How? can we recognize a perfect square??

           The following post on Facebook of Mahan Gholami caught my attention.

                The first relationship that helps us is

$$\alpha^3=1+2\cdot \alpha-\alpha^2.\tag{$\alpha^3$}$$

Then we write $\alpha^3+\alpha^2-2\alpha=1\;\;\Leftrightarrow\;\;\alpha \cdot (\alpha^2+\alpha-2)=1$ and we get

$$\frac{1}{\alpha}=\alpha^2+\alpha-2. \tag{$\alpha^{-1}$}$$

We are still calculating for the future $\alpha^4=\alpha \cdot \alpha^3 \underset{(\alpha^3)}{=}\alpha \cdot (1+2\alpha-\alpha^2)=\alpha +2\alpha^2-\alpha^3 \overset{(\alpha^3)}{=}\alpha +2\alpha^2-(1+2\alpha-\alpha^2)$ so

$$\alpha^4=3\alpha^2-\alpha-1.\tag{$\alpha^4$}$$

          From the first formula of Vieta $\alpha+\beta+\gamma=-1$ we get

$$\beta+\gamma=-1-\alpha. \tag{1}$$

And from the third $\alpha \cdot \beta \cdot \gamma=1$ we deduce $\beta \cdot \gamma=\frac{1}{\alpha}$ therefore (see relation $(\alpha^{-1})$)

$$\beta \cdot \gamma=\alpha^2+\alpha-2.\tag{2}$$

          Knowing the sum and the product, we calculate from relations (1) and (2):

$(\beta-\gamma)^2=(\beta+\gamma)^2-4 \cdot \beta \gamma=(-1-\alpha)^2-4(\alpha^2+\alpha-2)$, hence

$$(\beta-\gamma)^2=9-2\alpha-3\alpha^2. \tag{3}$$

The next delicate part is to be able to express the member on the right in (3) as a perfect square.

          Let's try to determine the numbers - preferably integers - $m$ and $n$ so that

$9-2\alpha-3\alpha^2=(3+m\cdot \alpha   +n \cdot \alpha^2)^2.\tag{4}$

 The right side is successively equal to

$9+m^2\alpha^2+n^2 \alpha^4+6m\alpha+6n\alpha^2+2mn\alpha^3=$

$\overset {(\alpha^4)}{\underset {(\alpha^3)}{=}}9+m^2\alpha^2+n^2 \cdot (3\alpha^2-\alpha-1)+6m\alpha+6n \alpha^2+2mn \cdot (1+2\alpha-\alpha^2)=$

$=(9-n^2+2mn)-\alpha \cdot (n^2-6m-4mn)-\alpha^2 \cdot (2mn-6n-3n^2-m^2).$

To fulfill the relationship (4), let's try to solve the equations

$\begin {cases}(i)\;9-n^2+2mn=9\\(ii)\;n^2-6m-4mn=2\\(iii)\;2mn-6n-3n^2-m^2=3\end {cases}$

It follows from (i) that $2mn-n^2=0$; but $n=0$ leads to the impossibility $(ii)\;-6m=2\;and\;(iii)\;-m^2=3$, hence $n=2m$. Then the remaining relations

$\begin{cases}4m^2-6m-8m^2=2\\4m^2-12m-12m^2-m^2=3\end{cases}$

provide the value $m=-1$. Therefore, relation (4) is written

$$9-2\alpha-3\alpha^2=(2\alpha^2+\alpha-3)^2.$$

          We then obtain from (3)

$$\beta-\gamma=\pm(3-\alpha-2\alpha^2). \tag{5}$$

We choose the minus sign in (5) and, combining with (1), we get $\beta=\alpha^2-2,\;\;\gamma=1-\alpha-\alpha^2.$ (The other choice reverses the values of $\beta$ and $\gamma$.)

$\blacksquare$

               Remark CiP

                     Trying to calculate $(\beta-\gamma)^2=(\beta+\gamma)^2-4\beta \gamma \overset{(1)}{=}$

$=(1+2\alpha +\alpha^2)-\frac{4}{\alpha}=\frac{\alpha^3+2\alpha^2+\alpha-4}{\alpha} \underset{(\alpha^3)}{=}\frac{(1+2\alpha-\alpha^2)+2\alpha^2+\alpha-4}{\alpha}=\frac{\alpha^2+3\alpha-3}{\alpha},$

we will write 

$$\frac{\alpha^2+3\alpha-3}{\alpha}=\frac{\alpha^3+3\alpha^2-3\alpha}{\alpha^2} \overset{(\alpha^3)}{=}\frac{(1+2\alpha-\alpha^2)+3\alpha^2-3\alpha}{\alpha^2}=\frac{2\alpha^2-\alpha+1}{\alpha^2}.$$

We would expect $2\alpha^2-\alpha+1$ to be a perfect square. This is true, but it is more difficult to recognize. The answer, not so obvious, is 

$$2\alpha^2-\alpha+1=(\alpha^2-\alpha-2)^2.$$

          Therefore, we repeat the question from the title, which we hope to answer someday: How can we recognize a perfect square ?

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