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A method used to answer the post Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? showed that the only positive integer solutions of the equations $$ x^4-3x^2y^2+3y^4=z^2$$ $$X^4+6X^2Y^2-3Y^4=Z^2,$$ are $(1,1,1)$ and $(1,1,2)$, respectively. Furthermore these solutions are related in that each can be deduced from the other.

In general, let $b,d,D$ be integers such that $b^2=4d+D$ and consider the pair of equations $$ x^4+bx^2y^2+dy^4=z^2\tag 1$$ $$X^4-2bX^2Y^2+DY^4=Z^2\tag 2$$ where we can assume that $(x,y)=(X,Y)=1$. Then for the method of the aforementioned post to produce an infinite sequence of solutions, we require a solution $(x,y,z)$ of $(1)$ to generate the solution $(\frac{z}{x},y,|\frac{x^4-dy^4}{x^2}|)$ of $(2)$ and we require a solution $(X,Y,Z)$ of $(2)$ to generate the solution $(\frac{Z}{2X},Y,|\frac{X^4-DY^4}{4X^2}|)$ of $(1)$.

I can prove that for this to occur $y=Y$ is odd. Also, since $x$ must be a factor of $z$, $x^2$ is a factor of $d$. Therefore there are only a finite number of possibilities for $x$ and so we are dealing with a finite loop of solutions. Beyond that, I only know that when solving specific equations these loops are rare and, when they occur, seem to have period $2$.

Questions

What periods are possible for such loops of solutions?

What else can be determined about them?

  • Can we prove that $\frac{z}{x}$ is always an integer or is it a hypothesis? – Alex Feb 13 '21 at 12:11
  • Yes, it can be proved. If you refer to the post referred to in the question, the loop requires this just like it requires $y$ to be odd. If either of these results were not true then $y$ would be further reduced. –  Feb 13 '21 at 12:15
  • Can you tell how you got that? I'm getting $$x^2 + by^2 + \dfrac{dy^4}{x^2} = \left(\dfrac{z}{x}\right)^2$$ and for $\left(\dfrac{z}{x} \right)$ to be an integer, $x^2 \mid d$ (since $\gcd(x,y) = 1$). I don't know how to get $z \mid x$ from here though. – Alex Feb 16 '21 at 14:37
  • Yes. In the post referred to above, the loop requires $tuX=1$. In particular $X=1$ and similarly, $x=1$. So the general form $ax^4+bx^2y^2+cy^4=z^2$ becomes $a+by^2+cy^4=z^2$. In the notation of this post we therefore have $x^2$ equal to $a$, a divisor of $d$. –  Feb 16 '21 at 14:52
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    Just for reference, this pair of equations is found in Page 22 (Theorem 4) of Mordell's book Diophantine Equation. – MathGod Jun 04 '21 at 23:22
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    Thanks for this reference @MathGod. –  Jun 06 '21 at 09:55

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SOME PROGRESS

Let the loop of solutions have period $2n$. We can suppose that successive values taken by $x$ and $X$ are $a_1,a_2,...a_n$ and $A_1,A_2,...A_n$, respectively, where for each $i$ there are integers $c_i$ and $C_i$ such that $d=a_i^2c_i$ and $D=A_i^2C_i$.

Then the conditions given in the post simplify to:- $$a_i^2+by^2+c_iy^4=A_i^2 \tag 1$$ $$A_i^2-2by^2+C_iy^4=4a_{i+1}^2 \tag 2$$ $$2a_{i+1}A_i=|a_i^2-c_iy^4| \tag 3$$ $$4a_{i+1}A_{i+1}=|A_i^2-C_iy^4| \tag 4$$ $y^2$ is a factor of $4^n-1$

Consider equations $(1)$ and $(2)$ modulo $y^2$. $$a_1^2\equiv A_1^2\equiv 4a_2^2\equiv 4A_2^2\equiv 16a_3^2\equiv 16A_3^2\equiv ... $$ Hence $a_1^2\equiv 4^na_1^2$ and so, since $x$ and $y$ are coprime, $y^2$ is a factor of $4^n-1.$

This significantly limits the values of $y$. For periods of less than or equal to $18$ the only possible values for $y$ are $1$ and $3$.

A general formula for loops of period $2$

If $n=1$, then $y=1$ and the equations further simplify to
$$a^2+b+c=A^2 \tag 1$$ $$A^2-2b+C=4a^2 \tag 2$$ $$2aA=|a^2-c| \tag 3$$ $$4aA=|A^2-C| \tag 4$$ From equation $(3)$ we have $c=a^2\pm 2aA$. Then $b=A^2\mp 2aA-2a^2$ and $C=A^2\mp4aA.$

The original equations are then $$ x^4+\left(A^2\mp 2aA-2a^2\right)x^2y^2+\left(a^4\pm2a^3A\right)y^4=a^2A^2$$ $$X^4-2\left(A^2\mp 2aA-2a^2\right)X^2Y^2+\left(A^4\mp4aA^3\right)Y^4=4a^2A^2,$$ with a loop of solutions $(a,1,aA)$ and $(A,1,2aA).$