Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$

Question

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 $\gcd(2n+1,9n+4)$,

$\frac{2n+1}{9n+4}$

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 $\gcd(2n+1, n)$,

$\frac{2n+1}{n}$

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

http://cms.math.ca/crux/v33/n5/public_page274-276.pdf

Am I doing this correctly?

Answer 1

One approach can be create a relation eliminating $n$ as follows :

$$S=9(2n+1)-2(9n+4)=1$$

If integer $d$ divides $\displaystyle 2n+1,9n+4$ it will divide $S=1$

Answer 2

Your approach is quite correct (As far as I think). Here is a similar but still different one.

Suppose that $\gcd(2n+1,9n+4)\ne1$ and there exist a $d$ such that $\gcd(2n+1,9n+4)=d$.

Now $d|9n+4$, and $d|2n+1$ $\implies d|9n+4- 4\times (2n+1)$.

$\implies d|n$.

$d|2n+1 - n\implies d|n+1$.

Again: $d|n+1-n\implies d|1$.

It is enough I think.

Answer 3

It can use this property: $$GCD(a,b)=GCD(a-qb,b) $$ So: $$ GCD(2n+1,9n+4)=GCD(2n+1,n)=GCD(1,n)=1$$

Answer 4

There is also the systematic matrix approach:

$$ \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 9 & 4 \end{pmatrix} \begin{pmatrix} n \\ 1 \end{pmatrix} \implies \begin{pmatrix} n \\ 1 \end{pmatrix} = \begin{pmatrix} -4 & \hphantom-1 \\ \hphantom-9 & -2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} $$ In particular, $1=9a-2b$, which implies $\gcd(a,b)=1$.

The key point here is that the first matrix has determinant $-1$ and so is invertible over the integers.