5. Solve the equation

$$
x^{4}+(x+2)^{4}=2 \text {. }
$$

## NATIONAL MATHEMATICS COMPETITION FOR HIGH SCHOOL STUDENTS, 11. 3. 2017.

## First Grade - A Category
5. By setting $x+1=t$, the given equation reduces to $(t-1)^{4}+(t+1)^{4}=2$, i.e.,

$$
2=\left((t-1)^{2}\right)^{2}+\left((t+1)^{2}\right)^{2}=\left(t^{2}-2 t+1\right)^{2}+\left(t^{2}+2 t+1\right)^{2}=2 t^{4}+12 t^{2}+2
$$

which is further equivalent to $0=t^{4}+6 t^{2}=t^{2}\left(t^{2}+6\right)$. The solutions to this equation are $t_{1}=t_{2}=0, t_{3 / 4}= \pm i \sqrt{6}$. By reversing the substitution $x=t-1$, we obtain the set of all solutions for $x: x \in\{-1,-1+i \sqrt{6},-1-i \sqrt{6}\}$.

## SOLUTIONS TO PROBLEMS FROM THE NATIONAL MATHEMATICS COMPETITION FOR HIGH SCHOOL STUDENTS, 11. 3. 2017.

## First Grade - A Category