Prove $\gcd(n, n + 1) = 1$ for any $n$ Let $n \in \mathbb Z$ be even. Then $n + 1$ is odd. So, $2$ doesn't divide $n + 1$. Thus there's no even number for which $\gcd(n, n+1)$ is not $1$. I am not sure how to show it for odd numbers. Is there a better way to prove the statement?

Related to 727 answer. 

Theorem: For any integers $x,a,b$, $(a,b) = (a,b+ax)$.

(See the proof in theorem 1.9 in Niven's "An introduction to the theory of numbers".)
Then, for the particular case that $a=n$ and $b=n+1$, choose $x=-1$ to get:
$$
(n,n+1)=(n,1)=1.
$$
