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?

Suppose $gcd(n,n+1)=d >0$ $$\to \left.\begin{matrix}
d|n\\ 
d|n+1 
\end{matrix}\right\}\Rightarrow  d|(n+1)-(n)\Rightarrow d|1\\\frac{1}{d} \in \mathbb{Z}\Rightarrow d=\pm1 \overset{d>0}{\rightarrow} d=1$$ 
