Let $x$ be a strictly positive real number such that $x+\frac{1}{x}=\sqrt{2020}$. What is the value of $x^{2}+\frac{1}{x^{2}}$?
To make $x^{2}+\frac{1}{x^{2}}$ appear, we will square the equality $x+\frac{1}{x}=\sqrt{2020}$. We obtain $2020=$ $x^{2}+\frac{1}{x^{2}}+2 x \times \frac{1}{x}=x^{2}+\frac{1}{x^{2}}+2$, thus $x^{2}+\frac{1}{x^{2}}=2018$.