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miércoles, 26 de abril de 2017

A play on the interplay 'twixt primes and polynomials

In what follows, we shall denote the set of positive prime numbers by \mathbf{P}.

Act I. It is more or less well-known that there does not exist a non-constant polynomial f \in \mathbb{Z}[x] such that f(n) \in \mathbf{P} for every n \in \mathbb{N}. This can be proven by reductio ad absurdum: if f(n) \in \mathbf{P} for every n \in \mathbb{N} and f(1) =: p, then p \mid f(1+kp) for every k \in \mathbb{N}; it follows that at least one of the equations f(x)=p or f(x)=-p has more solutions than \deg(f), Q. E. A. This result is typically attributed to Christian Goldbach: W. Narkiewicz, on page 25 of his "The Development of Prime Number Theory", even mentions that it can be found in a letter from Goldbach to Euler written on September 28th, 1743. Luckily for us, Springer-Verlag published two years ago a translation into English of the correspondence of L. Euler with C. Goldbach edited and commented by F. Lemmermeyer and M. Mattmüller.

Act II. Several years ago, while perusing a very interesting article on primes in arithmetic progressions by M. R. Murty, I learned the notion of prime divisor of a polynomial: if p \in \mathbf{P} and f \in \mathbb{Z}[x], we say that p is a prime divisor of f if p \mid f(n) for some n \in \mathbb{Z}. According to Murty, the basic theorem on the set of prime divisors of a non-constant f \in \mathbb{Z}[x] can be traced back (at least) to a 1912 paper of I. Schur. The theorem can be proven emulating the celebrated proof of Eucl. IX-20.

Theorem. If f is a non-constant polynomial of integer coefficients, then its set of prime divisors is infinite.

Proof. If f(0)=0, then every p \in \mathbf{P} is a prime divisor of f. If f(0) = c \neq 0, then f has at least one prime divisor as it can take on the values \pm 1 only finitely many times. Given any finite set \mathcal{P}_{k}:=\{p_{1}, \ldots, p_{k}\} of prime divisors of f, we are to show that we can always find another prime divisor of f which does not belong to \mathcal{P}_{k}. Let A := p_{1} \cdots p_{k} and consider the equality f(Acx)=cg(x) where g \in \mathbb{Z}[x] is a polynomial of the form 1+c_{1}x+c_{2}x^{2}+\cdots where every c_{i} is a multiple of A. Since g is a non-constant polynomial whose constant term is different from 0, g has at least one prime divisor p. Clearly enough, this prime number p is also prime divisor of f which does not belong to \mathcal{P}_{k}. Q.E.D.

Act III. Resorting to the ideas in the previous paragraphs plus Dirichlet's glorious theorem on primes in arithmetic progressions, we are going to determine all the non-constant polynomials f \in \mathbb{Z}[x] such that f(\mathbf{P}) \subseteq \mathbf{P}.

- If f is one such polynomial and f(0)=0, then f(x)=xg(x) for some some g \in \mathbb{Z}[x]. Given that p \cdot g(p) =f(p) \in \mathbf{P} for every p \in \mathbf{P}, it follows that g(p) = \pm 1 for every p \in \mathbf{P}; therefore, in this case we obtain that either g(x)=1 and f(x) =x or g(x)=-1 and f(x)=-x.
- Let us assume now that f is one such polynomial and f(0)=c \neq 0. By the above theorem, we may fix a prime divisor q of f which is greater than |c|. If n \in \mathbb{Z} is a witness of the fact that q is a prime divisor of f, then q \nmid n. Thus, if p_{1} < p_{2} < p_{3} < \ldots are all the positive primes in the arithmetic progression whose first term is n and whose common difference is q, we have that f(p_{i}) \equiv f(n) \equiv 0 \pmod{q} for every i \in \mathbb{N}, which is decidedly absurd because f can assume the values \pm q only finitely many times.

Hence, f(x)=x is the only non-constant polynomial with integer coefficients which sends \mathbf{P} to one of its subsets.

THE END.

2 comentarios:

Octavio dijo...

Wow, bro! I liked this one very much.

José Hdz. Stgo. dijo...

I am very glad you liked it! By the way, I recently left a comment for you in the entry on the Aristippus anecdote.

With warm regards,

J.H.S.