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
