Problem E:15900, not from AMERICAN MATHEMATICAL MONTHLY

It is from the Romanian magazine "Gazeta Matematica", No. 2, 2021, at page 105, proposed for 8th grade.

 In the translation of Miss Google:

          "Solve the equation in the set of integers

$$\left\{\frac{10x+4}{x+1}\right\}=\left\{\frac{7x+8}{x+2}\right\}.$$

     We have noted $\{a\}$ the fractional part of the real number $a$."

Answer CiP

$$x\in \{-4,\;-3,\;0,\;1\;\}$$

Solution CiP

     We are looking for the solutions of the equation in the set $\mathbb {Z} \setminus\{-2,\;-1\;\}.$  

      If we denote by $[ a ]$ the integer part of the real number $a$, we have

$$a=[ a ]+\{ a \},\;\;\;[ a ] \in \mathbb{Z},\;\{ a \} \in [ 0,\;1).$$

      The equation is written successively

$$\frac{10x+4}{x+1}-\left [ \frac{10x+4}{x+1} \right ]=\frac{7x+8}{x+2}-\left [ \frac{7x+8}{x+2} \right ]$$

$\Leftrightarrow \;\;\frac{10x+4}{x+1}-\frac{7x+8}{x+2}=\left [ \frac{10x+4}{x+1} \right ]-\left [ \frac{7x+8}{x+2} \right ]$

$\Leftrightarrow \;\;\left ( 10-\frac{6}{x+1} \right )-\left (7-\frac{6}{x+2} \right )=\left [ \frac{10x+4}{x+1} \right ]-\left [ \frac{7x+8}{x+2} \right ]$

$\Leftrightarrow \;\;\frac{6}{x+1}-\frac{6}{x+2}=3-\left [ \frac{10x+4}{x+1} \right ]+\left [ \frac{7x+8}{x+2} \right ]$

$\Leftrightarrow \;\;\frac{6}{(x+1)(x+2)}=3-\left [ \frac{10x+4}{x+1} \right ]+\left [ \frac{7x+8}{x+2} \right ]. \tag{1}$

In the relation (1) the right side is an integer. So we will have solutions only if 

$$\frac{6}{(x+1)(x+2)} \in \mathbb{Z}. \tag{2}$$

The integers $x$ that satisfy relation (2) are determined from the condition  

$(x_+1)(x+2)\in \{\pm 1,\;\pm 2,\;\pm 3,\;\pm 6\}$ - the integer divisors of the number 6.

 Moreover, since $(x+1)(x+2)$ is an even number, we only have to consider the cases

$(x+1)(x+2)=\pm 2$ and $(x+1)(x+2)=\pm 6.$

 The first equation provides the integer solutions $0,\;-3$ and the second - $1,\;-4$.

     All the found values check the equation:

$\left \{\frac{10 \cdot 0+4}{0+1}\right \}=\{4\}=0;\;  \left \{\frac{7\cdot 0+8}{0+2} \right \}=\left \{\frac{8}{2}\right \}=\{4\}=0$

 $\left \{ \frac{10\cdot (-3)+4}{-3+1} \right \}=\left \{ \frac{-26}{-2} \right \}=\{ 13 \}=0;\; \left \{ \frac{7 \cdot(-3)+8}{-3+2} \right \}=\{13 \}=0$

$\left \{ \frac{10\cdot 1 +4}{1+1} \right \}=\left \{ \frac{14}{2} \right \}=\{7\}=0;\;\left \{ \frac{7 \cdot 1+8}{1+2} \right \}=\left \{ \frac{15}{3} \right \}=\{5\}=0$

$\left \{ \frac{10 \cdot (-4)+4}{-4+1} \right \}=\left \{ \frac{-36}{-3} \right \}=\{12\}=0;\;\left \{ \frac{7\cdot (-4)+8}{-4+2} \right \}=\left \{ \frac{-20}{-2} \right \}=\{10 \}=0.$

$\blacksquare$