Two consecutive numbers are relatively prime

**dwsmith** · June 20th 2010, 11:19 AM

Two consecutive numbers are relatively prime.

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

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

**chiph588@** · June 20th 2010, 11:20 AM

Originally Posted by dwsmith

Two consecutive numbers are relatively prime.

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

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

$\alpha=-1,\; \beta=1$

**dwsmith** · June 20th 2010, 11:23 AM

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?

**chiph588@** · June 20th 2010, 01:05 PM

Originally Posted by dwsmith

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?

Let $g=(n,n+1)$

$g\mid n$ and $g\mid n+1 \implies g\mid (n+1)-n = 1 \implies g=\pm1$ . But since $g>0$ , we get $g=1$ .

**melese** · June 20th 2010, 02:35 PM

Originally Posted by dwsmith

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?

The answer is yes. In general: Let $a$ and $b$ be integers. There exists two integers $x$ and $y$ such that $ax+by=1$ if and only if $(a,b)=1$ . In your case, $a=n$ , $x=-1$ and $b=n<br /> +1$ , $y=1$ .

However, the reply you got from "chiph588" using the greatest common divisor ( $g=(n,n+1)$ ) seems to me more elegant.

**chiph588@** · June 20th 2010, 02:50 PM

Originally Posted by melese

The answer is yes. In general: Let $a$ and $b$ be integers. There exists two integers $x$ and $y$ such that $ax+by=1$ if and only if $(a,b)=1$ . In your case, $a=n$ , $x=-1$ and $b=n<br /> +1$ , $y=1$ .

This is a special case of Bézout's identity.

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