On the composite values of an Irreducible Polynomial

Transcription

1 On the composite values of an Irreducible Polynomial Luca Goldoni Liceo scientifico A.F. Formiggini Sassuolo,Italy Abstract Some remarks on the set of composite values of an irreducible Polynomial P (x) Z[x]. We ll get a necessary and sufficient condition on n N to let P(n) a non prime number.in this way we will able to sieve-out all the composite values of P(x) even if we still far to prove that the set of prime values of P(x) is infinite. It s well known the topics of primes among the values of a given polynomial.we can get a summarize as follow: No polynomial Z[x] can takes only prime values. Every polynomial of first degree P (x) = ax + b with a and b coprimes, takes infinitely many prime values.( Dirichlet theorem). If a polynomial takes infinitely many prime values then it is irreducible on Z[X] Every reducible polynomial takes at most finitely many prime values. there are irreducibles polynomials witch never takes prime values, because there is a prime p such that p P (x) x Z (for instance P (x) = x 2 + x + 4). So we can asking what is the situation if : 1. P (x) Z[x] is irreducible; 2. p P such that p P (x) x Z 1

2 From long time it has been conjectured that in this case the polynomial takes infinitely many prime values,but until now none was able to prove this even for the simplest polynomial: P (x) = x We will see in this paper that,on the contrary, it s quite easy to get a necessary and sufficient condition to find the set A of all the composite values of the polynomial in terms of his primes divisors.one can think that in this way the problem is near to be solved, but it s not like this.infact it s seems beyond the resources of the Mathematics of our time to prove that the complementary set of A.In any case it seems interesting to study the set A because it s possible to develop an Erathosten sieve alike able to get us the primes in a finite interval. [0, n] n N No lover-bound for the amount of primes in such interval is possible by means of this method, so we can t say anything about the asymptotic behavior of P(n), if P(n) is the number of such primes in the quoted interval,but in any case it s maybe interesting to analyze the situation even this seems only an algorithm. We start to observe the following property: Theorem 1 Let P(x) Z[x]then for infinitely many x Z P (x) / P. Proof If P (x) is reducible or if there is a prime q such that q P (x) x Z the theorem is trivial.if it is no so, we proceed as follow: let let let A = {p P : n N : P (n) = p} (1) B = {p P : n N P (n) p n N : P (n) 0 mod p} (2) If p A then the equation U = A B (3) P (x) = p (4) has at least a solution n by definition of A.So if we consider x p,k = pk + n with k 1, we have P (x p,k ) = a h (n + pk) h (5) 2

3 so remembering that a h (a + b) m h = a h a m h I m 1 + m h a h C m h,j a m h j b j II (6) and that b II,with a = n and b = pk and C m h,j = P (x p,k ) = but by hypothesis a h (n) m h I m 1 + m h a h (m h)! j!(m h j)! C m h,j (n) m h j (pk) j II (7) a h (n) m h = p (8) so p P (x p,k ) and so P (x p,k ) is composite for infinitely many integer values of x. If p B in the same way we are able to build an integer sequence witch let P(x) composite too.of course U so we have proved the theorem. Theorem 1 suggest how to implement an Erathosten sieve-alike process.let p U.If p A, then we solve the equation and we set: P (x) = p (9) Γ p = {x Z : P (x) = 0 P (x) 0} (10) In short words (10) is the set of distinct integral roots of P (x).we can write the elements of (10) as: x 1 < x 2 < x sp (11) with s p m,variable with p.so we can build s sequences: x 1 p,k = kp + x 1 k Z {0}. x sp p,k = kp + x s k Z {0} (12) and we have that on the points of these sequences P (x) is not prime. If p B,then we solve the congruence: P (x) 0 mod p (13) 3

4 From a well known theorem of Lagrange, we have that there are at most m solutions for (13) and so we can set Γ p as the set of distinct roots of (13).So in this case we can consider the sequences: x 1 p,k = kp + x 1. x sp p,k = kp + x s k Z k Z where, of course s p is the number of elements in Γ p.even in this case, we have that P (x) is no prime over the points of these sequences. we define now,for every p U (14) Λ p,j = { { x j p,k, k Z {0}} if p A { x j p,k, k Z} if p B (15) and then Ω p = s p Λ p,j (16) We can call Ω p as the set of integers x associated with p which let P (x) not prime. Now we set: C = p U Ω p (17) and Of course and it s quite easy to show that D = {x Z : P (x) / P} (18) C D (19) D C (20) Indeed if x D then P (x) = p 1 p 2 p r so for instance, x Ω p1.so we have that: Theorem 2 For every x Z, P (x) is prime if and only if x / C Remark: By means of this result it is possible develop an algorithm quite similar to the Erathosten sieve to find the prime values of the form n Indeed with the previously determined arithmetic sequences it s possible to delete every non prime values of n from 1 to M where M is a fixed natural number. 4

5 References [1] G.H Hardy and E.M Wright An Introduction to the Theory of numbers fifth edition 1979 Oxford Science Pubblications [2] O. Ore Number Theory and Its History Dover reprint 1976 [3] W. Sierpinski Elementary theory of numbers Monografie Matematyczne Tom 42 Warszawa 1964 electronic edition 5