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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.

Servaes
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Kieren MacMillan
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    how do you know the ratio in (1) is an integer? It is for your family in the earlier question... – Will Jagy Mar 11 '20 at 22:35
  • If you can prove that the ratio is always an integer, please post a proof. This might give more insight to the problem and could help to find the complete solution. Also context in questions is always welcome. – Peter Mar 12 '20 at 10:32
  • @WillJagy and Peter: See edit. – Kieren MacMillan Mar 12 '20 at 13:18
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    I'm also still (or again?) working on that same question; I just stumbled upon this follow-up. With the added constraint that the ratio in $(1)$ is an integer, say $x$, I can show that the ratio is an odd integer and at least $3$. Of course, because the integral solutions to $x^3=y^2+2$ are precisely $(x,y)=(3,\pm5)$ it seems to follow from your assumptions that the ratio must be $3$. Did I understand your assumptions correctly? – Servaes Aug 16 '20 at 17:35

2 Answers2

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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 $$\frac{9t^2+6t+3}{3t^2+2t+1}=\frac{27t^2+18t+9}{9t^2+6t+3}=3$$

Ataulfo
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  • Do you have a rigorous proof that $(a,b,c) = (5t+1,3t+1,t+2)$, plus the solutions given by the involution $(a,b,c) \to (a,-b,-c)$, give the complete solution of $(\star)$ in integers? – Kieren MacMillan Mar 13 '20 at 13:50
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    Suddenly I feel like maybe not. But if your identity does not come from modular arithmetic reasoning (in which case, it is clear that the obtained parameterization refers only to a class of numbers), then in principle it does provide all the solutions. Anyway, if not then there is a solution $(a,b,c)$ such that $\dfrac{a-1}{5},\space\space\dfrac{b-1}{3}$ and $c-2$ are nor equal (to a particular value of $t$) then you have proved that not. I bet it is not possible; (suite) – Ataulfo Mar 14 '20 at 16:57
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    for example $\dfrac{a-1}{5}=\dfrac{b-1}{3}\iff 5b-3a=2$ is always possible because of $(3,5)=1$ so $3\mathbb Z+2\mathbb Z=\mathbb Z$ (theorem of Bézout, in France called Lemma of Bézout), – Ataulfo Mar 14 '20 at 17:17
  • The transformation of the original equation ($x^3=y^2+2$) into $(\star)$ does not come from modular arithmetical reasoning — in fact, I applied Bézout’s theorem to obtain the definitions of $a$, $b$, and $c$. – Kieren MacMillan Mar 14 '20 at 22:27
  • Obviously, fixing any one of the values (e.g., $c := t+2$) shows that the identity is valid for all $c$. But what isn’t clear [to me] is that there don’t exist other values or parameterizations for $(a,b)$ — say, $(a,b) := (24c-11,10c+1)$ — that satisfy $(\star)$. I believe it’s true, but would love to prove it rigorously. – Kieren MacMillan Mar 14 '20 at 22:33
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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 $y$.

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 $$(a,b,c)=(\mp5c-9,3c\pm5,c).$$

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 $(a,b,c)\ \longmapsto\ (a,-b,-c)$.

Servaes
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