Prove that any two consecutive terms of the Fibonacci sequence are relatively prime

Congratulations, you have solved it. You have used the fact that gcd(a+b,b)=gcd(a,b)

Your proof is good. For reference, I have a proof without induction that uses Cassini's identity, $$f_{n-1}f_{n+1} - f_nf_n = (-1)^n,$$ which is proved directly at that Wikipedia page and in another direct way at Show $F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n$ for all $n \in \mathbb{N}$.

According to whether $n$ is even or odd, we have $$f_{n-1}f_{n+1} - f_nf_n = 1 \qquad \text{or} \qquad f_nf_n - f_{n-1}f_{n+1} = 1.$$

Now, the gcd of $f_n$ and $f_{n+1}$ may be defined alternatively and equivalently as the least positive integer that can be written in the form $pf_n + qf_{n+1}$ where $p$ and $q$ are integers. Because the coefficients of $f_n$ and $f_{n+1}$ in that pair of equations are Fibonacci numbers, hence integers, and because there is no positive integer less than $1$, gcd$(f_n, f_{n+1}) = 1$. Thus, any two consecutive terms of the Fibonacci sequence are relatively prime.