# Bunyakovsky conjecture

The Bunyakovsky conjecture (or Bouniakowsky conjecture) stated in 1857 by the Russian mathematician Viktor Bunyakovsky, asserts when a polynomial $f(x)$ in one variable with positive degree and integer coefficients should have infinitely many prime values for positive integer inputs. Three necessary conditions are

1. the leading coefficient of $f(x)$ is positive,
2. the polynomial is irreducible over the integers, and
3. as $n$ runs over the positive integers the numbers $f(n)$ should not share a common factor greater than 1.

Bunyakovsky's conjecture is that these three conditions are sufficient: if $f(x)$ satisfies the three conditions then $f(n)$ is prime for infinitely many positive integers $n$.

We need the first condition because if the leading coefficient is negative then $f(x) < 0$ for all large $x$, and thus $f(n)$ is not a prime number for large positive integers $n$. If we allow negative primes like -2, -3, -5,... to count as prime numbers then this first condition can be dropped; the second and third necessary conditions are more substantial, as we will see below, since they imply $f(n)$ can be prime only finitely many times for reasons that are more serious than a sign problem.

We need the second condition because if $f(x) = g(x)h(x)$ where the polynomials $g(x)$ and $h(x)$ have integral coefficients and are not $\pm 1$ then we have $f(n) = g(n)h(n)$ for all integers $n$, so $f(n)$ is composite for all large $n$ (because $g(x)$ and $h(x)$ take the values 0 and $\pm 1$ only finitely many times).

The third condition, that the numbers $f(n)$ have gcd 1, is the most technical sounding and is best understood by an example where it does not hold. Consider the polynomial $f(x) = x^2 - x + 2$. It has a positive leading coefficient and is irreducible, but $f(n)$ is even for all integers $n$, so the values of this polynomial are prime only finitely many times on the positive integers (namely when it takes the value 2, which is actually only at $n = 1$ among positive integers).

In practice, the easiest way to verify the third condition for a polynomial $f(x)$ is to find one pair of positive integers $m$ and $n$ such that $f(m)$ and $f(n)$ are relatively prime: when this happens no integer greater than 1 can divide all values of $f(x)$ on the positive integers because it would have to divide $f(m)$ and $f(n)$.

An example of Bunyakovsky's conjecture is the polynomial f(x) = x2 + 1, for which some of the prime values that it has on positive integers are listed below.

 x x2 + 1 1 2 4 6 10 14 16 20 24 26 36 2 5 17 37 101 197 257 401 577 677 1297

That $n^2+1$ should be prime infinitely often is a problem first raised by Euler, and it is also the fifth Hardy–Littlewood conjecture.

The third condition in Bunyakovsky's conjecture is saying that the set of integers $f(1),f(2), f(3),\dots$ has gcd 1. It is a surprise to most people at first that this is not the same as saying the coefficients of $f(x)$ are relatively prime together, but the example of $x^2 - x + 2$ shows this. If the third condition in Bunyakovsky's conjecture holds then necessarily the coefficients of the polynomial are relatively prime (in fact, if the second condition holds then also the coefficients are relatively prime, since a common factor of the coefficients that is greater than 1 would mean the polynomial is reducible over the integers), but the converse is not true.

As noted above, a practical way to prove the numbers $f(1),f(2), f(3),\dots$ have gcd 1 is to find a single pair of values that are relatively prime. A way of calculating the gcd of all the numbers $f(n)$ when $n \geq 1$, even in the case of this number being greater than 1, is to rewrite $f(x) = c_0 + c_1x + \cdots + c_dx^d$ as a linear combination of the binomial coefficient polynomials $\binom{x}{k}$: $f(x) = a_0 + a_1\binom{x}{1} + \cdots + a_d\binom{x}{d}$. If each $c_i$ is an integer then each $a_i$ is an integer and $\gcd\{f(n) : n \geq 1\} = \gcd(a_0,a_1,\dots,a_d).$ For example, $x^2 - x + 2 = 2\binom{x}{2} + 2$, and the coefficients in the second formula have gcd 2, which is related to the fact that $x^2 - x + 2$ has even values on the integers. Using this gcd formula it can be proved $\gcd\{f(n) : n \geq 1\}$ is 1 if and only if there is some pair of positive integers $m$ and $n$ such that $f(m)$ and $f(n)$ are relatively prime.

To date, the only case of Bunyakovsky's conjecture that has been proved is polynomials of degree 1. This is Dirichlet's theorem, which states that when $a$ and $m$ are relatively prime integers there are infinitely many prime numbers $p \equiv a \!\! \pmod m$. This is Bunyakovsky's conjecture for $f(x) = a + mx$ (or $a - mx$ if $m < 0$). The third necessary condition in Bunyakovsky's conjecture for a linear polynomial $mx + a$ is equivalent to $a$ and $m$ being relatively prime. Not a single case of Bunyakovsky's conjecture for degree greater than 1 is proved, although numerical evidence in higher degree is consistent with the conjecture.