Find solutions of $f(2020)x+f(2019)y=1$ where $f$ is Fibonacci sequence

Question

I need to find at least one solution of $$f(2020)x+f(2019)y=1$$ with $x,y\in\mathbb{Z}$, where $f(n)$ is the $n^{th}$ Fibonacci number, starting at $f(0)=0$, so that: $$f(0)=0,\qquad f(1)=1,\qquad f(n)=f(n-1)+f(n-2).$$

Answer 1

Hint: For every natural number $n$ you have $$f(n+1)^2-f(n)f(n+2)=(-1)^n.$$