# Can a primitive integral polynomial represent integers prime to any number?

Let $m\in\mathbb N$ be any natural number and $f(X)=aX^2+bX+c$ be a polynomial with coefficients $a,b,c\in\mathbb Z$ such that $gcd(a,b,c)=1$.

Is there an $R\in\mathbb Z$ so that $f(R)$ is prime to $m$?

Intuitively, the answer clearly seems yes. But I'm having a hard time proving it. Sure, if $gcd(c,m)=1$, we can simply put $R=0$. My idea would be looking into any prime factor $p^\alpha$ of $m$ and somehow produce simultaneous congruences such that $f(R)\equiv 1 \mod p^\alpha$. But that didn't work out.

• Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$? – packetpacket Jul 23 '18 at 17:01
• Ah yes. Thanks for noting. – David Bernstein Jul 23 '18 at 17:06
• What about $X^2+X$ and $m=2$? – Lord Shark the Unknown Jul 23 '18 at 17:14
• $\mathbf {Bouniakowsky's Conjecture}$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far. – Piquito Jul 23 '18 at 17:27

As asnwered by Shark, it is not true in general. In fact in only has problems with the even numbers $m$.
What you need first is that for any prime $p$ dividing $m$, there exists some $x_p$ such that $f(x_p) \ne 0 \pmod{p}$. Then you just take $R$ such that $R \equiv x_p \pmod{p}$ for any $p$ dividing $m$.
The existence of such a $x_p$ is guaranteed if $p>2$, since a polynomial of degree $d\le 2$ has at most $d$ roots. But for $p=2$ and $f(X)\equiv X^2+X \pmod{2}$, it does not exists.
• If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime. – quantum Aug 4 '18 at 10:51