# What are $x$ and $y$ in $xF_n$ + $yF_{n-1}$ = $1$?

We know that the $$\gcd$$ of consecutive Fibonacci numbers is $$1$$. But while finding the coefficients $$x$$ and $$y$$ in using euclidean algorithm in reverse direction I am not able to find any pattern so that I can write $$x,y$$ in terms of n.

You can show by induction that: $$F_{n+1}F_{n-1}-F_nF_n=(-1)^{n-1}$$ then your Bezout coefficient $$a,b$$ can be $$(a,b)=(F_{n-1},F_n)$$

These are the extended euclidean algorithm outputs for consecutive pairs of Fibonacci numbers:

 (1, 0, 1),
(1, 1, 0),
(1, -1, 1),
(1, 2, -1),
(1, -3, 2),
(1, 5, -3),
(1, -8, 5),
(1, 13, -8),
(1, -21, 13),
(1, 34, -21),
(1, -55, 34),
(1, 89, -55),
(1, -144, 89),
(1, 233, -144),
(1, -377, 233),
(1, 610, -377),
(1, -987, 610),


Since $$\left( {\matrix{ {F_{n + 1} } \cr {F_n } \cr } } \right) = \left( {\matrix{ 1 & 1 \cr 1 & 0 \cr } } \right)\left( {\matrix{ {F_n } \cr {F_{n - 1} } \cr } } \right)$$ then $$1 = \left( {\matrix{ {x_{n + 1} } & {y_{n + 1} } \cr } } \right)\left( {\matrix{ {F_{n + 1} } \cr {F_n } \cr } } \right) = \left( {\matrix{ {x_{n + 1} } & {y_{n + 1} } \cr } } \right)\left( {\matrix{ 1 & 1 \cr 1 & 0 \cr } } \right)\left( {\matrix{ {F_n } \cr {F_{n - 1} } \cr } } \right) = \left( {\matrix{ {x_n } & {y_n } \cr } } \right)\left( {\matrix{ {F_n } \cr {F_{n - 1} } \cr } } \right)$$
Hint  Take the determinant of the following matrix form of the fibonacci recurrence $$\left[\begin{array}{ccc} \,1 & 1 \\\ 1 & 0 \end{array}\right]^{\large n} = \left[\begin{array}{ccc} F_{\large n+1} & F_{\large n} \\\ F_{\large n} & F_{\large n-1} \end{array}\right]$$