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.
 A: 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 $(a,b)=(F_{n-1},F_n)$
A: 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),

Hopefully, you can get your answer from here.
A: Hint: 
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
$$
1 = \left( {\matrix{
   {x_{n + 1} } & {y_{n + 1} }  \cr 
 } } \right)\left( {\matrix{
   {F_{n + 1} }  \cr 
   {F_n }  \cr 
 } } \right) = \left( {\matrix{
   {x_{n + 1} } & {y_{n + 1} }  \cr 
 } } \right)\left( {\matrix{
   1 & 1  \cr 
   1 & 0  \cr 
 } } \right)\left( {\matrix{
   {F_n }  \cr 
   {F_{n - 1} }  \cr 
 } } \right) = \left( {\matrix{
   {x_n } & {y_n }  \cr 
 } } \right)\left( {\matrix{
   {F_n }  \cr 
   {F_{n - 1} }  \cr 
 } } \right)
$$
A: Hint $ $ Take the determinant of the following matrix form of the fibonacci recurrence
$$  \left[\begin{array}{ccc} \,1 & 1 \\\
1 & 0 \end{array}\right]^{\large n} = \left[\begin{array}{ccc}
F_{\large n+1} & F_{\large n} \\\
F_{\large n} & F_{\large n-1} \end{array}\right] $$
