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.
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.
This is a special case of Bézout's identity.