Prove that $(n, n + 1) = 1$ for all $n \gt 0$. [closed]

Question

I was thinking of doing this by contradiction. So by supposing: $$(n, n + 1) \neq 1$$

Then trying to to show that $(n, n + 1) \gt 1$ or $(n, n + 1) \lt 1$. But I'm not sure how I can accomplish this.

I also thought about the fact that $n$ and $n + 1$ are always of different parity. So if $n$ is even, then $n + 1$ is odd. But I don't think that helps me.

Answer 1

HINT: If $d\mid a$ and $d\mid b$, then $d \mid a-b$.

Answer 2

If $d$ divides $n$ and $d$ divides $n+1$, then $d$ divides $(n+1)-n=1$.

Answer 3

Hint $\ $ Common multiples $\rm\,n\ne m\,$ of $\rm\,d\,$ are no closer than $\rm\,d\:$(why?) $ $ Thus $\rm\,m = n\!+\!1\:\Rightarrow\: d \le \,\ldots$