Fibonacci Numbers $F_n$ and $F_{n + 1}$ are relatively prime for all $n \geq 0$.
Solution 1:
The idea is fine, but you really ought to use more words; it makes the argument much clearer and much more readable. Here’s how I would present the same argument.
Assume as induction hypothesis that $\gcd(F_n,F_{n+1})=1$. Then there are integers $s$ and $t$ such that
$$\begin{align*} 1&=sF_n+tF_{n+1}\\ &=s(F_{n+2}-F_{n+1})+tF_{n+1}\\ &=sF_{n+2}+(t-s)F_{n+1}\;, \end{align*}$$
and it follows that $\gcd(F_{n+1},F_{n+2})=1$ as well. The result now follows by induction.