. The p n are distinct for if not, then p m = p n for some m = n and so p m = gcd ( p m , p n ) divides gcd ( a m , a n ) = 1 , a contradiction. As the a n are distinct, any given value (in particular, − , and ) canbe attained at most once, and therefore all but at most three of the a n have absolutevalue > , and so have a prime factor p n .To construct such a sequence by modifying Euclid’s proof, let E = 2 and E n = E E · · · E n − + 1 for each n ≥ .If m < n , then E m divides E E · · · E n − = E n − (as E m is one of the terms inthe product) and so gcd ( E m , E n ) divides gcd ( E n − , E n ) = 1 , which implies thatgcd ( E m , E n ) = 1 . Therefore, if p n is a prime divisor of E n for each n ≥ , then p , p , . . . is an infinite sequence of distinct primes.The Fermat numbers , F , F , . . . , defined by F n = 2 n + 1 for each n ≥ , are amore familiar sequence of pairwise coprime integers. Fermat had actually conjecturedthat the F n are all primes. His claim starts off correct: , , , , are allprime, but is false for F = 641 × , as Euler famously noted. It is an openJanuary 2014] DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 2 question as to whether there are infinitely many primes of the form F n . Nonethelesswe can prove that the F n are pairwise coprime in a similar way to the E n , using theidentity F n = F F · · · F n − + 2 for each n ≥ and the fact that the F n are all odd. We deduce from Proposition 1 that if p n is aprime divisor of F n for each n ≥ , then p , p , . . . is an infinite sequence of distinctprimes. For any sequence of integers ( a n ) n ≥ we call p n a private prime factor of a n if itdivides a n and no other a m . The primes p n constructed in Proposition 1 are thereforeeach private prime factors of the a n .
2. A SIMPLER FORMULATION.
In Section 1, the sequences ( E n ) n ≥ and ( F n ) n ≥ were constructed in a similar way, multiplying all of the terms of the se-quence so far together and adding a constant. We would like a simpler, unified way toview these two sequences, with an eye to generalization. This is not difficult becausethe recurrence for the E n can be rewritten as E n +1 = E E · · · E n − · E n + 1 = ( E n − E n + 1 = f ( E n ) , where f ( x ) = x − x + 1 . Similarly the F n -values can be determined by F n +1 = (2 n + 1)(2 n −
1) + 2 = F n ( F n −
2) + 1 = f ( F n ) , where f ( x ) = x − x + 2 . So they are both examples of sequences ( x n ) n ≥ forwhich x n +1 = f ( x n ) for some polynomial f ( x ) ∈ Z [ x ] . The terms of the sequence are all given by therecursive formula, so that x n = f ( f ( . . . f | {z } n times ( x ))) = f n ( x ) , where the notation f n denotes the polynomial obtained by composing f with itself n times (which is definitely not the n th power of f ). Any such sequence ( x n ) n ≥ iscalled the orbit of x under the map f , since the sequence is completely determinedonce one knows x and f . We sometimes write the orbit as x → x → x → · · · . The key to the proof that the E n ’s are pairwise coprime is that E n ≡ E m ) whenever n > m ≥ ; and the key to the proof that the F n ’s are pairwise coprime isthat F n ≡ F m ) whenever n > m ≥ . These congruences are not difficultto deduce using the following lemma from elementary number theory. The only Fermat numbers known to be primes have n ≤ . We know that the F n are composite for ≤ n ≤ and for many other n besides. It is always a significant moment when a Fermat number is factoredfor the first time. It could be that all F n with n > are composite, or they might all be prime from somesufficiently large n onwards, or some might be prime and some composite. Currently, we have no way ofknowing which is true. Goldbach was the first to use the prime divisors of the Fermat numbers to prove that there are infinitelymany primes, in a letter to Euler in late July 1730. (cid:13)
THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 3
Lemma 1.
Let g ( x ) be a polynomial with integer coefficients. For any integers a = b ,their difference a − b divides g ( a ) − g ( b ) . Moreover, if m is an integer for which a ≡ b (mod m ) , then g ( a ) ≡ g ( b ) (mod m ) . For f ( x ) = x − x + 1 , the orbit of under the map f is → → → · · · . Therefore if n > m ≥ , then we have E n = f n − m ( E m ) ≡ f n − m (0) = 1 (mod E m ) , and so ( E n , E m ) = (1 , E m ) = 1 . (For clarity, we have used an “=” sign when twonumbers are equal despite working with a congruence. Typically we use the notation ( a, b ) rather than gcd ( a, b ) .) Similarly if f ( x ) = x − x + 2 , then the orbit of under the map f is → → → · · · and so F n = f n − m ( F m ) ≡ f n − m (0) = 2(mod F m ) , which implies that if n > m , then ( F n , F m ) = (2 , F m ) = 1 .This reformulation of two of the best-known proofs of the infinitude of primes hintsat the possibility of a more general approach.
3. DIFFERENT STARTING POINTS.
We just saw that the orbit of under themap x → x − x + 1 is an infinite sequence of pairwise coprime integers, → → → → → · · · . What about other orbits? The orbit of is also an infinitesequence of pairwise coprime integers beginning → → → → · · · ,as is the orbit of which begins → → → · · · . The same proof as beforeyields that no two integers in a given orbit have a common factor.The orbit of under the the map x → x − x + 2 yielded the Fermat numbers → → → → · · · , but starting at we get → → → → · · · . These are obviously not pairwise prime as every number in the orbit is even, but if wedivide through by , then we get → → → → · · · which are pairwise coprime. To prove this, note that x = 4 and x n +1 = x n − x n + 2 for all n ≥ , and so the above proof yields that if m < n then ( x m , x n ) =( x m ,
2) = 2 . Therefore, taking a n = x n / for every n we deduce that ( a m , a n ) =( x m / , x n /
2) = ( x m / ,
1) = 1 . This same idea works for every orbit under thismap: we get an infinite sequence of pairwise coprime integers by dividing through by1 or 2, depending on whether x is odd or even.However, things can be more complicated: Consider the orbit of x = 3 under themap x → x − x − . We have → − → → → → · · · . Here x n is divisible by if n is even, and is divisible by if n is odd. If we let a n = x n / when n is even, and a n = x n / when n is odd, then one can show theterms of the resulting sequence, → − → → → → · · · , are indeed pairwise coprime.Another surprising example is given by the orbit of under the map x → x ( x − x − . Reducing the elements of the orbit mod we find that → → → → → → → → · · · (mod 7) , January 2014]
DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 4 as x ( x − x − ≡ x ( x −
1) (mod 7) , which is ≡ x − if x by Fermat’s little theorem. So x n is divisible by for every n ≥ , but forno smaller n , and to obtain the pairwise coprime a n we let a n = x n for n ≤ , and a n = x n / once n ≥ .It starts to look as though it might become complicated to formulate how to definea sequence ( a n ) n ≥ of pairwise coprime integers in general; certainly a case-by-casedescription is unappealing. However, there is a simpler way to obtain the a n : In theselast two examples we have a n = x n / ( x n , and then a n = x n / ( x n , , respectively,for all n ≥ , a description that will generalize well.
4. DYNAMICAL SYSTEMS AND THE INFINITUDE OF PRIMES.
One mod-els evolution by determining the future development of the object of study from itscurrent state (more sophisticated models incorporate the possibility of random muta-tions). This gives rise to dynamical systems , a rich and bountiful area of study. Onesimple model is that the state of the object at time n is denoted by x n , and given aninitial state x , one can find subsequent states via a map x n → f ( x n ) = x n +1 forsome given function f ( . ) . Orbits of linear polynomials f ( . ) are easy to understand, but quadratic polynomials can give rise to evolution that is very far from what onemight naively guess (the reader might look into the extraordinary Mandelbrot set).This is the set-up that we had above! For us, f ( x ) is a polynomial with integercoefficients and x an integer. It will be useful to use dynamical systems terminology.If f n ( α ) = α for some integer n ≥ , then (the orbit of) α is periodic , and thesmallest such n is the exact period length for α . The orbit begins with the cycle α, f ( α ) , . . . , f n − ( α ) of distinct values, and repeats itself, so that f n ( α ) = α, f n +1 ( α ) = f ( α ) , . . . , and,in general f n + k ( α ) = f k ( α ) for all k ≥ .The number α is preperiodic if f m ( α ) is periodic for some m ≥ , and is strictlypreperiodic if α is preperiodic but not itself periodic. In all of our examples so far, has been strictly preperiodic. In fact, if any two elements of the orbit of α are equal, say f m + n ( α ) = f m ( α ) , then f k + n ( α ) = f k − m ( f m + n ( α )) = f k − m ( f m ( α )) = f k ( α ) for all k ≥ m , so that α is preperiodic.Finally, α has a wandering orbit if it is not preperiodic, that is, if its orbit neverrepeats itself so that the { f m ( α ) } m ≥ are all distinct. Therefore, we wish to start onlywith integers x that have wandering orbits.We now state our general result for constructing infinitely many primes from orbitsof a polynomial map. Theorem 1.
Suppose that f ( x ) ∈ Z [ x ] , and that is a strictly preperiodic point ofthe map x → f ( x ) . Let ℓ ( f ) = lcm [ f (0) , f (0)] . For any integer x that has a wan-dering orbit, ( x n ) n ≥ , let a n = x n gcd ( x n , ℓ ( f )) for all n ≥ . The ( a n ) n ≥ are an infinite sequence of pairwise coprime integers, and if n ≥ then a n has a private prime factor. For example, ℓ ( f ) = 1 , , and , respectively, for the four polynomials f ( x ) inthe examples of Section 3. One can verify: if f ( t ) = at + b , then x n = x + nb if a = 1 , and x n = a n x + b/ (1 − a ) if a = 1 . (cid:13) THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 5
Proof.
Suppose that k = n − m > . Then, by Lemma 1, x n = f k ( x m ) ≡ f k (0) (mod x m ) , and so gcd ( x m , x n ) divides gcd ( x m , f k (0)) , which divides f k (0) . But this divides L ( f ) := lcm [ f k (0) : k ≥ , which is the lcm of a finite number of nonzero integers, as is preperiodic. Therefore ( x m , x n ) divides L ( f ) , and so ( x m , x n ) divides both ( x m , L ( f )) and ( x n , L ( f )) .This implies that A m := x m / ( x m , L ( f )) divides x m / ( x m , x n ) , and A n divides x n / ( x m , x n ) . But x m / ( x m , x n ) and x n / ( x m , x n ) are pairwise coprime which there-fore implies that ( A m , A n ) = 1 .We will show below that L ( f ) = ℓ ( f ) , that is, the lcm of all the elements of theorbit of is the same as the lcm of the first two terms. This then implies that a n = A n for all n .One can also prove that | a n | > for all n ≥ (see the discussion of one case inSection 9), and all such n must have a private prime factor. The result follows.The example f ( x ) = 3 − x ( x − with → → has a wandering orbit → → − → · · · , so that if x = 2 then x = 1 and x = − . Therefore we cannot ingeneral improve the lower bound, n ≥ , in Theorem 1.We now determine all polynomials f that satisfy the hypothesis of Theorem 1.
5. POLYNOMIAL MAPS FOR WHICH IS STRICTLY PREPERIODIC.
There are severe restrictions on the possible exact periods. If is strictly preperiodic, then its exact period length is either one or two. Wehave already seen the examples x − x + 1 for which → → , and x − x − for which → − → → − , where is preperiodic with period length one andtwo, respectively. In fact exact periods cannot be any larger: Lemma 2.
Let f ( x ) ∈ Z [ x ] . If the orbit of a is periodic, then its exact period lengthis either one or two.Proof. Let N be the exact period length so that a N = a , and then a N + k = f k ( a N ) = f k ( a ) = a k for all k ≥ . Now assume that N > so that a = a . Lemma 1 im-plies that a n +1 − a n divides f ( a n +1 ) − f ( a n ) = a n +2 − a n +1 for all n ≥ . There-fore, a − a divides a − a , which divides a − a , . . . , which divides a N − a N − = a − a N − ; and this divides a − a N = a − a ,the nonzero number we started with. We deduce that | a − a | ≤ | a − a | ≤ · · · ≤ | a j +1 − a j | ≤ · · · ≤ | a − a | , and so these are all equal. Therefore there must be some j ≥ for which a j +1 − a j = − ( a j − a j − ) , else each a j +1 − a j = a j − a j − = · · · = a − a and so a N − a = N − X j =0 ( a j +1 − a j ) = N − X j =0 ( a − a ) = N ( a − a ) = 0 , January 2014]
DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 6 a contradiction. Therefore a j +1 = a j − , and we deduce that a = a N +2 = f N +1 − j ( a j +1 ) = f N +1 − j ( a j − ) = a N = a , as desired.To determine which polynomials can be used in Theorem 1, we need to understandthe possible orbits of . A classification of when is a strictly preperiodic point. There are just four possi-bilities for the period length and the preperiod length of the orbit of . We now giveexamples of each type. For some nonzero integer a , the orbit of is one of: → a → a → · · · for x → x − ax + a (e.g., ( a n ) n ≥ and ( F n ) n ≥ ); or → − a → a → a → · · · for a = 1 or , with x → x − or x − ,respectively; or → − → a → − → · · · with f ( x ) = x − ax − ; or → → → − → → − → · · · with f ( x ) = 1 + x + x − x ; andall other possible orbits are obtained from these, by using the observation that if → a → b → · · · is an orbit for x → f ( x ) , then → − a → − b → · · · is an orbit for x → g ( x ) where g ( x ) = − f ( − x ) .It is not difficult to determine all polynomials f for which has one of these orbits.For example, in the first case above we simply need to find all f for which f (0) = f ( a ) = a . One example is f ( x ) = a so that f is another example if and only if f − f has roots at and a . That is, f ( x ) − f ( x ) equals x ( x − a ) g ( x ) for some g ( x ) ∈ Z [ x ] . The analogous approach works in all of the other cases.Proving these are the only possible orbits is not difficult using Lemma 1, but a littletedious. To give an example, let’s determine the possible orbits of with two elementsin the preperiod and period length ; that is, of the shape → b → a → a → · · · fordistinct nonzero integers a and b . Now b = b − divides f ( b ) − f (0) = a − b andso b divides a . Moreover, a = a − divides f ( a ) − f (0) = a − b and so a divides b . Therefore | b | = | a | , and so b = − a as a and b are distinct.We make several deductions from analyzing the four cases of the classification:1. is strictly preperiodic if and only if f (0) = f (0) = 0 ;2. L ( f ) = lcm [ f (0) , f (0)] =: ℓ ( f ) , as claimed in the proof of Theorem 1;3. x has a wandering orbit if and only if x = x : To see this, note that if x isperiodic, then x n +2 = x n for all n ≥ as the period length is either one or two.Moreover, if x is strictly preperiodic, then is strictly preperiodic for the map x → f ( x + x ) − x , with orbit → b → b → · · · where b n = x n − x . Inall four cases of our classification b = b , and so x = x .
6. PRIVATE PRIME FACTORS WHEN IS A PERIODIC POINT.
We nowturn our attention to maps x → f ( x ) for which is not strictly preperiodic, beginning6 c (cid:13) THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 7 with when is periodic with f of degree > . We have seen that the period must havelength one or two. When is periodic with period length 1. We must have f ( x ) = x r g ( x ) for some g ( x ) ∈ Z [ x ] with g (0) = 0 , and r ≥ . We will assume that g ( x ) has degree ≥ ,else the integers in the orbit of x contain only the prime factors of x and of g (0) , sodo not help us to construct infinitely many primes. Theorem 2.
Suppose that f ( x ) = x r g ( x ) for some nonconstant g ( x ) ∈ Z [ x ] with g (0) = 0 , and r ≥ . For any given wandering orbit ( x n ) n ≥ with x ∈ Z , define a n +1 := g ( x n ) gcd ( g ( x n ) , g (0)) for all n ≥ . The ( a n ) n ≥ are an infinite sequence of pairwise coprime integers and, once n issufficiently large, each a n has a private prime factor.Proof. Given a wandering orbit ( x n ) n ≥ we see that x m divides x m +1 = x rm g ( x m ) for all m ; and so x m divides x n whenever m ≤ n . Now a n +1 is a divisor of x n +1 forall n ≥ . If m < n , then ( g ( x m ) , g ( x n )) divides ( x m +1 , g ( x n )) , which divides ( x n , g ( x n )) ,which divides ( x n , g (0)) , which divides g (0) .Therefore ( a m +1 , a n +1 ) = 1 , that is, the ( a n ) n ≥ are pairwise coprime, and so a n hasa private prime factor provided | a n | > . Now a n = 0 , else x n +1 = 0 , which wouldcontradict the assumption that the orbit of x is wandering. If | a n | = 1 , then g ( x n ) isa divisor d , of g (0) , that is, x n must be a root of the polynomial Y d is a divisor of g (0) ( g ( x ) − d ) . But this has finitely many roots so, as the orbit of x is wandering, there can only befinitely many n for which | a n | = 1 . Therefore a n must have a private prime factor forall sufficiently large n . When is periodic with period length 2. We rule out the polynomials f ( x ) = a − x for any constant a , else there are no wandering orbits. Theorem 3.
Suppose that f ( x ) ∈ Z [ x ] with f ( x ) + x nonconstant, such that isperiodic under the map x → f ( x ) , with period length 2. We write f ( x ) = x r G ( x ) with G (0) = 0 and r ≥ . For any given wandering orbit ( x n ) n ≥ with x ∈ Z , define a n +2 := G ( x n ) gcd ( G ( x n ) , G (0) f (0)) for all n ≥ . The ( a n ) n ≥ are an infinite sequence of pairwise coprime integers and, once n issufficiently large, each a n has a private prime factor.Proof. We begin by showing that G ( x ) cannot be a constant, c . If so, then f ( x ) = cx r but f (0) = 0 . Therefore, f ( x ) has just one root, as f ( x ) has at least as manydistinct roots as f . Writing f ( x ) = c ( x − a ) d with a = 0 , the roots of f ( x ) are theroots of c ( x − a ) d − a , each repeated d times. This implies that d = 1 as the roots ofJanuary 2014] DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 8 c ( x − a ) d − a are distinct, and so f ( c ) = c x − ac ( c + 1) and so c = − ; that is, f ( x ) = a − x , which we have already ruled out.Now, by definition, a n +2 is a divisor of x n +2 for all n ≥ . Suppose that n >m ≥ . If n − m is even, then ( G ( x m ) , G ( x n )) divides G (0) , by the argument of theprevious subsection, and so ( a m +2 , a n +2 ) = 1 . If n − m is odd, then G ( x m ) , divides x m +2 and so divides x n +1 . On the other hand, G ( x n ) divides x n +2 = f ( x n +1 ) , andso ( G ( x m ) , G ( x m )) divides ( x n +1 , f ( x n +1 )) = ( x n +1 , f (0)) which divides f (0) ,and so ( a m +2 , a n +2 ) = 1 .Therefore the ( a n ) n ≥ are pairwise coprime, and so a n has a private prime factorprovided | a n | > . We then show, just as in the previous subsection, that a n = 0 , andthere are only finitely many n for which | a n | = 1 .We can combine the first three theorems to obtain the following result: Corollary 1.
Suppose that f ( x ) ∈ Z [ x ] is not of the form cx d or a − x , for anyconstants a or c , and that is preperiodic for the map x → f ( x ) . Then, for any givenwandering orbit ( x n ) n ≥ with x ∈ Z , each x n has a private prime factor for allsufficiently large n . We now turn our attention to the more difficult case of those f for which itselfhas a wandering orbit.
7. POLYNOMIAL MAPS WHERE HAS A WANDERING ORBIT
Henceforthwe will suppose that has a wandering orbit under the map x → f ( x ) , for somepolynomial f ( x ) ∈ Z [ x ] . We will investigate whether the terms of wandering integerorbits all have private prime factors from some point onwards. Where better to beginthan with the orbit that begins with ? Orbits that begin with are divisibility sequences. If n = m + k , then x n = f k ( x m ) ≡ f k (0) = x k (mod x m ) , and so ( x n , x m ) = ( x k , x m ) . Subtracting m several times from n , we deduce that if n = mq + r then ( x n , x m ) = ( x r , x m ) ,which mirrors that ( n, m ) = ( r, m ) , the key step in the Euclidean algorithm. There-fore, by applying the same steps as in the Euclidean algorithm we deduce that ( x m , x n ) = x ( m,n ) . Therefore ( x n ) n ≥ is a strong divisibility sequence ; for example, x n = 2 n − , theorbit of under the map x → f ( x ) = 2 x + 1 . Other examples of strong divisibil-ity sequences include the Fibonacci numbers and any second order linear recurrencesequence starting with , and the denominators of multiples of a given point on anelliptic curve.A divisibility sequence cannot have private prime factors, because if prime p divides x m then it divides x n for every n divisible by m . Instead one looks for primitive primefactors , that is, primes p n that divide x n but do not divide x m for any m, ≤ m < n .It is known that for any second order linear recurrence sequence beginning with a , and for any elliptic divisibility sequence, x n has a primitive prime factor for allsufficiently large n . The two key ideas in the proof are to show that the x n are fast-growing and that if p is prime factor of x n , but not a primitive prime factor, then p butnot p divides x n /x n/p . In particular, one deduces that x n is too large for all its primepower factors to be powers of imprimitive primes.In the example x n = 2 n − (the orbit of − under the map x → x + 3 ), and forgeneral wandering orbits, one cannot be so precise about the power to which imprim-itive primes p divide x n , and so it is an open question as to whether the x n all have8 c (cid:13) THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 9 primitive prime factors for sufficiently large n . Nonetheless, we expect that this is trueand, in the next subsection, we will consider this question following roughly the sameoutline. The growth of the x n and ( x m , x n ) . The x n grow very fast, and we can gain a quiteprecise understanding: If f ( x ) ∈ Z [ x ] has degree d > and if x is an integer whoseorbit is wandering, then there exist real numbers ≥ α > and β , which depend onlyon f , for which | x n | is the integer nearest to ατ d n + β, when n is sufficiently large , (1)where τ > is a constant that depends on both f and x . (We will sketch a proof of(1) in Section 8.) In the most interesting case, the orbit of , we write τ = τ .Since α ≤ , (1) implies that | x r | ≤ τ d r and | f r (0) | ≤ τ d r if r is sufficientlylarge. Suppose that ≤ m ≤ n − . Since ( x m , x n ) divides f n − m (0) as well as | x m | ,we deduce that ( x m , x n ) ≤ min {| x m | , f n − m (0) } ≤ τ d n/ ∗ (2)if n is sufficiently large, where τ ∗ = max { τ, τ } .We can therefore deduce that the x n grow so fast that the product of all of thegcd ( x m , x n ) with m < n is far smaller than x n , so it seems very likely that x n hasprimitive prime factors. However, this argument cannot be made into a proof sincesome of the primes that divide the x m with m < n might divide x n to an extraordi-narily high power. Ruling out too many large prime power divisors.
The abc conjecture can be used tobound solutions to equations in which all of the variables are divisible by surprisinglyhigh powers of primes. The idea is to study pairwise coprime integer solutions to theequation a + b = c. The abc -conjecture states, in a quantitative form, that a, b and c cannot all be divisibleonly by large powers of primes. For example if we have a putative solution to Fermat’slast theorem, like x n + y n = z n we can take a = x n , b = y n and c = z n , and the abc -conjecture should rule out any possible solution, at least once the numbers involvedare sufficiently large. One does so by bounding the size of a, b , and c in terms of theproduct of the distinct primes that divide a, b , and c . To put this in quantitative formrequires some ǫ ’s and that sort of thing.Fix ǫ > (which should be thought of as being small). There exists a constant κ ǫ > (which depends on ǫ , and chosen so it works for all examples), such that if a More generally, the height , h ( r ) , of a rational number r = p/q is given by max {| p | , | q |} ; this equals | r | if r is a nonzero integer. For any orbit { x n } n ≥ of a rational function φ ( t ) = f ( t ) /g ( t ) , with f ( t ) , g ( t ) ∈ Z [ t ] of degree d , there exists a constant τ > (the exponential of the dynamical canonical height, as in [ ] orSection 3.4 of [ ]), and constants c , c > such that c τ d n < h ( x n ) < c τ d n . Mochizuki has recently claimed to have proved the abc -conjecture, but the experts have as yet struggledto agree, definitively, that Mochizuki’s extraordinary ideas are all correct.
January 2014]
DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 10 and b are coprime positive integers with c = a + b , then c ≤ κ ǫ Y p prime p divides abc p ǫ . This is the abc -conjecture , one of the great questions of modern mathematics. In thespecial case that b = 1 we have c = a + 1 , and so a ≤ κ ǫ Y p prime p divides a ( a +1) p ǫ . The product here is over all primes p that divide the polynomial f ( x ) = x + x eval-uated at x = a . Langevin [ ] proved that a far-reaching generalization, which appliesto the prime divisors of any given polynomial, is actually a consequence of the abc -conjecture. Consequence of the abc -conjecture : Suppose that f ( x ) ∈ Z [ x ] has at least d distinctroots, for some d ≥ . For any fixed ǫ > , there exists a constant κ ǫ,f > , whichdepends on both ǫ and f , such that for any integer a we have | a | ≤ κ ǫ,f Y p prime p divides f ( a ) p / ( d − ǫ . We will apply this to our polynomial f with ǫ = 1 , writing κ for κ ,f for x n = f ( x n − ) . Therefore, if f has at least two distinct roots, then the abc -conjecture im-plies that | x n − | ≤ κ Y p prime p divides x n p . (3) The product of the primes if x n has no new prime factors. If every prime factor of x n already divides a previous term in the sequence, then we have that Y p prime p divides x n p divides n − Y m =0 Y p prime p divides ( x m ,x n ) p, which divides n − Y m =0 ( x m , x n ) . Taking the product of (2) over all m, ≤ m ≤ n − , we deduce that Y p prime p divides x n p ≤ n − Y m =0 ( x m , x n ) ≤ n τ nd n/ ∗
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 11 once n is sufficiently large. Substituting this into (3), gives | x n − | ≤ κ n τ nd n/ ∗ . Finally, comparing this to (1), taking logarithms, and dividing through by log τ , wesee that there exists a constant γ > (which can be given explicitly in terms of τ and τ ∗ ) such that, for large enough n , d n ≤ γnd n/ . This is false once n is sufficiently large, since d n grows much faster than nd n/ as n increases. Therefore we have proved the following conditional result: Theorem 4.
Assume the abc -conjecture. Suppose that f ( x ) ∈ Z [ x ] has at least twodistinct roots, and that and x both have wandering orbits under the map x → f ( x ) .Then x n has a primitive prime factor for all sufficiently large n . Note that this includes the possibility that x = 0 ; and so f n (0) has a primitiveprime factor for all sufficiently large n , assuming the abc -conjecture. 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 r , we define r n +1 = φ ( r n ) for all n ≥ , and write r n = a n /b n where gcd ( a n , b n ) = 1 and b n ≥ .Suppose that f (0) = 0 but f ( t ) is not of the form at d . Ingram and Silverman [ ]showed that if the orbit of r is not eventually periodic, then a n has a primitive primefactor for all sufficiently large n . It is believed that some result of this sort should holdwhether or not f (0) = 0 , and indeed this has been confirmed by Gratton, Nguyen, andTucker ([ ]) under the assumption of the abc -conjecture.
8. ASYMPTOTIC GROWTH OF WANDERING ORBITS.
Fix an integer d > .Let ( y n ) n ≥ be any sequence of positive integers for which y dn < y n +1 < ( y n + 1) d for all n ≥ . If ℓ n = y /d n n and u n = ( y n + 1) /d n , then ℓ n < ℓ n +1 < · · · < u n +1 ≤ u n . The ℓ n are therefore an increasing bounded sequence, so must tend to a limit, call it τ . Then ℓ n < τ ≤ u n for all n and so y n < τ d n ≤ y n + 1 , with equality in the upper bound ifand only if y n = τ d n − for all n . Therefore y n is the largest integer < τ d n . In fact y n = ⌊ τ d n ⌋ , other than in the special case in which τ is an integer and y n = τ d n − for all n .Mills [ ] observed, in 1947, that advances in our understanding of the distributionof primes imply that if N is sufficiently large, then there is a prime between N and ( N + 1) . So taking p = y to be any prime greater than N , and then selecting p n +1 to be any prime in ( p n , ( p n + 1) ) for each n ≥ , one obtains a Mills’ constant , τ ,such that ⌊ τ n ⌋ is the prime p n for every integer n ≥ .This proof yields a lot of flexibility in constructing τ with this property (as we expectmany primes in any interval of the form ( N , ( N + 1) ) ), so it seems unlikely thatthese numbers τ are anything special beyond the beautiful property identified by Mills. An analogous argument is used to bound the imprimitive prime factors of recurrence sequences, and in arelated context in [ ]. January 2014]
DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 12 Sketch of a modification to prove (1) . For any polynomial f ( x ) = ax d + bx d − + · · · with a > , the polynomial α − ( f ( x ) − β ) − ( α − ( x − β )) d , where α = a − / ( d − and β = − b/ ( da ) , has degree ≤ d − . Given an integer orbit ( x n ) n ≥ , let y n = α − ( x n − β ) , so that y n +1 − y dn can be written as a polynomial in y n of degree ≤ d − (where the coefficients of the polynomial are independent of n ).This implies that, for any ǫ > , if n is sufficiently large, then ( y n − ǫ ) d + ǫ < y n +1 < ( y n + ǫ ) d − ǫ. One can then modify the proof above to show that ( y n − ǫ ) /d n tends to a limit τ andthat y n − ǫ < τ d n < y n + ǫ . This implies that x n = αy n + β ∈ (cid:18) ατ d n + β − , ατ d n + β + 12 (cid:19) taking ǫ = 1 / α , and so x n is the nearest integer to ατ d n + β as claimed.There is a power series P ( T ) = c T + c + c − T + c − T + · · · which satisfies thefunctionalequation P ( x d ) = f ( P ( x )) . To determine the c i , we compare the coefficients on both sides of the equation: Com-paring the coefficients of T d implies that c = α ; comparing the coefficients of T d − implies that c = β . For each j ≥ , the coefficients of T d − − j involve polynomialsin the c i with i > − j as well as a linear term in c − j , which allows us to determine c − j .Assuming that P ( T ) absolutely converges once | T | is sufficiently large, if x m islarge enough there is a unique positive real t for which P ( t ) = x m . Then P ( t d ) = f ( P ( t )) = f ( x m ) = x m +1 , and P ( t d ) = f ( P ( t )) = f ( x m +1 ) = x m +2 , etc., so that P ( t d k ) = x m + k for all k ≥ . Taking τ = t /d m we deduce that x n = P ( τ d n ) for all sufficiently large n .One can sidestep the convergence issue by working with finite length truncations of P ( T ) . Thus if P k ( t ) = P j =1 − k c j t j , one can prove there exists a constant C k > such that | x n − P k ( τ d n ) | ≤ C k /τ kd n for all n ≥ .
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 13
The mysteries of the numbers τ ? Although this last subsection allows one to showthe existence of the constant τ , it gives no hint as to what kind of number τ is. The τ corresponding to the Fermat numbers is . But what kind of number is τ for Euclid’ssequence ( E n ) n ≥ ? Is it rational? Irrational? Related to π ? I have no idea. Also, ingeneral, is there anything special about the numbers τ , the τ -values that arise out ofthe wandering orbits of ? For a given f ( x ) , what does the set of τ -values look likeas we vary over all wandering orbits? Is each value isolated, or does the set have limitpoints? If we also vary over all f of given degree d , we have a countable subset of R > . Are the τ dense in R > ? Or are there holes in the spectrum? If so, why? Someof these questions are addressed in section 3.4 of [ ], and there are some particularlyinteresting results and conjectures there on lower bounds for values of τ that are > in terms of the given map and the field of definition. Gaining a better understanding ofthe numbers τ seems like a subject that is ripe for further and broader study.
9. APPENDIX: PRIVATE PRIME FACTORS FOR ALL N ≥ . One can im-prove Theorem 1 by noting that if | a n | = 1 (that is, x n is a divisor of L ( f ) ) then n = 0 , , or . Our proof is given by a tedious case-by-case analysis. For example,when → a → a with a > we proceed as follows.Suppose that x n +2 = d is a divisor of a , for some n ≥ , so that x n → x n +1 → d where x n , x n +1 , x n +2 = d, a are distinct. Now x ( x − a ) divides f ( x ) − a andso x n +1 ( a − x n +1 ) divides a − x n +2 = a − d , and x n ( a − x n ) divides x n +1 − a .Moreover, as d divides a and d = a we have ≤ a − d ≤ a .Let T = min { x n +1 , a − x n +1 } so that T ( a − T ) = x n +1 ( a − x n +1 ) divides a − d . Now T ≥ − , else a + 2) ≤ | T ( a − T ) | ≤ a . If T = − , then a + 1 di-vides a − d , so that d = − , and therefore x n +1 = a + 1 as x n +1 = T or a − T and x n +1 = d . Finally, x n ( a − x n ) divides x n +1 − a = 1 and so a = 2 and x n = 1 . Thatis, a = 2 and we have → → − .We may now assume that T ≥ , so that a − ≤ | x n ( a − x n ) | = a − x n +1 ≤ a − T ≤ a − . This implies that x n +1 = T = 1 and x n = a − as x n = x n +1 ,and a ≥ as x n = 0 , x n +1 . Moreover, a − divides a − d and d = 1 or a , so d =1 + k (1 − a ) for some k ≥ . If k ≥ , then | k (1 − a ) | > a unless k = 2 and a = 3 , which gives the example → → − . For k = 1 , we have − a divides a only if − a divides and so a = 3 or . The a = 3 case gives rise to the example → → − . The a = 4 case does not work, else x n − x n +1 would divide f ( x n ) − f ( x n +1 ) = x n +1 − x n +2 = x n +1 − d = 3 .Now n must be in all of these three examples else x n − → x n → x n +1 → d for some n ≥ . But a = 3 with x n − → → is impossible else x n − − divides − , so that x n − = t or x n − = a . And a = 2 with x n − → → → − is impossible else x n − − divides − and x n − − divides − ( −
1) = 2 ,which together imply that x n − = 2 = a . ACKNOWLEDGMENTS.
This article was first conceived of in 1997 following invigorating conversationsabout arithmetic dynamics with Mike Zieve. My understanding of arithmetic dynamics subsequently greatlybenefitted from discussions with Jeff Lagarias and Xander Faber. Thanks to them all. I would also like to thankthe referees for their many helpful suggestions.
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Compos. Math. (1993)163–205.2. A. W. F. Edwards, Infinite Coprime Sequences, The Mathematical Gazette, (1964) 416–422 January 2014]
DYNAMICAL SYSTEMS AND PRIMES athematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 143. C. Gratton, K. Nguyen, T. J. Tucker, ABC implies primitive prime divisors in arithmetic dynamics, Bull.Lond. Math. Soc. (2013) 1194–1208.4. P. Ingram, J. H. Silverman, Primitive divisors in arithmetic dynamics, Math. Proc. Cambridge Philos.Soc (2009) 289–302.5. M. Langevin, Partie sans facteur carr´e de F ( a, b ) (modulo la conjecture abc , in S´eminaire de Th´eorie desNombres (1993/94), Publ. Math. Univ. Caen6. W. H. Mills, A prime-representing function,
Bull. Amer. Math. Soc. (1947) 604.7. B. Rice, Primitive prime divisors in polynomial arithmetic dynamics, Integers (electronic), (2007),16 pages.8. J. H. Silverman,
The arithmetic of dynamical systems , Graduate Texts in Mathematics , Springer, NewYork, 2007.
ANDREW GRANVILLE ([email protected]) specializes in understanding the distribution of primes.He is co-author of the soon-to-appear graphic novel,
MSI: Anatomy and Permutations (P.U. Press)
D´epartement de math´ematiques et de statistique, Universit´e de Montr´eal, CP 6128 succ. Centre-Ville,Montr´eal, QC H3C 3J7, [email protected]; andDepartment of Mathematics, University College London, Gower Street, London, WC1E 6BT, United [email protected]
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