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Related to Hilbert's Tenth problem.

Is there polynomial with integer coefficients $P(a,x_1,...,x_n)$ such that $P(A,X_i)=0$ has rational solutions $X_i$ iff $A$ is not the square of integer (or as another question not the square of rational)?

We think if $P$ is homogeneous and ask about integer solutions, scaling the solution might cause problems: $A^2,A X_1, A X_2, ...$

Over the integers solution is trivial via Pell equation:

$$ (2+x_1^2+x_2^2+x_3^2+x_4^2)^2- a x_5^2=1 $$

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    $\begingroup$ I guess the intended meaning is "Is there $n$ and $P\in\mathbf{Z}[a,x_1,\dots,x_n]$ such that for every $A\in\mathbf{Q}$, there exists $X\in\mathbf{Q}^n$ such that $P(A,X)=0$ iff $A$ is the square of an integer"? $\endgroup$
    – YCor
    Commented Oct 15, 2019 at 11:54
  • $\begingroup$ @YCor No, I don't ask about this. The part after "iff" is A is NOT the square of integer. The rest of the comment is correct. $\endgroup$
    – joro
    Commented Oct 15, 2019 at 12:05
  • $\begingroup$ OK thanks (and sorry, actually I focussed on the first quantifiers). $\endgroup$
    – YCor
    Commented Oct 15, 2019 at 12:11
  • $\begingroup$ How about some easier sets? Like "$A>5$? $\endgroup$
    – user6976
    Commented Oct 15, 2019 at 21:07
  • $\begingroup$ $A>5$ is too easy: $P = Q^2 + R^2$ where $$ Q = (A-5) (x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_5^2 + x_6^2 + x_7^2 + x_8^2) $$ and $R = (A-5) x_9 - 1$. $\endgroup$
    – Noam D. Elkies
    Commented Oct 15, 2019 at 21:40

1 Answer 1

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The set $A$ of non-squares (of rationals) is Diophantine in $\mathbb{Q}$ by [1]. The set $B:=\mathbb{Q}\smallsetminus\mathbb{Z}$ is also Diophantine by [2]. The set of non-squares of integers is equal to $A\cup B$, hence Diophantine.

For a generalization of [1], see also [3].

[EDIT] The paper [1] treats arbitrary (non-)$n$-th powers, but the case of (non-)squares was proved earlier by Poonen [4].

[1] Colliot-Thélène, Jean-Louis; van Geel, Jan, Le complémentaire des puissances $n$-ièmes dans un corps de nombres est un ensemble diophantien, Compos. Math. 151, No. 10, 1965-1980 (2015). ZBL1346.14066..

[2] Koenigsmann, Jochen, Defining $\mathbb Z$ in $\mathbb Q$, Ann. Math. (2) 183, No. 1, 73-93 (2016). ZBL1390.03032..

[3] Dittmann, Philip, Irreducibility of polynomials over global fields is diophantine, Compos. Math. 154, 761-772 (2018). ZBL06861881.

[4] Poonen, Bjorn, The set of nonsquares in a number field is diophantine, Math. Res. Lett. 16, No. 1, 165-170 (2009). ZBL1183.14031.

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  • $\begingroup$ Thank you. This appears to solve an open problem: mathoverflow.net/questions/199191/… $\endgroup$
    – joro
    Commented Oct 16, 2019 at 13:50
  • $\begingroup$ @joro Yes, very good point! $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 16, 2019 at 13:54
  • $\begingroup$ A comment in the other question doubts correctness about the other question, is it valid? $\endgroup$
    – joro
    Commented Oct 16, 2019 at 15:35

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