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I am interested in finding members of $\mathbb{Z}[x]$ that have roots modulo every integer but which have no integer roots.

Up until today, I was not able to find any which didn't have a factor of $x$ factorisable from them, such as $x(x+1)$ or $x^k$ or so forth. I am aware of this paper on the topic but the condition it gives there is not so easy to comprehend (i.e. it's not obvious how to produce a nice, large family of such polynomials given the condition).

Is there any literature or things you know that could provide a list of examples or some more easily computable sufficient condition for a polynomial to have this property?

Isky Mathews
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    I think that paper is likely to be the easiest thing you'll find on the topic. The notion of multiplying a bunch of quadratics to cover all the bases is pretty robust. I would not expect there to be a simple characterization of all such polynomials. indeed, for non-quadratics, it's not so easy to resolve the problem $\pmod p$ for all $p$. – lulu Dec 22 '21 at 21:39
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    Note: Non-zero constant term surely isn't what you meant to ask about. The polynomial $x-1$ has a root $\pmod n$ for every $n$ (because it has an integer root) but has a non constant term. i assumed you meant to ask about polynomials with no integer root but which do have a root $\pmod n$ for all $n$. – lulu Dec 22 '21 at 21:42
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    Can you edit your post for clarity? As I said, non-zero constant term surely isn't what you meant to ask about. – lulu Dec 22 '21 at 21:54
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    The paper you linked seems to give an example of such a polynomial: $p(x)=(x^2-13)(x^2-17)(x^2-221)$. Is this what you where referring to when you said "up until today"? – QC_QAOA Dec 22 '21 at 22:05
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    $(x^3+x+1)(x^2+31)$ – reuns Dec 23 '21 at 06:57
  • What are the downvotes for on this question? I apologize for not changing my question earlier, I was ill and had to be away from maths for a day. Is that it - the delay? When normally I browse and enjoy the site daily? I thought this was a fairly interesting and reasonable question, especially now with the improvement by @lulu. – Isky Mathews Dec 24 '21 at 11:25
  • @QA_QAOA precisely. They do have a small discussion of it in the paper but again it would be practically more useful to me to have some statement like "the pronic numbers have this property" rather than their Theorem 1. I'm also more generally interested in what literature has been written elsewhere on this topic, even just questions on MSE. – Isky Mathews Dec 24 '21 at 11:26
  • @lulu: even if an elegant characterisation is far from possible, I still hold out some hope that people might have a few more papers written on the topic, perhaps giving simple sufficient conditions or algorithms to check if a polynomial is of this type... – Isky Mathews Dec 24 '21 at 11:47
  • As i have said, the matter is very difficult for irreducible polynomials of degree $>2$. I do not believe that any simple tests of the form you want exist (or at least, I don't believe any such tests are known). For quadratics, we have reciprocity which lets us analyze individual primes. – lulu Dec 24 '21 at 11:50
  • One of the oldest incarnations of this question. I won't initiate duplicate closure because A) I answered that one, and B) I have a dupehammer so others would not get to vote at all. – Jyrki Lahtonen Dec 24 '21 at 16:20
  • All, please check out Rosie F's answer for more theory and a link to another paper. – Jyrki Lahtonen Dec 24 '21 at 16:24

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