PRIME-GENERATING QUADRATIC POLYNOMIALS

. Let a,b,c be integers. We provide a necessary condition for the function | ax 2 + bx + c | to generate only primes for consecutive integers. We then apply this criterion to give sufficient conditions for the real quadratic field K = Q ( √ d ), d ∈ N , to have class number one, in terms of prime-producing quadratic polynomials.


Introduction
For centuries there has been a fascination with prime-producing quadratic polynomials. The best-known polynomial that generates (possibly in absolute value) many consecutive primes is x 2 − x + 41, due to Euler, which gives distinct primes for the 40  . Polynomials like this, which generate long strings of primes, are called primegenerating quadratic polynomials. Is there any incredible math hidden within these polynomials? In fact, there is a strong relationship between these polynomials and factorization in quadratic fields. At the 1912 International Congress of Mathematicians, Rabinowitsch gave a proof of the following for imaginary quadratic fields: n 2 + n + q is prime for n = 0, 1, . . . , q − 2 iff the imaginary quadratic field Q( √ 1 − 4q) has class number equal to 1. By the Heegner-Baker-Stark theorem, one now knows that there are exactly nine complex quadratic fields with class number one. They are Q( √ d) for d ∈ {−1, −2, −3, −7, −11, −19, −43, −67, −163}. For surveys of these and related results see, for instance, Mollin [4] and Ribenboim [10].
In 1980, applying Rabinowitsch's method, Kutsuna obtained the following for real quadratic fields: If −n 2 + n + q is prime for all positive n < √ q − 1, then the class number of the field Q( √ 1 + 4q) must necessarily be one. Subsequently, there have been many investigations of prime-producing polynomials and their connection to the structure of real quadratic fields. For this matter, we suggest reading Mollin's book [3]. In recent years, Byeon, Lee, and Mollin [2,5] proved an analogue statement to the Rabinowitsch result for real quadratic fields. Let m be a 188 VÍCTOR JULIO RAMÍREZ VIÑAS positive integer and let f m (x) be a polynomial of the form f m (x) = x 2 + x − m. We call a polynomial f m (x) a Rabinowitsch polynomial if, for t = ⌊ √ m⌋ and consecutive integers x = x 0 , x 0 + 1, . . . , x 0 + t − 1, |f m (x)| is either 1 or a prime for some integer x 0 . They proved the following theorem: f m (x) is a Rabinowitsch polynomial iff m ∈ {1, 2, 3, 4, 5, 7, 9, 13, 17, 19, 23, 25, 43, 49, 73, 103, 109, 169, 283}. Let d > 1 be an integer; We give elementary proofs of the following criteria for unique factorization in Z[α]. We remind the reader that if R is a domain then an irreducible element in R is a nonzero, nonunit element q ∈ R that cannot be written as the product of two non-units in R. A nonzero, nonunit element π ∈ R is called a prime if π | αβ, where α, β ∈ R, implies that π | α or π | β. The ring R is said to have the unique factorization property, or to be a unique factorization ring (unique factorization domain, abbreviated UFD), if every nonzero, nonunit, element in R can be expressed as a product of irreducible elements in exactly one way (where two factorizations are counted as the same if one can be obtained from the other by rearranging the order in which the irreducibles appear and multiplying them by units).
The distinction between primes and irreducibles is needed, because in a ring which is not a UFD these notions are not equivalent. Prime elements are always irreducible; the converse holds in a UFD, but not in general. Let Ω be the set of all primes p ∈ N satisfying p | a. Suppose that b 2 − 4ac = u 2 d for some integer u ≥ 1. Also suppose that, for every p ∈ Ω, the equation 4p = |x 2 − dy 2 | is solvable in integers x, y. If |an 2 + bn + c| is 1 or a prime for every integer n with ] is a unique factorization domain.
is solvable in integers x, y. If |an 2 + bn + c| is 1 or a prime for every integer n with is a unique factorization domain.

Some preliminaries
and p ∤ a, then there exists n ∈ Z such that and Now, we distinguish these two cases: p ̸ = 2 and p = 2. In the first case we get Combining (2.2) and (2.5), we get In the second case, when p = 2, we get that a is odd. As p is not prime in Z[α] and p = 2, by [8, Lemma 2.3] we deduce that d ̸ ≡ 5 (mod 8).
As for every q ∈ Ω the equation 4q = |x 2 − dy 2 | is solvable in integers x, y, we get that the equation q = |N (z)| is solvable in Z[α]. Proceeding by induction we deduce, by [ We claim that p ∤ a. Indeed, suppose that p ∈ Ω. Then according to our hypothesis there exist integers r and s such that Now, let u = r 2 and v = s 2 . Then, since d ≡ 2, 3 (mod 4), from (4.2) we get that u, v ∈ Z and p = |N (γ)|, where γ = u + vα, which is impossible because p is irreducible in Z[α]. So p ∤ a.

Applications
To show that our conditions are not impossible to use, we present easy proofs of the following seven propositions, which generalize and refine Mollin-Williams' [7,Theorems 3.2,3.3,and 3.4] and [6,Conjectures 2 and 4]. We remind the reader that the class number of a number field is by definition the order of the ideal class group of its ring of integers. Let K be a number field which is a finite extension field of the rational field Q. If an element α of C satisfies an algebraic equation α n + a n−1 α n−1 + · · · + a 1 α + a 0 = 0, where a 0 , a 1 , . . . , a n−1 ∈ Z, then α is called an algebraic integer. The set of all algebraic integers in K forms a ring, which is called the ring of integers in K. In general, it is not a unique factorization domain. Now let h be the class number of the number field K. It is now well known (see [1,Theorem 12.1.1,p. 300]) that the condition h = 1 is equivalent to the unique factorization property on the ring R of integers in K. Proof. Clearly d ≡ 2 (mod 4). Also we shall prove that the equation is solvable in integers x, y. Since |2n 2 −u 2 q| is 1 or a prime number for every integer n with x 0 ≤ n ≤ x 0 + 2 d 5 − 1, by Theorem 1.2 we conclude that the class number of the real quadratic field Q( √ d) is equal to 1. Now, since q is a prime congruent to 3 (mod 4), according to Walsh [11,Lemma 2.3] we get that the equation is solvable in integers x, y. Multiplying both sides of (5.1) by 2 we get and therefore the assertions above follow. □