Two consecutive numbers are relatively prime.

$\displaystyle (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1$

I am not sure what to do next to prove this.

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- Jun 20th 2010, 11:19 AMdwsmithTwo consecutive numbers are relatively prime
Two consecutive numbers are relatively prime.

$\displaystyle (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1$

I am not sure what to do next to prove this. - Jun 20th 2010, 11:20 AMchiph588@
- Jun 20th 2010, 11:23 AMdwsmith
I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?

- Jun 20th 2010, 01:05 PMchiph588@
- Jun 20th 2010, 02:35 PMmelese
The answer is yes. In general: Let $\displaystyle a$ and $\displaystyle b$ be integers. There exists two integers $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle ax+by=1$ if and only if $\displaystyle (a,b)=1$. In your case, $\displaystyle a=n$, $\displaystyle x=-1$ and $\displaystyle b=n

+1$, $\displaystyle y=1$.

However, the reply you got from "chiph588" using the greatest common divisor ($\displaystyle g=(n,n+1)$) seems to me more elegant. - Jun 20th 2010, 02:50 PMchiph588@
This is a special case of Bézout's identity.