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.

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.
