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In my research I faced with an intricate construction of an algebraic number with some properties.

Problem. For which classes of polynomials $P(X,Y)\in \mathbb{Z}[X,Y]$, we have the following property. Let $\theta$ be an algebraic number, then there exist an $n$-degree real algebraic number $\gamma$ for which $$ P(r\gamma,\alpha)\neq \theta, $$ for all (but finitely many) rational numbers $r$ and all (but finitely many) algebraic numbers $\alpha$ with degree at most $n-1$.

I can prove this for polynomials of the form $P(X,Y)=F(X)G(Y)$ or $F(X)+G(Y)$, but not for other classes. I believe that this holds for any non-constant (in both variables) polynomial $P(X,Y)$.

Any suggestions?

Thanks.

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  • $\begingroup$ Suppose that $P$, $\alpha$, $\gamma$, and $\theta$ are such that $P(r\gamma,\alpha)=\theta$ for infinitely many $r$, that would mean that $P(X,\alpha) = \theta$ as polynomials (since only a constant polynomial can take the same value infinitely many times). This implies that $Y-\alpha \mid P(X,Y)-\theta$. This can only happen for finitely many $\alpha$ by the unique factorization property for polynomials. $\endgroup$
    – R.P.
    Commented Feb 28, 2023 at 14:09
  • $\begingroup$ @R.P. Thank you for your reply, but in the first part you are considering that there exists infinitely many $r$ for a same $\alpha$. This is not necessary, so maybe we have infinitely many distinct pairs $(r_i,\alpha_i)$ for which $P(r_i\gamma,\alpha_i)=\theta$, for all $i$. Am I missing something? $\endgroup$
    – Jean
    Commented Feb 28, 2023 at 14:18
  • $\begingroup$ My apologies, then I have misunderstood the question. There are a lot of quantifiers in your question, it looks like I didn't process them correctly. :-) $\endgroup$
    – R.P.
    Commented Feb 28, 2023 at 14:20
  • $\begingroup$ No problem, thank you very much for the interest. Maybe I could write it better :-) $\endgroup$
    – Jean
    Commented Feb 28, 2023 at 14:24
  • $\begingroup$ Compare math.stackexchange.com/questions/4646448/… $\endgroup$
    – Franz Lemmermeyer
    Commented Feb 28, 2023 at 14:25

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