SUPLIMENTUL cu EXERCIȚII al GMB N0 4/2023

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Vezi ERATA

The Equation of a Straight Line in different Analytical Geometries

          We denote with $\mathbb{R}$ the set of real numbers. Sometimes it means the field of reals , with the usual operations of addition $"+"$ and multiplication $"\cdot"\;$; we note its elements with small Greek letters $\;\alpha,\;\beta,\;\gamma \dots$

          With $\mathbb{C}$ we denote the set of complex numbers $z=\alpha +\imath \beta$; its elements are usually denoted by small Latin letters $\;a,\;b,\;c\dots$ The complex number $\alpha -\imath \beta$, written $\bar{z}$ is the complex conjugate of   $z$. We denote $\alpha=\Re(z)$ and $\beta=\Im(z)$ respectively the real part and the imaginary part of $z$.

          By $\mathbb{A}^2$ we denote the real 2-dimensional affine space or plane ; in it we consider a Cartesian coordinate system (affine frame) $\mathscr{R}=(O;\;\mathbf{e}_1,\mathbf{e}_2)$, where $O$ is a fixed point and $\;\mathbf{e}_1,\mathbf{e}_2$ a base of (real) vectorial space $\mathbb{R}^2$. Endowed with a dot product he becomes the Euclidean plane $\mathbb{E}^2$; in this particular case we prefer the base vectors to be orthonormal.

               In the Proposition below, $\mathbb{C}$ is considered as the 2-dimensional real affine space $\mathbb{R}\times \mathbb{R}$ with the affine frame $\mathscr{S}=(O;\;1,\imath)$. The point $O$ is the same as the one in $\mathscr{R}$ above.

           %1.     PROPOSITION  (i)  The straight line in $\mathbb{E}^2$ that corresponds in the

                               frame $\mathscr{R}$ to the equation

$$\alpha \cdot x+\beta \cdot y+\gamma=0\;\;\;(\alpha,\beta)\neq (0,0) \tag{1}$$

                                has in the frame $\mathscr{S}$ of $\mathbb{C}$ the equation

$$\frac{\alpha+\imath \beta}{2}\cdot z+\frac{\alpha-\imath \beta}{2} \cdot \bar{z}+\gamma=0.\tag{2}$$

                                  (ii) Converselly, to any equation in the frame $\mathscr{S}$ of the form 

$$a\cdot z+b \cdot \bar{z}+c=0,\;\;0\neq b=\bar{a},\;c=\bar{c} \tag{3}$$

                                corresponds a straight line which in the frame $\mathscr{R}$ has the equation

$$(2\cdot \Re\;a)\cdot x+(2\cdot \Im\;a)\cdot y +c=0.\tag{4}$$

          The most general case of an equation representing a straight line is given in the following theorem. ($|a|=\sqrt{\alpha^2+\beta^2}$ means the modulus of the complex number $a=\alpha+\imath \cdot \beta$.)

               %2. THEOREM  The equation

$$a\cdot z+b \cdot \bar {z}+c=0,\;\;\;(a,b)\neq (0,0) \tag{5}$$

           represents a straight line if and only if

 $$|a|=|b|\;\;and\;\;\bar {a} \cdot c=b \cdot \bar{c}. \tag{6}$$

               %3 CORROLARY  (i) If  $c=0$ then equation (5) represents a line iff

 $|b|=|a|. \tag{6i}$

                                                  (ii) If $c \neq 0$ then equation (5) represents a line iff

 $b \cdot \bar{c}=\bar{a} \cdot c.\tag{6ii}$

                        Indeed, if $c=0$, then the second condition in (6) is automatically satisfied. If $c\neq 0$, then $|b|=\left |\frac{\bar{a}\cdot c}{\bar{c}} \right |=\frac{|\bar{a}|\cdot |c|}{|\bar{c}|}=|\bar{a}|=|a|$, and the conditions (6) are both fulfilled.

$\square$

      Some authors prefer to write the equation of a line in the form

$b \cdot z-\bar{b}\cdot \bar{z}+c=0$, $c-$purely imaginary. $\tag{7}$

     The equation (3) is called the self-adjoint form. Since we can multiply equation (5) by an arbitrary non-zero complex number, any of the forms is equally justified when $c \neq 0$.

     Indeed, given a straight line (6) with $c\neq 0$

$$a\cdot  z+b\cdot \bar{z}+c=0$$

$$\underset{(6ii)}{\Rightarrow}\;\;a \cdot z+\frac{\bar{a}c}{\bar{c}} \cdot \bar{z}+c=0\;\;\Rightarrow$$

$$\Rightarrow\;\;a\bar{c}\cdot z+\bar{a}c \cdot \bar{z}+c\bar{c}=0;$$

because $c\bar{c}=|c|^2$ is real number, this is the self-adjoint form. Further, multiplying the last equation by $\imath$

$$\imath a\bar{c} \cdot z+\imath \bar{a}c \cdot \bar{z}+\imath |c|^2=0$$

and writing $\;b:=\imath a\bar{c}$, because $\imath \bar{a}c=(-\bar {\imath})\overline{(\bar{a})}\bar{c}=-\overline{\imath \bar{a}c}$, we get the form (7).

Let's note that, also in the case $c=0$, an equation $az+b\bar{z}=0$ in which the numbers $a$ and $b$ seem arbitrary (satisfying the condition (6i) $|a|=|b|$), can be brought to the self-adjoint form. If we multiply this equation by $\bar{a}+\bar{b}$, we get the equation 

$$a_1 \cdot z +b_1 \cdot \bar{z}=0 \tag{8}$$

in which $a_1=a(\bar{a}+\bar{b})$ and $b_1=b(\bar{a}+\bar{b})$. But the condition (6i) is also expressed as $a \cdot \bar{a}=b \cdot \bar{b}$ and then we have

$\bar{a_1}=\overline {a\bar{a}+a\bar{b}}=\bar{a}a+\bar{a}b=\bar{b}b+\bar{a}b=b(\bar{b}+\bar{a})=b_1.$

     Example: The equation of the straight line $z+\imath \cdot \bar{z}=0$, after multiplying by $1-\imath$, turns into $(1-\imath)z+(\imath +1)\bar{z}=0$, obviously a self-adjoint form (3).

          So (3) is the general form of the equation of a straight line.

{Last edit: 7/29/2023}

Text_ciorna :

$\mathscr{R}$

$\mathbf{e}$

and the FFF conditions are both fulfilled.

End_ciorna

एक पत्रिका जो दुनिया भर में घूमती है // A magazine that travels the world

           I mentioned this magazine before in the post of Wednesday, January 25, 2023.

           I received an email in the last days from a professor from Calcutta (India), interested in some Romanian magazines. He also asked me some details about "Revista Matematica din Timisoara". Unfortunately, I haven't scanned many numbers yet, but I hope that in the future I will be able to post the entire collection I have.

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 ?

{end Rem}

        

RATIONAL NUMBERS of the FORM $\frac{an+b}{cn+d}$ : the Problem of SIMPLIFICATION

               We are investigating some problems related to rational numbers of the form

$$\frac{a \cdot n+b}{c \cdot n+d} \tag{@}$$

 where $a,b,c,d$ are given integer constants, and $n$ is variable in the set of integers.

              We will present some theoretical generalities in the following $nr^{$}$. In this 

framework, we will permanently note 

$$\delta :=a\cdot  d-b \cdot c\;. \tag{$\beta$}$$

          $1^{$}.$     $\delta =0\;\Leftrightarrow\;\frac{an+b}{cn+d}=\;const.$

          $2^{$}.$     $\delta \neq 0\;\Rightarrow\;\frac{an_1+b}{cn_1+d} \neq \frac{an_2+b}{cn_2+d}$   for  $n_1 \neq n_2.$

          $3^{$}.$      The fraction $\frac{an+b}{cn+d}$ can be simplified only with a divisor of $\frac{\delta}{(a,c)}$

                      (reciprocal is not true); we noted $(a,c)=$Greatest Common Divisor 

                      - abbreviated GCD - of the numbers $a$ and $c$.

          $4^{$}.$        The integer values $k$ that the fraction $\frac{an+b}{cn+d}$ can take are among the

                        solutions in $\mathbb{Z}\times \mathbb{Z}$ of the equation with the unknowns $k$ and $n$

$$(c\cdot k-a)(c \cdot n+d)=b \cdot c-a \cdot d \tag{$\gamma$}$$

$$\Leftrightarrow \;\;\left (\frac{c}{(a,c)}\cdot k-\frac{a}{(a,c)}\right )\cdot \left (\frac{c}{(c,d)} \cdot n+\frac{d}{(c,d)}\right )=-\frac{\delta}{(a,c)\cdot (c,d)} \tag{$\gamma$'}.$$

          The proofs of these $nr^{$}$s are elementary, but we present them below.

   

               SECTION I: proofs 

          Proofs of $1,2^{$$}.$  The equality  $\frac{an_1+b}{cn_1+d}=\frac{an_2+b}{cn_2+d}$  is equivalent to $$(an_1+b)(cn_2+d)=(an_2+b)(cn_1+d)$$ 

$$\Leftrightarrow\;(b \cdot c-a \cdot d)(n_2-n_1)=0. \tag{I.1}$$

     If $\delta =0$ then relation (I.1) occurs for any numbers $n_1,\;n_2$, so $\frac{an_1+b)}{cn_1+d)}=\frac{an_2+b}{cn_2+d}$ for any $n_{1,2}$, that is $\frac{an+b}{cn+d}=const.$ Conversely, if $\frac{an+b}{cn+d}=const$ then for any $n_1 \neq n_2$ we have $\frac{an_1+b}{cn_1+d}=\frac{an_2+b}{cn_2+d}$, so relation (I.1) occurs, which implies $\delta=0$; this proves $1^{$}$.

     If $\delta \neq 0$, then for $n_1 \neq n_2$ the negation of the relation (I.1) is satisfied, which is equivalent to the negation of the equality $\frac{an_1+b}{cn_1+d}=\frac{an_2+b}{cn_2+d}$; this proves $2^{$}$.

          Proof of $3^{$}.$  If  $\frac{an+b}{cn+d}^{(s}$ i.e. the fraction can be simplified with the integer $s$

$$\Leftrightarrow \;\;\begin{cases}s \mid an+b\\s \mid cn+d \end{cases}$$

$$\overset{\color{Red}!}{ \Rightarrow}\;s \;\Bigg \vert \frac{[a,c]}{a}(an+b)-\frac{[a,c]}{c}(cn+d)=\frac{[a,c](-\delta)}{ac}=\frac{[a,c](-\delta)}{[a,c](a,c)}=-\frac{\delta}{(a,c)};$$ we noted $[a,c]=$the least Common Multiple - abbreviated lCM - of the numbers $a$ and $c$ and used the well-known formula $[a,c]\cdot (a,c)=a \cdot b\;$. Of course "$\mid$" means "divides".

          i) Example  The fraction $\frac{6n-1}{3n-1}$ can be simplified only with a divisor of $\pm \frac{6\cdot (-1)-(-1)\cdot 3}{(6,3)}=\pm\frac{-3}{3}=\pm 1$; so it is irreducible.

          ii) Counterexample  The fraction $\frac{2n-1}{2n+1}$ can be simplified only with a divisor of

$\pm \frac{2\cdot 1-2 \cdot (-1)}{(2,2)}=\pm \frac{4}{2}=\pm 2$. But it is impossible for the fraction to be simplified by 2, the numerator and denominator being odd numbers; so the fraction is irreducible.

          Proof of $4^{$}. $ We have $\frac{an+b}{cn+d}=k\;\;\Leftrightarrow\;an+b=k \cdot (cn+d)\;\Leftrightarrow$

$$\Leftrightarrow\;\;c \cdot kn-an+dk=n\;\;\;| \times \;c\;\;\;\Leftrightarrow$$

$$\Leftrightarrow\;\;c^2 \cdot kn-ac \cdot n+cd \cdot k=bc\;\;\;|-ad\;\;\;\Leftrightarrow$$

$$\Leftrightarrow\;\;cn \cdot (ck-a)+d\cdot (ck-a)=bc-ad$$

             and we get the equation $(\gamma).$

               SECTION II: exercises   

                   1. Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.

                    Solution CiP  If $s$ is a natural number by which the given fraction can be

                    simplified, then $\begin{cases}s \mid 21n+4\\s \mid 14n+3 \end{cases}$. But then

$$s \mid (-2)\cdot ({\color{Red}{21n+4}})+3 \cdot({\color{Red}{14n+3}})\;\Leftrightarrow\; s \mid 1$$

                     so $s=1$. Hence, the fraction is irreducible.(See remark $1^R$ in Sect.III.)

$\blacksquare$

                     Remark CiP For the given fraction we have from the  formula $(\beta)$

                     $\delta = 21 \cdot 3-14 \cdot 4=7$ so the fraction can be simplified, according to 

                      $3^{$}$, only by 1 or 7. But neither the numerator nor the denominator are 

                        divisible by 7. They give the remainders 4 and 3 respectively when 

                         dividing by 7.

{end Rem}

                    2. Show that for any $n \in \mathbb{N}$ the following fractions are irreducible:

                           $a)\frac{2n+1}{5n+3}\; ,\;b)\frac{5n+3}{8n+5}\;,\;c)\frac{33n+4}{22n+3}\;$ (see remark $2^R$ in Sect.III).

                    Solution CiP  a) We have       $5\cdot({\color{Red}{2n+1}})-2\cdot ({\color{Red}{5n+3}})=-1.\;\tag{2.1}$

                    If $2n+1$ and $5n+3$ have a common divisor $s$ then $s$ also divides the

                    left combination of $(2.1)$. So $s\mid -1$ and then it can only be $\pm1$; the

                    fraction is irreducible.

                        b) As in a), starting from the identity

$$8\cdot (5n+3)-5\cdot(8n+5)=-1.$$

                        c) As in a), starting from the identity

$$2 \cdot(33n+4)-3 \cdot(22n+3)=-1.$$

                      Remark CiP For the given fractions we have from the  formula $(\beta)$

                    a) $\delta = 2 \cdot 3-5 \cdot 1=1$ 

b) $\delta=5 \cdot 5-8 \cdot 3 =1$

c) $\delta=33 \cdot 3-22 \cdot4=11$

                        In all cases $\frac{\delta}{(a,c)}=1$ and we apply the property $3^{$}$.

                             WARNING !!!  There are cases when we CANNOT find a 

                         combination between the numerator and the denominator 

                        WITH INTEGER COEFFICIENTS , which gives the result $\pm 1$,

                        even though they are relatively prime numbers. See the 

                         counterexample from the proof of Property $3^{$}.$

{end Rem}

$\blacksquare$

                    3. Determine the integer $k$ so that the fraction $\frac{1980k+1}{1981k+2}$ is irreducible.

                          (See remark $2^R$ in Sect.III.)

                    Solution CiP     Answer : $k \neq 1979 \cdot p -1,\;p \in \mathbb{Z}\;\;\;\square$

                                   Let $s$ be a number by which the given fraction can be simplified.

                     We have $$\begin{cases} s \mid 1980k+1\;\;\;\;\;\;(3.1)\\s \mid 1981k+2\;\;\;\;\;\;(3.2)\end{cases}$$ 

                     From here $s \mid 1981 \cdot (1980k+1)-1980 \cdot (1981k+2)=-1979$; but

               1979 being a prime number, we can only have $s=\pm 1$ or $s=\pm 1979$. 

                The case when $s=\pm 1979$ gives us reducible fractions; then, substracting 

                (3.2)-(3.1) we get $\pm 1979 \mid (1981k+2)-(1980k+1)=k+1$. 

                 Hence $k+1=1979p,\;p \in \mathbb{Z}$ and we get the answer.

                           Remark CiP For $k=1979p-1$ the fraction becomes
$$\frac{1980 \cdot (1979p-1)-1}{1981 \cdot 1979p-1)+2}=\frac{1980\cdot 1979 \cdot p-1979}{1981 \cdot 1979 \cdot p-1979}=\frac{1980p-1}{1981p-1}.$$

                  The last fraction is irreducible because if $s \mid 1980p-1$  and

                  $s \mid 1981p-1$ then $s \mid 1981p-1-(1980p-1)=p$; and then 

                  $s \mid 1980p-1-1980 \cdot p=-1.$

$\blacksquare$

                    4. (i) Find the integer numbers $n$ so that the fraction $\frac{2n+3}{3n+2}$

 a) is reducible; b) is irreducible.

                        (ii) The same problem for  1) $\frac{6n+5}{3n-1}$;  2) $\frac{6n+1}{3n-2}.$ 

                    Solution CiP   (i)  Answer :   a) $n=5k+1$ :  the fraction

                                                                                              can be simplified by $5$;

                                                                            b) $n \neq 5k+1; \;\;\;k \in \mathbb{Z}$        $\square$

                               a) The fraction is reducible $\Rightarrow$

$\Rightarrow\;$  there is $s>1$ a common divisor of the numerator and the denominator

      $\Rightarrow\;\;s \mid 2n+3$ and $s \mid 3n+2 \tag{4.1}$

$\Rightarrow\;s \mid 3\cdot (2n+3)-2 \cdot (3n+2)\;\Rightarrow\;s \mid 5.$ So $s=5.$

                                 Then the formulas (4.1) are written

$5 \mid 2n+3\;$ and $5 \mid 3n+2,\;\;\Rightarrow$

$\Rightarrow\;5 \mid (3n+2)-(2n+3)=n-1$. So $n-1=5k,\;k \in \mathbb{Z}$

                                and we get the answer. The fraction is in this case

$$\frac{2(5k+1)+3}{3(5k+1)+2}=\frac{{\color{Red}{5}}\cdot (2k+1)}{{\color{Red}{5}} \cdot (3k+1)}=\frac{2k+1}{3k+1}.$$

                          The last form is irreducible, as shown with the techniques used in

                    Exerc's 1, 2 .

                          b) In the other cases, $n \neq 5k+1$, we have the possibilities

$$n=5k,\;\;5k+2,\;\;5k+3,\;\;5k+4,\;\;k \in \mathbb{Z}$$

                     and are obtained respectively the fractions 

$$\frac{10k+3}{15k+2},\;\frac{10k+7}{15k+8},\;\frac{10k+9}{15k+11},\;\frac{10k+11}{15k+14}$$,

                     which are also shown to be irreducible (see Exercises 1 and 2)  

                      Remark CiP The fraction $\frac{2n+3}{3n+2}$ takes the integer values

                      $-1$ and $+1$ when $n=-1$ and $n=+1$ respectively.

                      (ii) Answer :  1) the fraction can be simplified by $7$ $$\;\Leftrightarrow\;n=7p-2,\;p \in \mathbb{Z};$$

                                             2) the fraction can be simplified by $5$ $$\;\Leftrightarrow\;n=5p-1,\;p \in \mathbb{Z}.$$

                        1) $s \mid 6n+5$ and $s \mid 3n-1\;\;\Rightarrow\;\;s \mid 6n+5-2\cdot (3n-1)=7.$

                    In case $s=7$, from $7 \mid 6n+5$ and $7 \mid 3n-1\;\;\Rightarrow$

$$7 \mid(6n+5)-(3n-1)=3n+6=3(n+2)\;\;\underset{7 \;\not{\mid \mid}\;3}{\Rightarrow}\;7 \mid n+2.$$

                    Hence $n+2=7p,\;p \in \mathbb{Z}.$

                           Remarks CiP  $1^r$ After simplifying with $\color{Red}7$ the fraction becomes

     $$\frac{6(7p-2)+5}{3(7p-2)-1}=\frac{42p-7}{21p-7}=\frac{\color{Red}7 \cdot (6p-1)}{\color{Red}7 \cdot (3p-1)}=\frac{6p-1}{3p-1}$$

                      wich is irreducible (see Exercises 1 and 2).

                                                  $2^r$ The fraction $\frac{6n+5}{3n-1}$ takes the integer values

                      $-5$ and $+1$ when $n=0$ and $n=-2$ respectively. 

{end Rem's}

                                  2) $s \mid 6n+1$ and $s \mid 3n-3\;\;\Rightarrow\;$

$\Rightarrow\;s \mid 6n+1)-2 \cdot (3n-2)=5.$

                         In case $s=5$, from $5 \mid 6n+1 =5 \cdot n+(n+1)\;\;\Rightarrow \;s \mid n+1.$

                          Hence $n+1=5p,\;p\in \mathbb{Z}.$

                                 After simplifying with $5$ the fractiion becomes

$$\frac{6(5p-1)+1}{3(5p-1)-2}=\frac{30p-5}{15p-5}=\frac{6p-1}{3p-1}$$

                            wich is again irreducible.

                        The fraction $\frac{6n+1}{3n-2}$ takes the integer values $7$ and $+1$ when $n=1$ and

                    $n=-1$ respectively.                         

$\blacksquare$

                    5.   Find all the integers $x$ for which the fraction $\frac{3x+2}{2x-2}$ is an integer.

                    Solution CiP              Answer : [mistake !!, for x=6 the corect value is 2]

                    

                          In case $2x-2=\pm1$ we do not have integer solutions.

                          The fraction $\frac{3x+2}{2x-2}$ can be simplified with an $s>1\;\Rightarrow$

$\Rightarrow\;\;s \mid 3x+2$ and $s \mid 2x-2\;\;\Rightarrow$

$\Rightarrow\;\;s \mid 2 \cdot(3x+2)-2(2x-2)\;\;\Rightarrow\;\;s \mid 10.$

                   So $s=2$(only) or $s=5$(only) or $s=10$(both).

                            If the fraction is simplified by only $2$, the result will be an integer if

 $2x-2=\pm2$; then $x=0$ and $x=2$.

                             If the fraction is simplified by only $5$, the result will be an integer if

 $2x-2=\pm5$; this has no integer solutions.

                             If the fraction is simplified by $10$, the result will be an integer if

 $2x-2=\pm10$; then $x=-4$ and $x=6$.

                             All values $x \in \{0,2,-4,6\}$ give integer values for fraction $\frac{3x+2}{2x-2}$.

                             We get the answer.

                           Remarks CiP  $1^R$ We can also apply rule $4^{$}$ (see the Introduction,

                     which precedes section I) We will solve in integers $(x,y) \in \mathbb{Z} \times \mathbb{Z}$ the

                       equation

$$\frac{3x+2}{2x-2}=y\;\;\Leftrightarrow$$

$ \Leftrightarrow\;\;3x+2=y \cdot (2x-2)\;\Leftrightarrow\;2xy-3x-2y=2\;\;|+3\;\;\Leftrightarrow$

$\Leftrightarrow\;x(2y-3)-(2y-3)=5\;\;\Leftrightarrow\;(2y-3)(x-1)=5.$

             (Compare this equation with the one given in the formula ($\gamma$) from the

              Introduction.)

                      All possibilities in integer numbers for the two factors are given in the

              table below (first two lines). On the last two lines are respectively the values

              of $x$ and the corresponding values $y$ of the fraction.

{end Rem 1}

                        $2^R$ Regarding the reducibility of the fraction $\frac{3x+2}{2x-2}$, we have:

the fraction is simplified by 10   iff   $x=10p-4\;\;p\in \mathbb{Z};$

the fraction is simplified only by 5   iff   $x=10p+1\;\;p \in \mathbb{Z};$

the fraction is simplified only by 2  iff   $x=10p,\;10p\pm2,\;10p+4\;\;p \in \mathbb{Z};$

the fraction is irreducible   iff    $x=10p-1,\;10p \pm 3,\;10p+5,\;\;p \in \mathbb{Z}.$

                (see Exerc 4.)

{end Rem 2}

{end Rem's}

$\blacksquare$

                 6. The same problem as in Exercise 5 for the fraction $\frac{5n+6}{3n+6}.$

                             Answer CiP :

                                                - for $n \in \{-6,-3,0\}$ the fraction take respectively

                                                    the values $2,\;3,\;1;$

                                                 - the fraction

                                                          is irreducible  iff   $n=12p\pm1,\;12p\pm5,$

                                                          is simplified by 2 only   iff   $n=12p\pm4,$

                                                          is simplified by 3 only    iff   $n=12p \pm3,$

                                                          is simplified by 4 only   iff   $n=12p\pm2,$

                                                          is simplified by 6 only   iff   $n=12p,$

                                                          is simplified by 12   iff   $n=12p+6;$ 

                                                         everywhere $p \in \mathbb{Z}.$

                              The solution imitates the one in Ex 5.

         $\blacksquare$

                 7.   List the elements of the following sets. 

 $$A=\left \{ x\in \mathbb{N}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{N}\right \};$$ 

 $$B=\left \{ x\in \mathbb{Z}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{Z}\right \}; $$

$$C=\left \{ x\in \mathbb{N}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{Z}\right \};$$

$$D=\left \{ x\in \mathbb{Z}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{N}\right \};$$

$$E=\left \{ x\in \mathbb{Z}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{Z}\setminus \mathbb{N}\right \};$$

$$F=\left \{ x\in \mathbb{N}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{Z}\setminus \mathbb{N}\right \};$$

$$G=\left \{ x\in \mathbb{Z}\setminus \mathbb{N}\;\bigg \vert x=\frac{2k+6}{k-2}\;,\;k \in \mathbb{Z}\setminus \mathbb{N}\right \}.$$

                       Write all the inclusions between them.(See remark $3^R$ in Sect.III.)

                           Answer CiP : $A=\{3,4,7,12\},\;\;B=\{-8,-3,0,1,3,4,7,12\}$,

                                                   $C=\{0,1,3,4,7,12\},\;\;D=\{-8,-3,3,4,7,12\}$,

                                                    $E=\{0,1\},\;\;F=\{0,1\},\;\;G=\varnothing.$

                    Solution CiP  Since here the form (@) has $c=1$, it can be solved more

                 simply by writing

$$\frac{2k+6}{k-2}=\frac{2k-4+10}{k-2}=2+\frac{10}{k-2}.$$

                 For the solution, we are looking for $k-2$ among the integer divisors of

                 the number $10\;:\;\pm1,\;\pm2,\;\pm5,\;\pm10.$ We get the beautiful table below.

$\blacksquare$

               SECTION III: remarks

                    $1^R$. Exercice 1 in Section II is from (First) International (Mathematical) Olympiad, 1959.

                    $2^R$. Exercises 2, 3 are taken from the book shown in the image, page 276.

Those who can read in Romanian can download it by clicking on the image.

                    $3^R$. Exercise 7 is ibidem exercise I.15, page 141 from the book mentioned in the previous remark

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