We $ $ have $\,n\mid \color{#c00}{\overbrace{n\!+\!1}^{\large m}}-1.\ $ Your (correct) proof easily generalizes widely as follows.
Generally $\,n\mid \,k\:\color{#c00}m-1\,\Rightarrow\, k\,m-j\,n = 1,\ $ so $\,\ d\mid m,n\,\Rightarrow\, d\mid 1,\ $ so $\ \gcd(m,n) = 1$
$ $ i.e. $\ {\rm mod}\,\ n\!:\,\ m^{-1}\,$ exists $\,\Rightarrow\, \gcd(m,n) = 1\,$ (and conversely by Bezout, see here and here)
Remark $\ $ We can also use (a single step of) the Euclidean algorithm
$$ \gcd(km,n) = \gcd(jn\!+\!1,n) = \gcd(1,n) = 1\,\Rightarrow\, \gcd(m,n) = 1$$
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.
- #2
Two consecutive numbers are relatively prime. \(\displaystyle \alpha=-1,\; \beta=1 \)
\(\displaystyle (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1\)
I am not sure what to do next to prove this.
- #3
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?
- #4
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 \(\displaystyle g=(n,n+1) \)
\(\displaystyle g\mid n \) and \(\displaystyle g\mid n+1 \implies g\mid (n+1)-n = 1 \implies g=\pm1 \). But since
\(\displaystyle g>0 \), we get \(\displaystyle g=1 \).
- #5
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 \(\displaystyle a\) 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.
- #6
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 \(\displaystyle b=n +1\), \(\displaystyle y=1\). This is a special case of Bézout's identity.