# Can second degree polynomials generate as many as we wish prime numbers in the way described?

While I was getting in my pyjamas, a few minutes ago, the Euler polynomial $n^2+n+41$ came into my mind. As you know, this polynomial is famous because the set $\{f(0),f(1),...f(39)\}$ consists of prime numbers only, so this polynomial takes for the first $40$ of its values on the set $\mathbb N_0$ only prime numbers.

So the natural question to ask is:

Is it true that for every $m \in \mathbb N$ there exists second degree real polynomial of a real variable $P$ and a number $k(m) \in \mathbb N$ such that all numbers in the set $\{P(k(m)), P(k(m)+1),...,P(k(m)+m-1)\}$ are prime numbers?

• Note that by argument translation you can assume $k(m)=0$ to simplify the statement. – lisyarus Apr 29 '16 at 21:52
• @lisyarus Thank you, but does it simplify it really or it makes it look simpler? – Farewell Apr 29 '16 at 21:53
• Great context to the question. +1 – M47145 Apr 29 '16 at 21:56

Yes. In fact, you don't even need the quadratic term: there are degree-$1$ polynomials giving arbitrarily long sequences of primes by the Green-Tao theorem. (Of course, you could also let $P$ be a constant prime).

• But I wrote second degree because I want them to be of second degree, not of degree one or degree zero. – Farewell Apr 29 '16 at 21:55
• @Farewell OK. Take a linear function $f(X)$, and replace $X \rightarrow X^2,$ if you really want to. – user334871 Apr 29 '16 at 21:56
• Replacing a linear with $X \to X^2$ is not the nicest solution, because people still do not know if there are infinitely many primes of the form $X^2+1$ for instance. I don't think any general quadratics of this form are known to have this property yet (though conjectured of course). – Supersingularity Apr 29 '16 at 22:04
• @Supersingularity The polynomials are not supposed to take infinitely many prime values, but rather arbitrarily long (finite) prime progressions. – user334871 Apr 29 '16 at 22:06
• @user334871. Re : Your 1st comment....At first that didn't seem right, but it is. As there are arbitrarily long arithmetic sequences of primes (Green & Tao,2004), if $a\ne 0$ and $aX +b=a(X-m)+(a m +b)$ is prime for $X=m,m+1,,...,m+n$ then $a Y^2 +a m+b$ is prime for $0\leq Y\leq \sqrt n.$ – DanielWainfleet Apr 29 '16 at 22:15

This follows by the Green-Tao theorem: there exist arbitrarily long arithmetic sequences of primes.
If $$ak+b,a(k+1)+b,\ldots,a(k+m^2)+b$$ are prime, we can take $P(n)=a(k+n^2)+b$, being prime for $n=0,\ldots,m$.

Note that the same works if we require $\deg P=1$.

• So, if I understand it right, this argument would work for polynomials of any degree? – Farewell Apr 29 '16 at 22:00
• Yes, indeed.${}$ – punctured dusk Apr 29 '16 at 22:01
• I suspect that answering the Q without using the Green-Tao theorem may be as difficult proving the Green-Tao theorem. – DanielWainfleet Apr 29 '16 at 22:18
• Is the Tao Ziegler paper not cleaner for polynomials? arxiv.org/abs/math/0610050 – abligh Apr 30 '16 at 14:55

If you allow non-monic polynomials the question is not very arithmetic in nature.

Any class of integer functions closed under the operations $f(n) \to Af(n)+B$ with integer $A,B$, is guaranteed to have this property for any set that contains long arithmetic progressions (such as the primes, by the Green/Tao theorem as people have mentioned). This includes degree $k$ polynomials for any $k$.

For monic non-linear polynomials, as long as $f(n)$ does not assume all values mod $p$ for some prime, then for all $k$ there is an integer $A$ such that $(f(x) + A)$ has no congruence obstruction to having a string of $k$ consecutive prime values. According to the the Hypothesis H conjectures (Schinzel, Bateman-Horn) such strings then occur infinitely often.