Submitted Successfully!
To reward your contribution, here is a gift for you: A free trial for our video production service.
Thank you for your contribution! You can also upload a video entry or images related to this topic.
Version Summary Created by Modification Content Size Created at Operation
1 handwiki -- 1238 2022-10-31 01:50:16

Video Upload Options

Do you have a full video?


Are you sure to Delete?
If you have any further questions, please contact Encyclopedia Editorial Office.
HandWiki. Bunyakovsky Conjecture. Encyclopedia. Available online: (accessed on 12 April 2024).
HandWiki. Bunyakovsky Conjecture. Encyclopedia. Available at: Accessed April 12, 2024.
HandWiki. "Bunyakovsky Conjecture" Encyclopedia, (accessed April 12, 2024).
HandWiki. (2022, October 31). Bunyakovsky Conjecture. In Encyclopedia.
HandWiki. "Bunyakovsky Conjecture." Encyclopedia. Web. 31 October, 2022.
Bunyakovsky Conjecture

The Bunyakovsky conjecture (or Bouniakowsky conjecture) gives a criterion for a polynomial [math]\displaystyle{ f(x) }[/math] in one variable with integer coefficients to give infinitely many prime values in the sequence[math]\displaystyle{ f(1), f(2), f(3),\ldots. }[/math] It was stated in 1857 by the Russian mathematician Viktor Bunyakovsky. The following three conditions are necessary for [math]\displaystyle{ f(x) }[/math] to have the desired prime-producing property: Bunyakovsky's conjecture is that these conditions are sufficient: if [math]\displaystyle{ f(x) }[/math] satisfies (1)-(3), then [math]\displaystyle{ f(n) }[/math] is prime for infinitely many positive integers [math]\displaystyle{ n }[/math].

bunyakovsky bouniakowsky sequencemath\displaystyle{

1. Discussion of Three Conditions

We need the first condition because if the leading coefficient is negative then [math]\displaystyle{ f(x) \lt 0 }[/math] for all large [math]\displaystyle{ x }[/math], and thus [math]\displaystyle{ f(n) }[/math] is not a (positive) prime number for large positive integers [math]\displaystyle{ n }[/math]. (This merely satisfies the sign convention that primes are positive.)

We need the second condition because if [math]\displaystyle{ f(x) = g(x)h(x) }[/math] where the polynomials [math]\displaystyle{ g(x) }[/math] and [math]\displaystyle{ h(x) }[/math] have integer coefficients, then we have [math]\displaystyle{ f(n) = g(n)h(n) }[/math] for all integers [math]\displaystyle{ n }[/math]; but [math]\displaystyle{ g(x) }[/math] and [math]\displaystyle{ h(x) }[/math] take the values 0 and [math]\displaystyle{ \pm 1 }[/math] only finitely many times, so [math]\displaystyle{ f(n) }[/math] is composite for all large [math]\displaystyle{ n }[/math].

The third condition, that the numbers [math]\displaystyle{ f(n) }[/math] have gcd 1, is obviously necessary, but is somewhat subtle, and is best understood by a counterexample. Consider [math]\displaystyle{ f(x) = x^2 + x + 2 }[/math], which has positive leading coefficient and is irreducible, and the coefficients are relatively prime; however [math]\displaystyle{ f(n) }[/math] is even for all integers [math]\displaystyle{ n }[/math], and so is prime only finitely many times (namely when [math]\displaystyle{ f(n)=2 }[/math], in fact only at [math]\displaystyle{ n =0,-1 }[/math]).

In practice, the easiest way to verify the third condition is to find one pair of positive integers [math]\displaystyle{ m }[/math] and [math]\displaystyle{ n }[/math] such that [math]\displaystyle{ f(m) }[/math] and [math]\displaystyle{ f(n) }[/math]are relatively prime. We describe a general way to calculate the gcd of [math]\displaystyle{ \{f(n)\}_{n\geq 1}. }[/math] Any integer-valued polynomial [math]\displaystyle{ f(x) = c_0 + c_1x + \cdots + c_dx^d }[/math]can be written in the basis of binomial coefficient polynomials:

[math]\displaystyle{ f(x) = a_0 + a_1\binom{x}{1} + \cdots + a_d\binom{x}{d} }[/math]

where each [math]\displaystyle{ a_i }[/math] is an integer, and [math]\displaystyle{ \gcd\{f(n)\}_{n \geq 1} \ = \ \gcd(a_0,a_1,\dots,a_d). }[/math]

For the above example, we have:

[math]\displaystyle{ x^2 - x + 2 = 2\binom{x}{2} + 2, }[/math]

and the coefficients in the second formula have gcd 2, which implies that [math]\displaystyle{ x^2 - x + 2 }[/math] has even values on the integers.

Using this gcd formula, it can be proved [math]\displaystyle{ \gcd\{f(n)\}_{n \geq 1} =1 }[/math] if and only if there are positive integers [math]\displaystyle{ m }[/math] and [math]\displaystyle{ n }[/math] such that [math]\displaystyle{ f(m) }[/math] and [math]\displaystyle{ f(n) }[/math] are relatively prime.

2. Examples

2.1. [math]\displaystyle{ f(x)=x^2+1 }[/math]

An example of Bunyakovsky's conjecture is the polynomial f(x) = x2 + 1, for which some prime values are listed below. (Values of x form OEIS sequence A005574; those of x2 + 1 form A002496)

x 1 2 4 6 10 14 16 20 24 26 36 40 54 56 66 74 84 90 94 110 116 120
x2 + 1 2 5 17 37 101 197 257 401 577 677 1297 1601 2917 3137 4357 5477 7057 8101 8837 12101 13457 14401

That [math]\displaystyle{ n^2+1 }[/math] should be prime infinitely often is a problem first raised by Euler, and it is also the fifth Hardy–Littlewood conjecture and the fourth of Landau's problems. Despite the extensive numerical evidence, it is not known that this sequence extends indefinitely.

2.2. Cyclotomic Polynomials

The cyclotomic polynomials [math]\displaystyle{ \Phi_k(x) }[/math]for [math]\displaystyle{ k=1,2,3,\ldots }[/math] satisfy the three conditions of Bunyakovsky's conjecture, so for all k, there should be infinitely many natural numbers n such that [math]\displaystyle{ \Phi_k(n) }[/math] is prime. It can be shown that if for all k, there exists an integer n > 1 with [math]\displaystyle{ \Phi_k(n) }[/math] prime, then for all k, there are infinitely many natural numbers n with [math]\displaystyle{ \Phi_k(n) }[/math] prime.

The following sequence gives the smallest natural number n > 1 such that [math]\displaystyle{ \Phi_k(n) }[/math] is prime, for [math]\displaystyle{ k=1,2,3,\ldots }[/math]:

3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 6, 2, 4, 3, 2, 10, 2, 22, 2, 2, 4, 6, 2, 2, 2, 2, 2, 14, 3, 61, 2, 10, 2, 14, 2, 15, 25, 11, 2, 5, 5, 2, 6, 30, 11, 24, 7, 7, 2, 5, 7, 19, 3, 2, 2, 3, 30, 2, 9, 46, 85, 2, 3, 3, 3, 11, 16, 59, 7, 2, 2, 22, 2, 21, 61, 41, 7, 2, 2, 8, 5, 2, 2, ... (sequence A085398 in the OEIS).

This sequence is known to contain some large terms: the 545th term is 2706, the 601st is 2061, and the 943rd is 2042. This case of Bunyakovsky's conjecture is widely believed, but again it is not known that the sequence extends indefinitely.

Usually, there is integer 2≤n≤φ(k) such that [math]\displaystyle{ \Phi_k(n) }[/math] is prime (note that the degree of [math]\displaystyle{ \Phi_k(n) }[/math] is φ(k)), but there are exceptions, the exception numbers k are

1, 2, 25, 37, 44, 68, 75, 82, 99, 115, 119, 125, 128, 159, 162, 179, 183, 188, 203, 213, 216, 229, 233, 243, 277, 289, 292, ...

3. Partial Results: Only Dirichlet's Theorem

To date, the only case of Bunyakovsky's conjecture that has been proved is that of polynomials of degree 1. This is Dirichlet's theorem, which states that when [math]\displaystyle{ a }[/math] and [math]\displaystyle{ m }[/math] are relatively prime integers there are infinitely many prime numbers [math]\displaystyle{ p \equiv a \pmod m }[/math]. This is Bunyakovsky's conjecture for [math]\displaystyle{ f(x) = a + mx }[/math] (or [math]\displaystyle{ a - mx }[/math] if [math]\displaystyle{ m \lt 0 }[/math]). The third condition in Bunyakovsky's conjecture for a linear polynomial [math]\displaystyle{ mx + a }[/math] is equivalent to [math]\displaystyle{ a }[/math] and [math]\displaystyle{ m }[/math] being relatively prime.

No single case of Bunyakovsky's conjecture for degree greater than 1 is proved, although numerical evidence in higher degree is consistent with the conjecture.

4. Generalized Bunyakovsky Conjecture

Given k ≥ 1 polynomials with positive degrees and integer coefficients, each satisfying the three conditions, assume that for any prime p there is an n such that none of the values of the k polynomials at n are divisible by p. Given these assumptions, it is conjectured that there are infinitely many positive integers n such that all values of these k polynomials at x = n are prime.

Note that the polynomials {x, x + 2, x + 4} do not satisfy the assumption, since one of their values must be divisible by 3 for any integer x = n. Neither do {x, x2 + 2}, since one of the values must be divisible by 3 for any x = n. On the other hand, {x2 + 1, 3x - 1, x2 + x + 41} do satisfy the assumption, and the conjecture implies the polynomials have simultaneous prime values for infinitely many positive integers x = n.

This conjecture includes as special cases the twin prime conjecture (when n = 2, and the two polynomials are x and x + 2) as well as the infinitude of prime quadruplets (when n = 4, and the four polynomials are x, x + 2, x + 6, and x + 8), sexy primes (when n = 2, and the two polynomials are x and x + 6), Sophie Germain primes (when n = 2, and the two polynomials are x and 2x + 1), and Polignac's conjecture (when n = 2, and the two polynomials are x and x + k, with k any even number). When all the polynomials have degree 1 this is Dickson's conjecture.

In fact, this conjecture is equivalent to the Generalized Dickson conjecture.

Except for Dirichlet's theorem, no case of the conjecture has been proved, including the above cases.

Subjects: Others
Contributor MDPI registered users' name will be linked to their SciProfiles pages. To register with us, please refer to :
View Times: 326
Entry Collection: HandWiki
Revision: 1 time (View History)
Update Date: 31 Oct 2022