Question: Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$
Attempt: I want to use Euclid's Algorithm because it seemed to be easier than what my book was doing which was manually finding the linear combination.
Euclid's Algorithm states that we let $a,b \in N $. By applying the Division Algorithm repeatedly...then $\gcd(a,b) = r_j$ will be the last non-zero remainder. By the Well Ordering Principle, there is a smallest element of the set of natural numbers less than or equal to $r_j$.
I have used long division, but since I can't get it to show up here, I will type what I've done.
Starting at $\gcd(2n+1,9n+4)$,
I can multiply $2n+1$ four times and I would have a remainder of $n$ because $(9n+4)-(8n+4) = 9n+4-8n-4=n$
so we have $4 \cdot (2n+1)+n$ if I apply Euclid's Algorithm.
For $\gcd(2n+1, n)$,
I can multiply $n$ 2 times and I will have only 1 as the remainder because $(2n+1)-(2n+0) = 2n+1-2n+0=1$
Therefore, we have $2 \cdot (n) +1$
and $\gcd(n,1)$ which is $1$
Since the end result is $1, \gcd(2n+1,9n+4)=1$
I followed an example from this link
Am I doing this correctly?