Answer
See explanation
Work Step by Step
Below is a standard proof that for every integer \(n \ge 0\),
\[
\gcd\bigl(F_{n+1}, F_n\bigr) \;=\; 1,
\]
where \(\{F_k\}\) is the Fibonacci sequence defined by
\[
F_0=0,\quad F_1=1,\quad \text{and for }k\ge2:\; F_k=F_{k-1}+F_{k-2}.
\]
---
## Proof via the Euclidean Algorithm
Recall that the Euclidean Algorithm for \(\gcd(a,b)\) repeatedly replaces the larger of \(a\) and \(b\) by its difference with the smaller until eventually one number is reduced to zero. The nonzero value at that stage is the greatest common divisor.
Consider \(\gcd(F_{n+1}, F_n)\). By definition of the Fibonacci numbers,
\[
F_{n+1} = F_n + F_{n-1}.
\]
Hence
\[
\gcd(F_{n+1}, F_n)
\;=\;
\gcd\bigl(F_n + F_{n-1},\,F_n\bigr)
\;=\;
\gcd\bigl(F_{n-1},\,F_n\bigr)
\quad
(\text{since }\gcd(a+b,b)=\gcd(a,b)).
\]
But again,
\[
\gcd(F_{n-1}, F_n)
\;=\;
\gcd\bigl(F_{n-1},\,F_{n-1}+F_{n-2}\bigr)
\;=\;
\gcd(F_{n-1}, F_{n-2}),
\]
and so on. Repeating this step by step “unwinds” the Fibonacci recurrence, reducing the indices by 1 each time:
\[
\gcd(F_{n+1},F_n)
\;=\;
\gcd(F_{n-1},F_n)
\;=\;
\gcd(F_{n-1},F_{n-2})
\;=\;
\cdots
\;=\;
\gcd(F_1, F_0).
\]
Since \(F_1 = 1\) and \(F_0 = 0\), \(\gcd(F_1, F_0) = \gcd(1,0) = 1.\)
Thus,
\[
\gcd(F_{n+1}, F_n) = 1
\quad
\text{for all integers }n \ge 0.
\]
This completes the proof.
