4. Solve the equation

$$
x^{2018}+\frac{1}{x^{2018}}=1+x^{2019}
$$
Answer: $x=1$.

Solution: $x^{2018}+\frac{1}{x^{2018}} \geq 2$ for $x \neq 0$, because $x^{4036}-2 x^{2018}+1=\left(x^{2018}-1\right)^{2} \geq 0 . x^{2019}=x^{2018}+\frac{1}{x^{2018}}-1 \geq 2-1=1 . x \geq 1$. If $\quad x>1, \quad x^{2019}+1>$ $x^{2018}+\frac{1}{x^{2018}}$, because $x^{2019}=x * x^{2018}>x^{2018}$ and $1>\frac{1}{x^{2018}}$. Contradiction, so $x=1, x<=1$, so $x=1$. $x=1$ is indeed a solution to the equation.