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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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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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  • $\begingroup$ 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 $n$). $\endgroup$ – Marcel Besixdouze Feb 4 '15 at 20:40
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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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one slightly easier question, is when can a integer polynomial be odd, the answer is when an odd number of odd coefficients (including constant term possibly, otherwise they might pair up for some x values above 0 ( note if an even number are not the constant ( leavign 1 to be the constant within the odd number of odds), then it's always odd, if not only half the time is it odd), also we know that there can't be more than the constant term of primes in a row because of the polynomial remainder theorem.

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