نه تنها خدا کمک نمی کند، بلکه خداوند شما را در تنگنا قرار نمی دهد. // Not only God does not help, but also Allah does not "put you in trouble"

          I have solved such a problem before, here.

          It's Problem #5 on page 256 of the book НЕСТРЕНКО Ю.[рий] В.[алентинович] ТЕОРИЯ ЧИСЕЛ

(Издательский центр "АКАДЕМИЯ", Москва, 2008)

In translation :

" Let $\alpha$ be a root of the polynomial $f(x)=x^3-9x-9$ and $\beta=6+\alpha-\alpha ^2\;,\;\gamma=-6-2\alpha+\alpha ^2.$ Prove that the numbers $\beta\;,\;\gamma$, are also roots of the polynomial $f(x)$."

As we have seen before, the main difficulty of the problem lies in writing certain expressions as perfect squares.

Solution CiP

Obvious

$$\alpha ^3=9\alpha+9. \tag{1}$$

From here we get, first multiplying (1) by $\alpha$

$$\alpha ^4=9\alpha ^2+9\alpha\;, \tag{2}$$

then, from $\alpha \cdot (\alpha^2-9)=9$, that

$$\frac {1}{\alpha}=\frac{1}{9}\cdot \alpha ^2-1. \tag{3}$$

From the first formula of Vieta $\alpha +\beta +\gamma=0$ we get

$$\beta+\gamma=-\alpha, \tag{4}$$

and from the third $\alpha \cdot \beta \cdot \gamma=9$ we deduce $\beta \cdot \gamma=\frac{9}{\alpha}$  therefore (see relation (3))

$$\beta \cdot \gamma=\alpha^2-9. \tag {5}$$

We calculate now, $(\beta -\gamma)^2=(\beta+\gamma)^2-4\cdot \beta \cdot \gamma \underset {(4)\;(5)}{==}$

$$=(-\alpha)^2-4(\alpha^2-9)=36-3\alpha^2. \tag{6}$$

Or we can, from $\beta \cdot \gamma=\frac{9}{\alpha}$, calculate the same way

$(\beta-\gamma)^2=(\beta+\gamma)^2-4\cdot \beta \gamma=(-\alpha)^2-4 \cdot \frac{9}{\alpha}=\frac{\alpha^3-36}{\alpha}\underset{(1)}{=} \frac{9\alpha+9-36}{\alpha}=$

$$=9 \cdot \frac{\alpha -3}{\alpha}=9\cdot \frac{\alpha^2-3\alpha}{\alpha^2}. \tag{7}$$

          It would be preferable to have perfect squares in relations (6) and/or (7). The method of undetermined coefficients did not help me this time. I don't know by what miracle I finally determined (which can be verified immediately by calculation) the following:

$36-3\alpha^2=(2\alpha^2-3\alpha-12)^2, $ and

$\alpha^2-3\alpha=(\alpha^2-2\alpha-6)^2.$

Now, from (6) it follows $\beta-\gamma=12+3\alpha-2\alpha^2$ (making one of the possible choices of $\pm$ signs) which combined with (4) leads to the desired values

 $$\beta=6+\alpha -\alpha^2\;, \;\gamma=-6-2\alpha+\alpha^2.$$

$\blacksquare$

A cute non-UFD ring

                You can quickly read about UFD here. (The author "Herstein", mentioned in the paragraph following Definition 4, seems to be I. N. HERSTEIN - Topics in Algebra - JOHN WILEY & SONS, 1975, Theorem at page 148)

               Let the ring $R=\mathbb{Z}_8[X]$. Units in $R$ are $\{\hat{1},\hat{3},\hat{5},\hat{7}\}$.

          The polynomial $X^2-\hat{1}$ admits two decompositions into irreducible factors:

$$X^2-\hat{1}\;=\;(X-\hat{1})(X+\hat{1})\;=\;(X-\hat{3})(X+\hat{3}). \tag{1}$$

The factor $X-\hat{1}$ is not associated with any of the factors $X\pm \hat{3}$. Neither does the other one $X+\hat{1}$. Indeed, $\hat{3}\cdot (X-\hat{1})=\hat{3}X-\hat{3}\neq X\pm \hat{3},\;\hat{5}\cdot (X-\hat{1})=-\hat{3}X+\hat{3}\neq X\pm\hat{3},\;$

$\hat{7}\cdot (X-\hat{1})=-X+\hat{1}\neq X\pm \hat{3}.$

          It can also be seen from (1) that the polynomial $X^2-\hat{1}$, although of degree two, has four roots in $\mathbb{Z}_8$.

               The example is taken from Michael ARTIN's book "ALGEBRA" (PRENTICE HALL, 1991) page 392, the example following Proposition (1.8).

QUO VADIS, Olimpiada Națională „Gazeta Matematică” ?

               In the past, it was written on the cover of GMB: "by solving problems from the math magazine, you also prepare for math competitions".

               The password to open the file is : ogeometrie . Just click on the image.

               Other information about the Mathematics Competitions in Romania can be found here.

               A collection of "Gazeta Matematica Seria B" can be found here.

MY ERRATA , MY FRIEND : the most harmless math mistakes

           I like to look for mistakes in other people's works. So, I don't have time to find my own mistakes.

           I was reading a wonderful book yesterday: 

My Numbers, My Friends (Popular Lectures on Number Theory) by the author  Paulo Ribenboim.

A first mistake, more of a haste than a typo, appears on page 10 (see marked text).

In the marked place, the correct text is :

$$...\;if\;p\;is\;a\;prime\;dividing\;both\;P,\;Q,\;then...$$

$\blacksquare$

Those who wish can read as much as they want by clicking on the image of the book cover.

GAZETA MATEMATICĂ Seria B N0 2/2024

 Click on the image to download.

For a wider collection click here. The password for opening encrypted files is    ogeometrie

SUPLIMENTUL cu EXERCIȚII al GMB N0 2/2024

 Click on the image to download.

For a wider collection click here. The password for opening encrypted files is    ogeometrie

It is not sure if I will be able to update ERATA with each issue of the magazine

           Nu este sigur daca voi reusi actualizarea ERATA la fiecare numar al revistei.