Possible triples

Let the triple be $\left \{\dfrac br, b, br \right \}$ , where $b$ is the second term and $r$ the common ratio of the geometric progression.

From the sum of the first and third terms equals to the square of the second term, we have $\dfrac br + br = b^2 \implies b = r + \dfrac 1r$ .

From the sum of squares of the first and second terms equals to the square of the third term, we have:

$\begin{aligned} \frac {b^2}{r^2} + b^2 & = b^2r^2 \\ \frac 1{r^2} + 1 & = r^2 \\ \implies r^4 - r^2 - 1 & = 0 \end{aligned}$

$\implies r = \sqrt \varphi$ . where $\varphi = \dfrac {1+\sqrt 5}2$ denotes the golden ratio . And $b = \sqrt \varphi + \dfrac 1{\sqrt \varphi} \approx \boxed{2.06}$ .

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Note: $b = \sqrt \varphi + \dfrac 1{\sqrt \varphi} = \dfrac {\varphi + 1}{\varphi} = \dfrac {\varphi^2}{\sqrt \varphi} = \varphi^\frac 32$ . Therefore, the triplet is $\left \{ \varphi, \varphi^\frac 32, \varphi^2 \right \}$ . Then $\varphi + \varphi^2 = 2\varphi + 1 = \varphi^3 = (\varphi^\frac 32)^2$ . And $\varphi^2 + \varphi^3 = \varphi^4 = (\varphi^2)^2$ .

Write the following three relations for z>y>x :

(1). y^2=z*x

(2). x+z=y^2

(3). x^2+y^2=z^2

Solve using WolframAlpha

y=√[2+√5]~=2.05817