Properties of three terms of a geometric series

Question

I’m [still!] working on the equation in this question, namely

$$(b^2+2)^2=(a^2+2c^2)(bc-a). \tag{$\star$}$$

where $a,b,c$ are integers. Evidently, $(\star)$ implies

$$\frac{b^2+2}{bc-a} = \frac{a^2+2c^2}{b^2+2}, \tag{1}$$

which is to say that $\{bc-a,b^2+2,a^2+2c^2\}$ are three consecutive terms of a geometric series.

QUESTION: Does that fact provide any information that would help in solving $(\star)$? i.e., are there properties of geometric series that can be brought to bear on the problem?

Each fraction in $(1)$ is actually an integer, in case that provides more leverage/structure.

EDIT: The reason I know this is that I derived this equation from the equation $x^3=y^2+2$, where $x=(b^2+2)/(bc-a)$ is a positive integer by assumption.

Answer 1

If I have understood the additional constraints to your original problem correctly, the problem is now to find all integral solutions to $$(b^2+2)^2=(a^2+2c^2)(bc-a), \tag{$\star$}$$ such that $$\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2},$$ and moreover $x:=\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2}$ is an integral solution to $x^3=y^2+2$ for some integer $y$.

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A standard argument in $\Bbb{Z}[\sqrt{-2}]$, which is a UFD, shows that then $x=3$. Hence we can express the problem as a system of two simultaneous diophantine equations: \begin{eqnarray} a^2+2c^2&=&3(b^2+2)\tag{1.1}\\ b^2+2&=&3(bc-a).\tag{1.2} \end{eqnarray} Equation $(1.2)$ shows that $3a=3bc-b^2-2$, and hence from $(1.1)$ we find that $$27(b^2+2)=9(a^2+2c^2)=(3a)^2+18c^2=(3bc-b^2-2)^2+18c^2.$$ Expanding and collecting like terms shows that this is equivalent to $$b^4-6b^3c+9b^2c^2-23b^2-12bc+18c^2-50=0,$$ which in turn we can rewrite as $$(b^2+2)(b-3c)^2=25(b^2+2).$$ Of course $b^2+2\neq0$ and so it follows that $(b-3c)^2=25$, or equivalently $$b=3c\pm5\qquad\text{ and hence }\qquad a=\frac{3bc-b^2-2}{3}=\mp5c-9.$$ This shows that every solution to your system of diophantine equations is of the form $$(a,b,c)=(\mp5c-9,3c\pm5,c).$$

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Note that these are precisely the solutions you already found in your original question, i.e. the parametric family of solutions $$(5d+1,3d+1,d+2)\qquad\text{ with }\qquad d\in\Bbb{Z},$$ along with their involutions given by $(a,b,c)\ \longmapsto\ (a,-b,-c)$.

Answer 2

Dear friend: Your parameterization $$a=5t+1\\b=3t+1\\c=t+2$$ is (very!) good and it really comes from an identity that you have established (by brute force you have said) and which is valid for all value, real or non-real, of the parameter $t$. $$\big((3t+1)^2+2\big)^2=\big((5t+1)^2+2(t+2)^2\big)\cdot\big((t+2)(3t+1)-(5t+1)\big)$$

Therefore your parameterization, likely those known for $x^2+y^2=z^2$ or $x^2+y^2=2z^2$ or $x^2+y^2=z^2+w^2$ and other ones, gives all the solutions (integer or not) of your equation. Any other parameterization of distinct form would be equivalent giving also all the solutions which can be clearly ilustrated by the parameterization of the linear equation $ax+by=c$ given by $$x=-bt+x_0\\y=at+y_0$$ where ($x_0,y_0)$ is an arbitrary particular solution.

►Regarding the geometric progression that you say, you have in fact one whose reason is unique and whose value is $3$. It is enough to use your parameterization from which you obtain $$\frac{9t^2+6t+3}{3t^2+2t+1}=\frac{27t^2+18t+9}{9t^2+6t+3}=3$$