Is my logic valid in showing that any two consecutive terms of the Fibonacci sequence are coprime Assume terms $a_n$ and $a_{n+1}$ share some common factor $x$ so that $a_n = xm$ for some integer $m$ and $a_{n+1} = xk$ for some integer $k$.
Since $a_{n-1}$ = $a_{n+1} - a_n = xk - xm = x(k-m)$, $a_{n-1}$ is a multiple of $x$ as well. Then $a_{n-2}$ must also be a multiple of $x$ because $a_{n-2} = a_{n} - a_{n-1} = xm - (xk - xm) = x(2m-k)$. Since each lower or upper term can be found by subtracting or adding one multiple of $x$ to another, we can say that every term in the sequence must then be a multiple of $x$.
Therefore, if any two consecutive terms share a common factor, then all terms must share a common factor. However, if $a_1=1$ we can see the counterexample of $a_3 = 2, a_4=3$ where $2$ and $3$ are consecutive terms with no common factors. Since not all terms then share a common factor, no two consecutive terms can share a common factor.
 A: Alternative:
One has $\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_{n} \\ f_{n} & f_{n-1} \end{pmatrix}$ and so taking the determinant $(-1)^n = f_{n-1}f_{n+1} - f_n^2$. This shows $1$ is a linear combination of $f_{n+1}$ and $f_n$ and they are therefore coprime.
A: What you have is good. It is essentially a reverse induction, showing that if two consecutive terms have a common factor, then the previous two consecutive terms also have that same common factor, and then work that all the way back to $F_1=F_2=1$ which have no common factor.
Another approach is to use the identity
$$
F_{n-1}F_{n+1}-F_n^2=(-1)^n\tag1
$$
which can be proven by induction since it holds for $n=1$:
$$
F_0F_2-F_1^2=-1\tag2
$$
and if it holds for $n-1$, it holds for $n$:
$$
\begin{align}
F_{n-1}F_{n+1}-F_n^2
&=F_{n-1}(F_n+F_{n-1})-F_n^2\tag{3a}\\[2pt]
&=F_{n-1}^2-(F_n-F_{n-1})F_n\tag{3b}\\
&=F_{n-1}^2-F_{n-2}F_n\tag{3c}\\
&=-\left(F_{n-2}F_n-F_{n-1}^2\right)\tag{3d}\\
&=-(-1)^{n-1}\tag{3e}\\[2pt]
&=(-1)^n\tag{3f}
\end{align}
$$
Explanation:
$\text{(3a)}$: $F_{n+1}=F_n+F_{n-1}$
$\text{(3b)}$: expand and refactor
$\text{(3c)}$: $F_{n-2}=F_n-F_{n-1}$
$\text{(3d)}$: factor out a $-1$
$\text{(3e)}$: apply case $n-1$
$\text{(3f)}$: arithmetic
Equation $(1)$ says that $(F_{n-1},F_n)=1$ (any common factor would have to divide $1$).
