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If $f(x)$ is a $4^{th}$ degree polynomial with integer coefficients, what is the largest set ${x_1, x_2, x_3, ...x_n}$ (where $x_i$ are integers) for which $|f(x_i)|$ is a prime number?

Things I have tried:

I tried to see how I can restrict the coefficients of $ax^4+bx^3+cx^2+dx+e$ by dividing by $fx+g$. If this worked I could've restricted the coefficients so that hopefully $f(x)$ isn't divisible by such polynomials, but it got really messy so I don't think that's the way to go. Another thing I came up with is that we would probably want this polynomial as $ax^4-bx^3+cx^2-d+e$ , so that when we have $ax^4-bx^3+cx^2-dx+e=-p_i$ , then by Descartes's rule of signs $ax^4-bx^3+cx^2-dx+(e+p_i)=0$ could potentially have up to $4$ solutions. I'm not sure if the problem said $x_i$ are integers, but I believe they have to be because otherwise we could guarantee a solution for every single prime number (therefore the set of $x_i$ would be infinitely large) if we just had one sign change, namely $-ax^4+bx^3+cx^2+dx+(e+p_i)=0$

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Infinitely often. – Andrés Caicedo Jul 5 '13 at 17:30
up vote 6 down vote accepted

It is an open problem. In fact, by the Dirichlet's theorem, linear polynomial $f(x)=ax+b$ takes infinitely many prime values, provided $(a,b)=1.$ For all other polynomials of degree $> 1$ it is not known whether there exists a polynomial which takes infinitely many primes. The conjecture is that even $f(x)=x^2+1$ can take infinitely many prime values. If you allow yourself to go to other higher dimensions, then $f(x,y)=x^2+y^2$ would take all prime valuse of the form $4k+1$ by the theorem of Fermat.

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This problem is weaker than showing that there exists a /single/ polynomial which takes arbitrarily many prime values. For example, the corresponding problem for quadratics was proven by Sierpenski in 1964 (that there is no bound on $n$). – Marcel T. Feb 4 '15 at 20:40

This is most likely an open problem.

There are two cases:

  • $f$ is reducible in $\mathbb Z[x]$: Then $f=gh$ for non-constant polynomials $g,h\in\mathbb Z[x]$. If $f(x)$ is a prime, then either $g(x)=1$ or $h(x)=1$. $g-1$ is not the zero polynomial, so it has at most $\deg g$ roots, so $g(x)=1$ for at most $\deg g$ many $x$. Same for $h$, so there are at most $\deg g+\deg h=\deg f$ values $x\in\mathbb Z$, such that $|f(x)|$ is prime.

  • $f$ is irreducible: The Bunyakovsky conjecture states, that $|f(x)|$ is prime for infinitely many $x\in\mathbb Z$, where $f$ is an arbitrary irreducible polynomial. For $\deg f=1$, this was in fact proven by Dirichlet, for $\deg f=4$, it seems to be unsolved.

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