# Diophantine equation that has an infinite number of positive integers solutions

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 $$q,n,m$$.

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.

• This is an amazing link between dynamical systems and number theory. Jun 17, 2020 at 7:25

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.

• Your polynomial is $q^8+q^6-2$ and it can be written as $X^4+X^3-2$ with $X=q^2.$ Jun 15, 2020 at 16:52
• @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. Jun 15, 2020 at 17:05
• Sorry,The true equation is: $$16(n+1)^2q^8+16(n+1)^2q^6+1=m^2$$ Jun 16, 2020 at 6:48