Solve the equation

$$
\frac{x^{3}-7 x+6}{x-2}=\frac{2 x+14}{x+2}
$$

in the set of integers.
I. solution. Conditions: $x \neq \pm 2, x \in \mathbb{Z}$. Since

$$
\begin{aligned}
x^{3}-7 x+6 & =x^{3}-4 x-3 x+6=x\left(x^{2}-4\right)-3(x-2)= \\
& =x(x-2)(x+2)-3(x-2)=(x-2)\left(x^{2}+2 x-3\right),
\end{aligned}
$$

we can rewrite the numerator on the left side of the equation as:

$$
\frac{(x-2)\left(x^{2}+2 x-3\right)}{x-2}=\frac{2 x+14}{x+2}
$$

Since

$$
\frac{2 x+14}{x+2}=\frac{2 x+4+10}{x+2}=2+\frac{10}{x+2}
$$

we get

$$
x^{2}+2 x-3=2+\frac{10}{x+2}, \quad \text { from which } \quad x^{2}+2 x-5=\frac{10}{x+2}
$$

Considering that $x \in \mathbb{Z} \backslash\{ \pm 2\}$, and thus $x^{2}+2 x-5 \in \mathbb{Z}$, it follows that $\frac{10}{x+2} \in \mathbb{Z}$ must be true, meaning $x+2$ must be a divisor of 10. Therefore, $x+2 \in\{ \pm 1 ; \pm 2 ; \pm 5 ; \pm 10\}$.

By substituting, it can be seen that none of them are solutions.

| $x+2=1$ | $x=-1$ | $1-2-5 \neq \frac{10}{1}$ |
| :---: | :---: | :---: |
| $x+2=-1$ | $x=-3$ | $9-6-5 \neq \frac{10}{-1}$ |
| $x+2=2$ | $x=0$ | $0-0-5 \neq \frac{10}{2}$ |
| $x+2=-2$ | $x=-4$ | $16-8-5 \neq \frac{10}{-2}$ |
| $x+2=5$ | $x=3$ | $9+6-5 \neq \frac{10}{5}$ |
| $x+2=-5$ | $x=-7$ | $49-14-5 \neq \frac{10}{-5}$ |
| $x+2=10$ | $x=8$ | $64+16-5 \neq \frac{10}{10}$ |
| $x+2=-10$ | $x=-12$ | $144-24-5 \neq \frac{10}{-10}$ |

István Molnár (Békéscsaba, Széchenyi István Secondary School, 11th grade)

II. solution. Interpret and eliminate the denominators: $x \neq \pm 2$.

$$
\begin{aligned}
\left(x^{3}-7 x+6\right)(x+2) & =(x-2)(2 x+14) \\
x^{4}+2 x^{3}-7 x^{2}-8 x+12 & =2 x^{2}+10 x-28 \\
x^{4}+2 x^{3}-9 x^{2}-18 x & =-40 \\
x^{3}(x+2)-9 x(x+2) & =-40 \\
x(x+2)(x-3)(x+3) & =-40
\end{aligned}
$$

Since we are working with integers, we need to find the divisors of -40, and the factors must match these divisors.

The divisors of -40 are: $1,2,4,5,8,10,20,40$ and their negatives. The numbers $x+2$ and $x+3$ are consecutive integers, so only the pairs 1,2; -2,-1; 4,5 and -5,-4 can be considered. In these cases, $x=-1 ;-3 ; 2 ;-6$. The numbers -3 and -6 are not valid because they are not divisors of -40, and 2 is not in the domain of the equation. Let's check if $x=-1$ is a solution:

$$
(-1) \cdot 1 \cdot(-4) \cdot 2 \neq-40
$$

Therefore, the equation has no integer solutions.

András Görcs (Somorja, Madách Imre Secondary School, 9th grade)

Note. Some students examined divisibility by 3: they rewrote the equation in the form $(x-1)(x+2)(x+3)=2(x+7)$, then checked for $x \geq 0$ and $x<0$ cases, what the remainder is on the left and right sides when $x$ gives a remainder of 0, 1, or 2 when divided by 3. Since the remainder does not match in any case, there are no solutions in the set of integers.