Using Dynamical Systems to Construct Infinitely Many Primes
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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 Construct Infinitely 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 that there are infinitely many primes, which appears in Euclid’s Elements, is a proof by contradiction. This proof does not indicate how to find primes, despite establishing that there are infinitely many of them. However a minor variant does do so. The key to either proof is 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 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, the prime factors of the an, as proved in the following result. 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. To construct such a sequence by modifying Euclid’s proof, let E0 = 2 and arXiv:1708.06953v1 [math.NT] 23 Aug 2017 En = E0E1 · · · En−1 + 1 for each n ≥ 1. If m<n, then Em divides E0E1 · · · En−1 = En − 1 (as Em is one of the terms in the product) and so gcd(Em, En) divides gcd(En − 1, En) = 1, which implies that gcd(Em, En) = 1. Therefore, if pn is a prime divisor of En for each n ≥ 0, then p0,p1,... is an infinite sequence of distinct primes. 2n The Fermat numbers, F0, F1,..., defined by Fn = 2 + 1 for each n ≥ 0, are a more familiar sequence of pairwise coprime integers. Fermat had actually conjectured that the Fn are all primes. 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 January 2014] DYNAMICAL SYSTEMS AND PRIMES 1 Mathematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 2 1 question as to whether there are infinitely many primes of the form Fn. Nonetheless we can prove that the Fn are pairwise coprime in a similar way to the En, using the identity Fn = F0F1 · · · Fn−1 + 2 for each n ≥ 1 and the fact that the Fn are all odd. We deduce from Proposition 1 that if pn is a prime divisor of Fn for each n ≥ 0, then p0,p1,... is an infinite sequence of distinct primes.2 For any sequence of integers (an)n≥0 we call pn a private prime factor of an if it divides an and no other am. The primes pn constructed in Proposition 1 are therefore each private prime factors of the an. 2. A SIMPLER FORMULATION. In Section 1, the sequences (En)n≥0 and (Fn)n≥0 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 to view these two sequences, with an eye to generalization. This is not difficult because the recurrence for the En can be rewritten as En+1 = E0E1 · · · En−1 · En +1=(En − 1)En +1= f(En), 2 where f(x)= x − x + 1. Similarly the Fn-values can be determined by 2n 2n Fn+1 = (2 + 1)(2 − 1)+2= Fn(Fn − 2)+1= f(Fn), 2 where f(x) = x − 2x + 2. So they are both examples of sequences (xn)n≥0 for which xn+1 = f(xn) for some polynomial f(x) ∈ Z[x]. The terms of the sequence are all given by the recursive formula, so that n xn = f(f(...f(x0))) = f (x0), n times | {z } where the notation f n denotes the polynomial obtained by composing f with itself n times (which is definitely not the nth power of f). Any such sequence (xn)n≥0 is called the orbit of x0 under the map f, since the sequence is completely determined once one knows x0 and f. We sometimes write the orbit as x0 → x1 → x2 →··· . The key to the proof that the En’s are pairwise coprime is that En ≡ 1 (mod Em) 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 deduce using the following lemma from elementary number theory. 1 The only Fermat numbers known to be primes have n ≤ 4. We know that the Fn are composite for 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 with n > 4 are composite, or they might all be prime from some sufficiently large n onwards, or some might be prime and some composite. Currently, we have no way of knowing which is true. 2Goldbach 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. 2 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 Mathematical 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 6= 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)= x2 − x + 1, the orbit of 0 under the map f is 0 → 1 → 1 → ··· . Therefore if n>m ≥ 0, then we have n−m n−m En = f (Em) ≡ f (0) = 1 (mod Em), and so (En, Em) = (1, Em) = 1. (For clarity, we have used an “=” sign when two numbers are equal despite working with a congruence. Typically we use the notation (a, b) rather than gcd(a, b).) Similarly if f(x) = x2 − 2x + 2, then the orbit of 0 n−m n−m under the map f is 0 → 2 → 2 → ··· and so Fn = f (Fm) ≡ f (0) = 2 (mod Fm), which implies that if n>m, then (Fn, Fm) = (2, Fm) = 1. This reformulation of two of the best-known proofs of the infinitude of primes hints at the possibility of a more general approach. 3. DIFFERENT STARTING POINTS. We just saw that the orbit of 2 under the map x → x2 − x + 1 is an infinite sequence of pairwise coprime integers, 2 → 3 → 7 → 43 → 1807 →··· . What about other orbits? The orbit of 4 is also an infinite sequence of pairwise coprime integers beginning 4 → 13 → 157 → 24493 →··· , as is the orbit of 5 which begins 5 → 21 → 421 →··· . The same proof as before yields that no two integers in a given orbit have a common factor. The orbit of 3 under the the map x → x2 − 2x + 2 yielded the Fermat numbers 3 → 5 → 17 → 257 →··· , but starting at 4 we get 4 → 10 → 82 → 6562 →··· . These are obviously not pairwise prime as every number in the orbit is even, but if we divide through by 2, then we get 2 → 5 → 41 → 3281 →··· 2 which are pairwise coprime. To prove this, note that x0 = 4 and xn+1 = xn − 2xn + 2 for all n ≥ 0, and so the above proof yields that if m<n then (xm,xn)= (xm, 2) = 2. Therefore, taking an = xn/2 for every n we deduce that (am, an) = (xm/2,xn/2) = (xm/2, 1) = 1. This same idea works for every orbit under this map: we get an infinite sequence of pairwise coprime integers by dividing through by 1 or 2, depending on whether x0 is odd or even. However, things can be more complicated: Consider the orbit of x0 = 3 under the map x → x2 − 6x − 1. We have 3 →−10 → 159 → 24326 → 591608319 →··· . Here xn is divisible by 3 if n is even, and is divisible by 2 if n is odd. If we let an = xn/3 when n is even, and an = xn/2 when n is odd, then one can show the terms of the resulting sequence, 1 →−5 → 53 → 12163 → 197202773 →··· , are indeed pairwise coprime. Another surprising example is given by the orbit of 6 under the map x → 7+ x5(x − 1)(x − 7). Reducing the elements of the orbit mod 7 we find that 6 → 5 → 4 → 3 → 2 → 1 → 0 → 0 →··· (mod 7), January 2014] DYNAMICAL SYSTEMS AND PRIMES 3 Mathematical Assoc. of America American Mathematical Monthly 121:1 August 24, 2017 12:20 a.m. MonthlyArticleDynamicsOnly.v5.tex page 4 as x5(x − 1)(x − 7) ≡ x6(x − 1) (mod 7), which is ≡ x − 1 (mod 7) if x 6≡ 0 (mod 7) by Fermat’s little theorem.