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

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.

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## closed as not a real question by 23rd, ncmathsadist, vonbrand, achille hui, Lord_FarinMay 10 '13 at 11:20

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Assuming that the parentheses denote $\text{gcd}$, how could it be less than 1? – Trevor Wilson Feb 7 '13 at 20:21

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

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Now that seemed elementary :). Thanks! – Tejen Shrestha Feb 7 '13 at 20:27
@icanc: You’re welcome! – Brian M. Scott Feb 7 '13 at 20:27

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

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

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