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Let us consider a sequence of continuous functions $g_{q}:ℝ^2\to ℝ^2$. Let $(A_{q})_{q\geq 1}$ be a sequence of compact sets in $ℝ^2$. Assuming that each function $g_{q}$ is topologically mixing in $A_{q}$ for all $q\geq 1$, i.e., for every open subsets $U,V$ of $ℝ^2$ such that $U\cap A_{q}$ and $V\cap A_{q}$ are non-empty, there exists $k_0=k_0(q,U,V)$ such that for every $k\geq k_0$ the set $g_{q}^{k}(U)\cap V\cap A_{q}$ is non-empty.

My problem is about finding a common set $A$ in which all the functions $g_{q}$ are topologically mixing in $A$. My approach based on several techniques (symbolic dynamics) leads to this diophantine equation:

$$16(n+1)^2q^8+16(n+1)^2q^6+1=m^2$$

and the main problem is equivalent to the fact that the above diophantine equation has an infinite number of positive integers solutions $q,n,m$. So, the question is how one can prove that the above diophantine equation has an infinite number of positive integers solutions $q,n,m$.

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It is just Pell's equation $m^2-Ny^2=1$, for $N=16q^6(q^2+1) $. Thus even for fixed $q$ it has infinitely many solutions, that was proved by Lagrange and you may find the proof in many textbooks.

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  • $\begingroup$ This is an amazing link between dynamical systems and number theory. $\endgroup$
    – Safwane
    Jun 17 '20 at 7:25
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Your curve can be written as $Y^2=X^4+X^3-2,$ where $Y=4n$ and $X=q^2.$ This Diophantine equation satisfies Runge's condition, so this is relatively easy to handle and one obtains that there are only finitely many integral solutions (see Poulakis-Quartic). You may also consider it as a genus 1 curve and there are techniques to determine all integral points on such curves (see Tzanakis-Quartic).

To make it a bit more explicit, let $P(X)=X^4+X^3-2, P_1(X)=4X^2+2X$ and $P_2(X)=4X^2+2X-1.$ Here we get that $16P(X)-P_1(X)^2=-4X^2-32$ and $16P(X)-P_2(X)^2=4X^2 + 4X - 33.$ Hence $$(4X^2+2X-1)^2<16P(X)=(4Y)^2<(4X^2+2X)^2$$ if $X\notin [-4..3].$ That is we have a contradiction, since $(4Y)^2$ is supposed to be between two consecutive squares. It remains to deal with the values $X\in [-4..3].$ The only solution is given by $$(X,Y)=(1,0).$$ Thus $n=0$ and $q=\pm 1$ (you look for positive solutions only, so $q=1$ remains). That was the Runge approach.

The elliptic curve part can be done by the program package Magma (see Magma), you simply type

IntegralQuarticPoints([1,1,0,0,-2],[1,0]);

and you get

 [
    [ 1, 0 ]
].

Here $[1,1,0,0,-2]$ comes from the degree 4 polynomial, these are the coefficients and $[1,0]$ is a point on the curve.

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    $\begingroup$ Your polynomial is $q^8+q^6-2$ and it can be written as $X^4+X^3-2$ with $X=q^2.$ $\endgroup$
    – castor
    Jun 15 '20 at 16:52
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    $\begingroup$ @Safwane Can you provide a quartic Diophantine equation for which every prime is a solution? :P The point here is that any integer solution to your equation would ALSO provide be a solution to the equation that castor treats here and shows to have only a finite number of solutions. $\endgroup$ Jun 15 '20 at 17:05
  • $\begingroup$ Sorry,The true equation is: $$16(n+1)^2q^8+16(n+1)^2q^6+1=m^2$$ $\endgroup$
    – Safwane
    Jun 16 '20 at 6:48

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