A linear congruence equation with Fibonacci numbers Let $(f_n)_{n\geq 0}$ be the Fibonacci sequence. Describe for each $n\in \mathbb{N}_0$ the solution of the linear congruence equation
$$ f_n x \equiv f_{n+1} \mod f_{n+2}$$ 
I know that two consecutive Fibonacci numbers are relatively prime but that doesn't seem to help.
 A: It's not hard to see all three numbers are coprime,so $f_{n}$ has an inverse $\bmod f_{n+2}$, from here:
$x\equiv f_{n+1}f_n^{-1}\equiv f_{n+1}(f_{n+2}-f_{n+1})^{-1}\equiv f_{n+1}(-f_{n+1})^{-1}\equiv -1 \bmod f_{n+2}$.
A: If you check the cases for first few n's, you see that
$$ x \equiv 1 \pmod 2 $$
$$ x \equiv 2 \pmod 3 $$
$$2x \equiv 3 \pmod 5 \Rightarrow x \equiv 4 \pmod 5$$
$$3x \equiv 5 \pmod 8 \Rightarrow x \equiv 7 \pmod 8$$
Here, you might have noticed that for each n's, it seems like that the solution tends to be $$x \equiv f_{n+2}-1 \pmod {f_{n+2}} $$ or instead, $$x \equiv -1 \pmod {f_{n+2}} \tag {1}.$$
Although you don't know yet whether this is true or not, you can assume so and try substituting in to the original equation (just to see if this makes anything clearer) $$f_nx \equiv f_{n+1} \pmod {f_{n+2}}$$ to get $$-f_n \equiv f_{n+1} \pmod {f_{n+2}} \tag 2$$
Well, is (2) a true statement? If you can prove so, you will be able to work your way backwards to prove (1) (it is then evident that (1) is a valid solution for the problem, and since gcd$(f_n,f_{n+2})=1$, we know it will be the unique solution as well).
It turns out that (2) is literally from the definition of Fibonacci numbers. Since $$f_n + f_{n+1} = f_{n+2},$$ we take $\pmod {f_{n+2}}$ to get $$-f_n +  \equiv f_{n+1} \pmod {f_{n+2}}. \blacksquare$$
