A sequence defined as $a(n)=n-a(a(n-1))$ $n\geq 1,\ a(0)=0$, how to prove that $a(n)=⌊(n+1)(-1+√5)/2⌋$

Question

$a(n)=n-a(a(n-1)), \ n \geq 1,\ a(0)=0$, to prove that $$ a(n)=⌊(n+1)\cdot \frac{\sqrt{5} - 1}{2}⌋. $$ This is an exercise of "Discrete Mathematics and Its Application".(Supplementary exercise 72 of chapter 5)
this unusual sequence is first defined in Douglas Hofstader's "Gödel, Escher, Bach".

And here is some Hint from the exercise:
Let $$u=(-1+\sqrt{5})/2$$ First show for all n>0 that $$(u*n-⌊u*n⌋)+(u^2*n-⌊u^2*n⌋)=1$$ Then show for all real numbers a with 0≤a<1 and a≠1-u that $$⌊(1+u)*(1-a)⌋+⌊a+u⌋=1 $$