TITLE: Does the $\gcd(2n-1,2n+1)=1?$
QUESTION [7 upvotes]: I am posting this to ask if my proof is correct as I haven't taken number theory in a year and I feel a bit rusty.  If it isn't correct, please tell me where I went wrong so I can fix it.
I want to prove that the $\gcd(2n-1,2n+1)=1$ for all $n$.

Using the Euclidean Algorithm, we have that
  $$
2n+1=(2n-1)\cdot(1)+2
$$
  $$
2n-1=2(n-1)+1
$$
  $$
2=1\cdot(2)+0
$$
  Therefore, $\gcd(2n-1,2n+1)=1$ for all $n$.

REPLY [10 votes]: Another way: if $d$ divides both $2n-1$ and $2n+1$ then it divides their difference, which is $2$. But $2n-1$ and $2n+1$ are odd and so $d$ cannot be $2$ and must be $1$.

REPLY [2 votes]: We can also prove like this :
suppose $gcd(2n-1,2n+1)=a$, then we have $$a|2n-1; a|2n+1$$. So there exists $t_1, t_2$ such that $2n-1=at_1$ and $2n+1=at_2$, so from this two equations we get $$at_1+1=at_2-1 \iff a(t_2-t_1)=2$$.
So, $a=1 or 2$, if $a=2$ it contradicts with $a|2n+1$.
So, $a=1$