Bunyakovsky conjecture

The first condition is necessary because if the leading coefficient is negative then ${\displaystyle f(x)<0}$ for all large ${\displaystyle x}$, and thus ${\displaystyle f(n)}$ is not a (positive) prime number for large positive integers ${\displaystyle n}$. (This merely satisfies the sign convention that primes are positive.)

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

The second condition also fails for the polynomials reducible over the rationals.

For example, the integer-valued polynomial ${\displaystyle P(x)=(1/12)\cdot x^{4}+(11/12)\cdot x^{2}+2}$ doesn't satisfy the condition (2) since ${\displaystyle P(x)=(1/12)\cdot (x^{4}+11x^{2}+24)=(1/12)\cdot (x^{2}+3)\cdot (x^{2}+8)}$, so at least one of the latter two factors must be a divisor of ${\displaystyle 12}$ in order to have ${\displaystyle P(x)}$ prime, which holds only if ${\displaystyle |x|\leq 3}$. The corresponding values are ${\displaystyle 2,3,7,17}$, so these are the only such primes for integral ${\displaystyle x}$ since all of these numbers are prime. This isn't a counterexample to Bunyakovsky conjecture since the condition (2) fails.

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

In practice, the easiest way to verify the third condition is to find one pair of positive integers ${\displaystyle m}$ and ${\displaystyle n}$ such that ${\displaystyle f(m)}$ and ${\displaystyle f(n)}$ are relatively prime. In general, for any integer-valued polynomial ${\displaystyle f(x)=c_{0}+c_{1}x+\cdots +c_{d}x^{d}}$ we can use ${\displaystyle \gcd\{f(n)\}_{n\geq 1}=\gcd(f(m),f(m+1),\dots ,f(m+d))}$ for any integer ${\displaystyle m}$, so the gcd is given by values of ${\displaystyle f(x)}$ at any consecutive ${\displaystyle d+1}$ integers.[1] In the example above, we have ${\displaystyle f(-1)=2,f(0)=2,f(1)=4}$ and so the gcd is ${\displaystyle 2}$, which implies that ${\displaystyle x^{2}+x+2}$ has even values on the integers.

Alternatively, when an integer polynomial ${\displaystyle f(x)}$ is written in the basis of binomial coefficient polynomials

$${\displaystyle f(x)=a_{0}+a_{1}{\binom {x}{1}}+\cdots +a_{d}{\binom {x}{d}},}$$

each coefficient ${\displaystyle a_{i}}$ is an integer and ${\displaystyle \gcd\{f(n)\}_{n\geq 1}=\gcd(a_{0},a_{1},\dots ,a_{d}).}$ In the example above, this is

$${\displaystyle x^{2}+x+2=2{\binom {x}{2}}+2{\binom {x}{1}}+2,}$$

and the coefficients in the right side of the equation have gcd 2.

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

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 ${\displaystyle a}$ and ${\displaystyle m}$ are relatively prime integers there are infinitely many prime numbers ${\displaystyle p\equiv a{\pmod {m}}}$. This is Bunyakovsky's conjecture for ${\displaystyle f(x)=a+mx}$ (or ${\displaystyle a-mx}$ if ${\displaystyle m<0}$). The third condition in Bunyakovsky's conjecture for a linear polynomial ${\displaystyle mx+a}$ is equivalent to ${\displaystyle a}$ and ${\displaystyle m}$ 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.

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

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

This conjecture includes as special cases the twin prime conjecture (when ${\displaystyle k=2}$, and the two polynomials are ${\displaystyle x}$ and ${\displaystyle x+2}$) as well as the infinitude of prime quadruplets (when ${\displaystyle k=4}$, and the four polynomials are ${\displaystyle x,x+2,x+6}$, and ${\displaystyle x+8}$), sexy primes (when ${\displaystyle k=2}$, and the two polynomials are ${\displaystyle x}$ and ${\displaystyle x+6}$), Sophie Germain primes (when ${\displaystyle k=2}$, and the two polynomials are ${\displaystyle x}$ and ${\displaystyle 2x+1}$), and Polignac's conjecture (when ${\displaystyle k=2}$, and the two polynomials are ${\displaystyle x}$ and ${\displaystyle x+a}$, with ${\displaystyle a}$ 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.

• Integer-valued polynomial
• Cohn's irreducibility criterion
• Schinzel's hypothesis H
• Bateman–Horn conjecture
• Hardy and Littlewood's conjecture F