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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_Farin May 10 '13 at 11:20

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Assuming that the parentheses denote $\text{gcd}$, how could it be less than 1? – Trevor Wilson Feb 7 '13 at 20:21
up vote 10 down vote accepted

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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